The first part follows from the definition of \(\mathcal{H}^{+}\)-norm.

For the second part, suppose \({(p,q)} \in {H_{a}^{+} \times H_{b}^{+}}\) and \({p + q'} \in W \in \tau_{a + b}\). Then there is \(r\mspace{7mu}{({> 0})}\) such that \({B{({p + q'},r)}} = {\{{z \in H_{a + b}^{+}}:{{\|{p + q' + z'}\|} < r}\}} \subseteq W\). Then \({{B{(p,{r/2})}} \times B}{(q,{r/2})}\) is an open set in \(H_{a}^{+} \times H_{b}^{+}\) and \({(x,y)} \in {{{B{(p,{r/2})}} \times B}{(q,{r/2})}}\) implies \({\|{p + q' + {({x + y'})}'}\|} = {\|{p + q' + 0_{x + y'} + y + 0_{x + y'} + x' + 0_{x + y'}}\|} = {\|{x' + 0_{x + y'} + p + q' + 0_{x + y'} + y + 0_{x + y'}}\|} = {\|{x' + p + q' + 0_{x + y'} + y + 0_{x + y'}}\|} = {\|{x' + p + q' + y}\|} \leq {{\|{x' + p}\|} + {\|{q' + y}\|}} = {{\|{p + x'}\|} + {\|{q + y'}\|}} < r\). Therefore, \(f_{a,b}:\underset{{(x,y)}\mapsto{x+y^{\prime}}}{\overset{{H_{a}^{+}\times H_{b}^{+}}\longrightarrow H_{a+b}^{+}}{}}\) is continuous at \((p,q)\). For the continuity of \(m_{a,b}:\underset{{(x,y)}\mapsto{x⁢y}}{\overset{{H_{a}^{+}\times H_{b}^{+}}\longrightarrow H_{a\cdot b}^{+}}{}}\), if \({p \cdot q} \in V\) for some \(V \in \tau_{a \cdot b}\), then there is \(r_{1}\mspace{7mu}{({> 0})}\) such that \({B{({p \cdot q},r_{1})}} = {\{{z \in H_{a \cdot b}^{+}}:{{\|{{p \cdot q} + z'}\|} < r_{1}}\}} \subseteq W\). Let \(t > 0\) be such that \({t^{2} + {t{({{\| p\|} + {\| q\|}})}}} < r_{1}\). Then \({{B{(p,t)}} \times B}{(q,t)}\) is an open set in \(H_{a}^{+} \times H_{b}^{+}\) and \({(x,y)} \in {{{B{(p,t)}} \times B}{(q,t)}}\) implies \({\|{{p \cdot q} + {({x \cdot y})}'}\|} = {\|{{({p \cdot q})}' + {x \cdot y}}\|} = {\|{{p' \cdot q} + {x \cdot y}}\|} = {\|{{{({x + p'})} \cdot {({y + q'})}} + {p \cdot {({y + q'})}} + {{({x + p'})} \cdot q}}\|} \leq {{{\|{x + p'}\|}{\|{y' + q}\|}} + {{\| p\|}{\|{y' + q}\|}} + {{\|{x + p'}\|}{\| q\|}}} < r_{1}\). This implies \(m_{a,b}:\underset{{(x,y)}\mapsto{x⁢y}}{\overset{{H_{a}^{+}\times H_{b}^{+}}\longrightarrow H_{a\cdot b}^{+}}{}}\) is continuous at \((p,q)\). Consequently, by [10, Theorem 2.18], it follows that if \(\tau_{a}\) is the topology on \(H_{a}^{+}\) induced by the norm, then \(S\) is a topological completely regular (algebraic) semiring with respect to the topology \(\tau\), where \(\tau\) is generated by \(\bigcup\limits_{a \in {E^{+}{(S)}}}\tau_{a}\).

Clearly, the norm topology on \(S\) is first countable. Suppose \(p \in H_{a}^{+}\) and \({\{ p_{n}\}}_{n}\) be a sequence in \(H_{a}^{+}\) converging to \(p\). Then as \(n\rightarrow\infty\), \({\|{p_{n} + p'}\|}\longrightarrow 0\). Now, for all \(n \in {\mathbb{N}}\), \({|{{\| p_{n}\|} - {\| p\|}}|} \leq {\|{p_{n} + p'}\|}\) implies that \({\| p_{n}\|}\longrightarrow{\| p'\|}\) as \(n\rightarrow\infty\). So, \(|| \cdot ||\) is continuous on \(H_{a}^{+}\). Since under the norm topology, \(\mathcal{H}^{+}\)-classes are disjoint open set in \(S\), it follows that \(|| \cdot ||\) is continuous on \(S\).

Since \(S\) is an idempotent semiring of Hausdorff topological rings, so for every \(a \in {E^{+}{(S)}}\), \(H_{a}^{+}\) is a Hausdorff topological ring. Let \(\tau\) be the topology on \(S\) and \(\tau_{a}\) be the subspace topology on \(H_{a}^{+}\). We may assume that \(H_{a}^{+}\) is a ring with unity (otherwise, we take \(\overset{\sim}{H_{a}^{+}} = {H_{a}^{+} \cup {\{ 1_{a}\}}}\) as the ring with unity \(1_{a}\) with the topology \(\tau_{a} \cup {\{{A \cup {\{ 1_{a}\}}}:{A \in \tau_{a}}\}}\)). Let \(Q_{a}\) (\(a \in {E^{+}{(S)}}\)) be the rings of all fractions \(\frac{x}{y}\) where \(x \in H_{a}^{+}\) and \(y\) is a cancellable element of \(H_{a}^{+}\) belonging to the center of \(H_{a}^{+}\) such that \(\psi:\underset{z\mapsto{z⁢y}}{\overset{H_{a}^{+}\longrightarrow H_{a}^{+}}{}}\) is an open mapping. Then the neighbourhoods of \(a\) in \((H_{a}^{+},\tau_{a})\) forms a fundamental system of neighbourhoods of \(0\), the zero element of \(Q_{a}\), making it a Hausdorff topological ring containing \(H_{a}^{+}\) as an open subring. Therefore, replacing \(H_{a}^{+}\) with \(Q_{a}\), if necessary, we may assume that each cancellable element \(c_{a}\) is invertible in \(H_{a}^{+}\). Let \(W\) be a bounded and symmetric neighbourhood of \(a\) (otherwise, we take \(W \cap W'\) in place of \(W\)) in \(H_{a}^{+}\). Let \(U = {\{{x \in H_{a}^{+}}:{{xW} \subseteq W}\}}\). As \(W\) is symmetric and bounded, \(U\) is also a symmetric neighbourhood of \(a\). Let \(V\) be a neighbourhood of \(a\). As \(W\) is bounded, there exists a neighbourhood \(W_{1}\) of \(a\) such that \({W_{1}W} \subseteq V\) and \({WW_{1}} \subseteq V\). Since \({\lim\limits_{n\rightarrow\infty}c_{a}^{n}} = a\), there exists a natural number \(k_{0}\) such that \(c_{a}^{k} \in W\) for all \(k \geq k_{0}\). This implies \(c_{a}^{k}W_{1}\) is a neighbourhood of \(a\), and \({{U \cdot c_{a}^{k}}W_{1}} \subseteq {WW_{1}} \subseteq V\) . Similarly, \({{c_{a}^{k}W_{1}} \cdot U} = {W_{1}Uc_{a}^{k}} \subseteq V\) (since \(c_{a}\) is in the center of \(H_{a}^{+}\)). Thus \(U\) is a bounded symmetric neighbourhood of \(a\) satisfying \({UU} \subseteq U\). Let \(U_{1}\)(\(\subseteq U\)) be a neighbourhood of \(a\) such that \({{U_{1}U} + {U_{1}U} + {U_{1}U}} \subseteq U\). Replacing \(c_{a}\) by a power of \(c_{a}\), if necessary, we may assume that \(c_{a} \in U_{1}\). Then \({{c_{a}U} + {c_{a}U} + {c_{a}U}} \subseteq U\). Now, \({({c_{a}^{k}U})}_{k \in {\mathbb{Z}}}\) is a decreasing sequence of neighbourhoods of \(a\) which is also a fundamental system of neighbourhoods of \(a\), for if \(Y\) is a neighbourhood of \(a\), then \(U\) being bounded, there exists a neighbourhood \(Z\) of \(a\) such that \({ZU} \subseteq Y\) and there exists \({t\mspace{10mu}{({\geq k_{0}})}} \in {\mathbb{N}}\) such that \(c_{a}^{t} \in Z\) (since \({\lim\limits_{n\rightarrow\infty}c_{a}^{n}} = a\)). This implies \({c_{a}^{t}U} \subseteq {ZU} \subseteq Y\) and hence in particular, \({\bigcap\limits_{t \geq 0}{c_{a}^{t}U}} = {\{ a\}}\).

For \(a \in {E^{+}{(S)}}\), let us now define \(g_{a}:{H_{a}^{+}\rightarrow{\mathbb{R}}}\) by \({g_{a}{(a)}} = 0\), and for each nonzero \(x \in H_{a}^{+}\) let \({g_{a}{(x)}} = 2^{- k}\), where \(k = {\max{\{{{j \in {\mathbb{N}}}:{x \in x \in {c_{a}^{j}U}}}\}}}\). If \(W\) is an additive subgroup, then \(U\) is a subring, and \(g_{a}\) is clearly a non-Archimedean norm defining the topology of \(H_{a}^{+}\).

Then the mapping \(g:{S\rightarrow{\mathbb{R}}}\) defined by : for all \(x \in S\), \({g{(x)}} = {g_{a}{(x)}}\), if \(x \in H_{a}^{+}\) is a non-Archimedean \(\mathcal{H}^{+}\)-norm on \(S\) defining the topology of \(S\).

In general, let \({f_{a}{(x)}} = {\inf\left\{ {\sum\limits_{i = 1}^{p}{g_{a}{(z_{i})}}}:{{z_{1} + z_{2} + \cdots + z_{p}} = x} \right\}}\), for all \(x \in H_{a}^{+}\). Since \({{c_{a}U} + {c_{a}U} + {c_{a}U}} \subseteq U\) and \({{c_{a}^{n + 1}U} + {c_{a}^{n + 1}U} + {c_{a}^{n + 1}U}} \subseteq {c_{a}^{n}U}\) for all \(n \in {\mathbb{N}}\), by [3, Proposition I.2.2], it follows that \(f_{a}\) is a norm defining the topology of the additive group \(H_{a}^{+}\). Since \(c_{a} \in U\) and \({UU} \subseteq W\), \({g_{a}{({xy})}} \leq {g_{a}{(x)}g_{a}{(y)}}\) for all \({x,y} \in H_{a}^{+}\), which implies \({f_{a}{({xy})}} \leq {f_{a}{(x)}f_{a}{(y)}}\) for all \({x,y} \in H_{a}^{+}\). This implies \(f_{a}\) is a norm defining the topology of the ring \(H_{a}^{+}\).

Then the mapping \(f:{S\rightarrow{\mathbb{R}}}\) defined by : for all \(x \in S\), \({f{(x)}} = {f_{a}{(x)}}\), if \(x \in H_{a}^{+}\) is an \(\mathcal{H}^{+}\)-norm on \(S\) defining the topology of \(S\).

First suppose that the topology \(\tau\) on \(S\) is defined by a natural \(\mathcal{H}^{+}\)-pseudo-valuation \(v\). Since \(S\) is a topological completely regular (algebraic) semiring with each \(H_{a}^{+}\) open, so by Theorem 1.6, it follows that each \(H_{a}^{+}\) is a topological ring and hence for each \(a \in {E^{+}{(S)}}\), \({v|}_{H_{a}^{+}}\) is the natural pseudo-valuation on \(H_{a}^{+}\) which defines the topology of the subspace \(H_{a}^{+}\). Therefore, by Theorem 2.13, we must have for each \(a \in {E^{+}{(S)}}\),

\((i)\) \(H_{a}^{+}\) is metrizable,

\(({ii})\) the open ideals of \(H_{a}^{+}\) form a fundamental system of neighbourhoods of \(a\)
and \(({iii})\) \(H_{a}^{+}\) has an open topologically nilpotent ideal.

Suppose for each \(a \in {E^{+}{(S)}}\), \(\tau_{a}\) be the subspace topology of \(\tau\) on \(H_{a}^{+}\) and \(d_{a}\) be the metric on \(H_{a}^{+}\) which defines \(\tau_{a}\). Let \(d:{{S \times S}\rightarrow{\mathbb{R}}_{\geq 0}}\) be defined by : for \({(x,y)} \in {S \times S}\),

One can easily prove that \(d\) is a metric on \(S\). Now, to complete the proof, it remains to show that \(\tau_{d} = \tau\). Suppose \(\mathcal{B}\) and \(\mathcal{B}_{1}\) be two bases for the topologies \(\tau\) and \(\tau_{d}\) respectively. Then for any \(x \in B \in \mathcal{B}\), there is some \(a \in {E^{+}{(S)}}\) such that \(x \in {B \cap H_{a}^{+}} \in \tau_{a}\), which implies there is some \(r\mspace{7mu}{({> 0})}\) such that \({B_{d_{a}}{(x,r)}} \subseteq {B \cap H_{a}^{+}} \subseteq B\), where \({B_{d_{a}}{(x,r)}} = {\{{y \in H_{a}^{+}}:{{d_{a}{(y,x)}} < r}\}}\). Let \(r_{1} = \frac{r}{1 + r}\). Now, if \(y \in {B_{d}{(x,r_{1})}}\), then \({d{(x,y)}} = \frac{d_{a}{(x,y)}}{1 + {d_{a}{(x,y)}}} < r_{1}\), which implies \({d_{a}{(x,y)}} < r\) and so \(y \in {B_{d_{a}}{(x,r)}}\). Thus, \(x \in {B_{d}{(x,r_{1})}} \subseteq {B_{d_{a}}{(x,r)}} \subseteq B\), showing that \(\tau \subseteq \tau_{d}\). Again, for any \(x \in B_{1} \in \mathcal{B}_{1}\), \(x \in {B_{d}{(x,t)}} \subseteq B_{1}\), for some \(0 < t < 1\). Also, there is some \(a \in {E^{+}{(S)}}\) such that \(x \in H_{a}^{+}\). Suppose \(t_{1} = \frac{t}{1 - t}\). Now, if \(z \in {B_{d_{a}}{(x,t_{1})}}\), then \({d_{a}{(x,z)}} < t_{1}\), which implies \(\frac{d_{a}{(x,z)}}{1 + {d_{a}{(x,z)}}} < \frac{t_{1}}{1 + t_{1}} = t\) and so \(z \in {B_{d}{(x,t)}}\). Thus, \(x \in {B_{d_{a}}{(x,t_{1})}} \subseteq {B_{d}{(x,t)}} \subseteq B_{1}\), showing that \(\tau_{d} \subseteq \tau\). Consequently, \(S\) is metrizable.

Conversely, suppose that \((S,\tau)\) is a topological completely regular (algebraic) semiring satisfying conditions \({(i)} - {({iii})}\) as stated in the statement. Then by Theorem 2.13, it follows that for every \(a \in {E^{+}{(S)}}\), there is a natural pseudo-valuation \(v_{a}\) which defines the subspace topology of \(H_{a}^{+}\). Suppose \(v:{S\rightarrow{{\mathbb{N}} \cup {\{\infty\}}}}\) is defined by : for \(x \in S\), \({v{(x)}} = {v_{a}{(x)}}\), if \(0_{x} = a\) for some \(a \in {E^{+}{(S)}}\). Then \(v\) is a natural \(\mathcal{H}^{+}\)-pseudo-valuation on \(S\). Also, if \(\tau_{v}\) is the topology on \(S\) defined by \(v\), then \(\tau_{v}\) is generated by the topologies on \(H_{a}^{+}\) which are defined by \(v_{a}\). But \(v_{a}\) defines the subspace topology of \(\tau\) on \(H_{a}^{+}\). Since each \(H_{a}^{+}\) is open in \((S,\tau)\), it follows that \(\tau_{v}\) = \(\tau\).

Since each \(\mathcal{H}^{+}\)-class is an open subspace of \(S\), for each \(a \in {E^{+}{(S)}}\), \(H_{a}^{+}\) is a topological ring whose topology is defined by a natural pseudo-valuation. So by Theorem 2.13, for each \(a \in {E^{+}{(S)}}\), conditions \((i)\) and \(({ii})\) hold and \(H_{a}^{+}\) is metrizable.

For \(({iii})\), let \(\tau_{a}\) be the subspace topology of \(\tau\) on \(H_{a}^{+}\) and \(d_{a}\) be the metric on \(H_{a}^{+}\) for which \(\tau_{a}\) = \(\tau_{d_{a}}\). Let \(d:{{S \times S}\rightarrow{\mathbb{R}}_{\geq 0}}\) be defined by : for \({(x,y)} \in {S \times S}\),

Then it is easy to prove that \(d\) is a metric on \(S\). The proof for \(\tau_{d} = \tau\) is similar to the proof of Theorem 2.14.

Suppose \((S,\tau)\) be a first countable idempotent semiring of topological rings. Then every \(\mathcal{H}^{+}\)-class is a first countable topological ring and thus for every \(a \in {E^{+}{(S)}}\), \((H_{a}^{+}, + )\) is a first countable topological group.

For each \(a \in {E^{+}{(S)}}\), let \(\tau_{a}\) be the subspace topology of \(\tau\) on \(H_{a}^{+}\). Then from Theorem 2.18, it follows that for each \(a \in {E^{+}{(S)}}\), \((H_{a}^{+},\tau_{a})\) is metrizable and \(\tau_{a}\) is induced by an invariant metric \(d_{a}\). Let \(d:{{S \times S}\rightarrow{\mathbb{R}}_{\geq 0}}\) be defined by : for \({(x,y)} \in {S \times S}\),

Then one can easily prove that \(d\) is a metric on \(S\) and similar to the proof of Theorem 2.14, it follows that \(\tau_{d} = \tau\).

Since \(S\) is an idempotent semiring of topological rings, for each \(a \in {E^{+}{(S)}}\), \(H_{a}^{+}\) is an open set in \(S\). Therefore, the quotient space \(S/\mathcal{H}^{+}\) is a discrete space and hence it is metrizable.

Since \(S/\mathcal{H}^{+}\) is discrete, each \(H_{a}^{+}\) is an open set in \(S\) and so \(S\) is an idempotent semiring of topological rings. Therefore, from Theorem 2.19, it follows that \(S\) is metrizable.

Since \(S\) is an idempotent semiring of topological rings, so by Theorem 2.29, it follows that \(S/I\) is also an idempotent semiring of topological rings. Again, since \(S\) is metrizable and \(\psi:{S\longrightarrow{S/I}}\) is a continuous surjective mapping, we have \(S/I\) is first countable. We now show that \(S/I\) is Hausdorff. For this, let \({a + I},{b + I}\) be two distinct elements in \(S/I\). Then either \(0_{a} \neq 0_{b}\) or \({a' + b} \notin I\). If \(0_{a} \neq 0_{b}\), then \({H_{a}^{+} \cap H_{b}^{+}} = \varnothing\). Clearly, then \(H_{a}^{+} + I\) and \(H_{b}^{+} + I\) are open sets in \(S/I\) containing \(a + I\) and \(b + I\) respectively such that \({{({H_{a}^{+} + I})} \cap {({H_{b}^{+} + I})}} = \varnothing\). On the other hand, if \({a' + b} \notin I\), then \(I\) being a closed ideal, there is an open set \(U\) in \(S\) such that \({a' + b} \in U \subseteq {S \smallsetminus I}\). Since \(S\) is an idempotent semiring of topological rings, so there are open sets \(V_{a}\)and \(V_{b}\) containing \(a\) and \(b\) respectively in \(S\) such that \({V_{a}' + V_{b}} \subseteq U\). Since \(\psi:{S\longrightarrow{S/I}}\) is open, it follows that \(V_{a} + I\) and \(V_{b} + I\) are open in \(S/I\) such that \({a + I} \in {V_{a} + I}\) and \({b + I} \in {V_{v} + I}\). Now, if \({{({V_{a} + I})} \cap {({V_{b} + I})}} \neq \varnothing\), then there is some \(x \in S\) such that \({x + I} \in {{({V_{a} + I})} \cap {({V_{b} + I})}}\). This implies \({x + I} = {v_{a} + I} = {v_{b} + I}\), for some \(v_{a} \in V_{a}\) and \(v_{b} \in V_{b}\). From these it follow that \({v_{a}' + v_{b}} \in I\), contradicting the fact that \({V_{a}' + V_{b}} \subseteq U \subseteq {S \smallsetminus I}\). Therefore, \({{({V_{a} + I})} \cap {({V_{b} + I})}} = \varnothing\) and hence \(S/I\) is Hausdorff. Consequently, using Theorem 2.19, it follows that \(S/I\) is metrizable.

Let \(\mathcal{U}\) be a family of open neighbourhoods of \(\Delta_{E^{+}{(S)}}\) in \({{E^{+}{(S)}} \times E^{+}}{(S)}\) such that \({card{(\mathcal{U})}} = {\Delta{({E^{+}{(S)}})}}\) with \({\cap \mathcal{U}} = \Delta_{E^{+}{(S)}}\) and \(\mathcal{V}\) be a family of open neighbourhoods of \(E^{+}{(S)}\) in \(S\) such that \({card{(\mathcal{V})}} = {\psi{({E^{+}{(S)}},S)}}\) with \({\cap \mathcal{V}} = {E^{+}{(S)}}\). Since \(S\) is a topological completely regular (algebraic) semiring, we must have that the mapping \(\phi:\underset{{(x,y)}\mapsto{({x+x^{\prime}},{y+y^{\prime}})}}{\overset{{S\times S}\longrightarrow{{{E^{+}⁢{(S)}}\times E^{+}}⁢{(S)}}}{}}\) is a continuous mapping. Now, for any \(U \in \mathcal{U}\), there is an open set \(W_{U} \in {S \times S}\) such that \({\phi{(W_{U})}} \subseteq U\). Also, the mapping \(\mu:\underset{{(x,y)}\mapsto{x+y^{\prime}}}{\overset{{S\times S}\longrightarrow S}{}}\) is continuous. So for any \(V \in \mathcal{V}\), there is an open set \(W_{V} \in {S \times S}\) such that \({\mu{(W_{V})}} \subseteq V\). Suppose \(W_{U,V} = {W_{U} \cap W_{V}}\). Then \(\mathcal{W} = {\{ W_{U,V}:{{U \in \mathcal{U}},{V \in \mathcal{V}}}\}}\) is a family of open subsets of \(S \times S\). Suppose \({(x,y)} \in {\cap \mathcal{W}}\). Since \({\cap \mathcal{U}} = \Delta_{E^{+}{(S)}}\) and \({\cap \mathcal{V}} = {E^{+}{(S)}}\), it follows that \({x + x'} = {y + y'}\) and \({x + y'} \in {E^{+}{(S)}}\). These imply that \(x = y\) and so \({(x,y)} \in \Delta_{S}\). Therefore, \({\cap \mathcal{W}} \subseteq \Delta_{S}\). Again, if \(x \in S\), then \({(0_{x},0_{x})} \in U\) and \(0_{x} \in V\) for every pair of \({U \in \mathcal{U}},{V \in \mathcal{V}}\) implies \({(x,x)} \in W_{U,V}\) and thus \(\Delta_{S} \subseteq {\cap \mathcal{W}}\). Therefore, \({\cap \mathcal{W}} = \Delta_{S}\) and hence \({\Delta{(S)}} \leq {card{(\mathcal{W})}} \leq {{{card{(\mathcal{U})}} \cdot c}ard{(\mathcal{V})}} = {{{\Delta{({E^{+}{(S)}})}} \cdot \psi}{({E^{+}{(S)}},S)}}\).

Since \(S\) is a locally compact topological completely regular (algebraic) semiring, so using Theorem 3.1 and Theorem 3.2, we have \({w{(S)}} \leq {{{{{l{(S)}} \cdot \Delta}{({E^{+}{(S)}})}} \cdot \psi}{({E^{+}{(S)}},S)}}\). Again, from definition, it follows that \({\Delta{({E^{+}{(S)}})}} \leq {w{({E^{+}{(S)}})}}\) and therefore, \({w{(S)}} \leq {{{{{l{(S)}} \cdot w}{({E^{+}{(S)}})}} \cdot \psi}{({E^{+}{(S)}},S)}}\). Suppose \(\mathcal{V}\) is a family of open neighbourhoods of \(E^{+}{(S)}\) in \(S\) such that \({card{(\mathcal{V})}} = {\psi{({E^{+}{(S)}},S)}}\) with \({\cap \mathcal{V}} = {E^{+}{(S)}}\) and \(\mathcal{B}\) be a base for open sets in \(E^{+}{(S)}\). Since \(S\) is a topological completely regular (algebraic) semiring, so the mapping \(\phi:\underset{{(x,y)}\mapsto{x+y^{\prime}}}{\overset{{S\times S}\longrightarrow S}{}}\) is a continuous mapping. Now, for any \(V \in \mathcal{V}\) and for any \(B \in \mathcal{B}\), there is an open set \(W_{V,B} \in {S \times S}\) such that \(B \subseteq {\phi{(W_{V,B})}} \subseteq V\). This implies \({\cap {\{ W_{V,B}:{{V \in \mathcal{V}},{B \in \mathcal{B}}}\}}} = \Delta_{S}\), where \(\{ W_{V,B}:{{V \in \mathcal{V}},{B \in \mathcal{B}}}\}\) is a family of open subsets of \(S \times S\). Thus, \({{{w{({E^{+}{(S)}})}} \cdot \psi}{({E^{+}{(S)}},S)}} \leq {\Delta{(S)}}\) and hence \({{{{{l{(S)}} \cdot w}{({E^{+}{(S)}})}} \cdot \psi}{({E^{+}{(S)}},S)}} \leq {{{\Delta{(S)}} \cdot l}{(S)}} = {w{(S)}}\). Consequently, \({w{(S)}} = {{{{{l{(S)}} \cdot w}{({E^{+}{(S)}})}} \cdot \psi}{({E^{+}{(S)}},S)}}\).

If \(S\) is metrizable, then \(E^{+}{(S)}\) is also metrizable. Moreover, since \(E^{+}{(S)}\) is a closed subset of the metric space \(S\), it is a \(G_{\delta}\)-set.

Conversely, we assume that \(E^{+}{(S)}\) is a metrizable \(G_{\delta}\)-set. Since \(E^{+}{(S)}\) is a countable intersection of open sets in \(S\), therefore, \({\psi{({E^{+}{(S)}},S)}} \leq \aleph_{0}\). This implies \({\Delta{(S)}} = {{{\Delta{({E^{+}{(S)}})}} \cdot \psi}{({E^{+}{(S)}},S)}} \leq \aleph_{0}\) and hence by Theorem 3.1, it follows that \(S\) is metrizable.

If \(S\) is metrizable, then \(E^{+}{(S)}\) is also metrizable. Also \(E^{+}{(S)}\) being a closed subset of the metric space \(S\), it is a \(G_{\delta}\) set. Moreover, by Theorem 3.9, it follows that \(S\) is an \(M\)-space.

Conversely, we assume that \(E^{+}{(S)}\) is a metrizable \(G_{\delta}\)-set and \(S\) is an \(M\)-space. Since \(E^{+}{(S)}\) is a \(G_{\delta}\)-set, we must have \({\psi{({E^{+}{(S)}},S)}} \leq \aleph_{0}\). Again, since \(E^{+}{(S)}\) is metrizable, it follows that \({\Delta{({E^{+}{(S)}})}} \leq \aleph_{0}\). Thus \({\Delta{(S)}} \leq {{{\Delta{({E^{+}{(S)}})}} \cdot \psi}{({E^{+}{(S)}},S)}} \leq \aleph_{0}\) and therefore, \(\Delta_{S}\) is a \(G_{\delta}\) set in \(S\). Hence by Theorem 3.9, it follows that \(S\) is metrizable.

Since \(S\) is an idempotent semiring of topological rings, each \(\mathcal{H}^{+}\)- class is a clopen subset of \(S\) and a topological ring. Therefore, for every \(a \in {E^{+}{(S)}}\), \(M \cap H_{a}^{+}\) is a compact subset of the topological commutative group \((H_{a}^{+}, + )\). Again, \(f\) being continuous on \(S\), for every \(a \in {E^{+}{(S)}}\), \(f_{a} = {f|}_{H_{a}^{+}}\) is a continuous function. Therefore, by Theorem 4.4, it follows that for every \(\epsilon\mspace{7mu}{({> 0})}\) and for every \(a \in {E^{+}{(S)}}\), there is an open set \(V\) containing \(a\) such that if \({x,y} \in {M \cap H_{a}^{+}}\) and \(x \in {y + V} = {V + y}\), then \({|{{f_{a}{(x)}} - {f_{a}{(y)}}}|} \leq \epsilon\). This implies if \({x,y} \in M\) with \(0_{x} = 0_{y} = a\) and \(x \in {y + V} = {V + y}\), then \({|{{f{(x)}} - {f{(y)}}}|} \leq \epsilon\). Consequently, \(f\) is uniformly continuous on \(M\).

Since \(S\) and \(B\) are both idempotent semirings of topological rings, so \(S\) and \(B\) are completely regular (algebraic) semirings and hence all the \(\mathcal{H}^{+}\)-classes of \(S\) and \(B\) are topological rings. For each \(z \in B\), let \({\overset{\sim}{H}}_{z}^{+}\) be the \(\mathcal{H}^{+}\)-classes of \(B\) containing the element \(z\).

\((i)\) Clearly, \({E^{+}{(S)}} \subseteq {E^{+}{(B)}}\). For the reverse inclusion, let \(p \in {E^{+}{(B)}}\). Then \({\overset{\sim}{H}}_{p}^{+} \cap S\) is a subring of \(S\). Since maximal subrings of \(S\) are precisely the \(\mathcal{H}^{+}\)-classes of \(S\), we must have \({{\overset{\sim}{H}}_{p}^{+} \cap S} \subseteq H_{b}^{+}\) for some \(b \in S\) and thus \(p = 0_{b} \in {E^{+}{(S)}}\). Therefore, \(E + {(B)} \subseteq E^{(}S)\) and hence \({E^{+}{(S)}} = {E^{+}{(B)}}\).

(ii) For every \(a \in {E^{+}{(S)}}\), let \(f_{a}:{H_{a}^{+}\rightarrow R}\) be the mapping defined by \(f_{a} = {f|}_{H_{a}^{+}}\). Then \(f_{a}\) is a continuous ring homomorphism form the topological ring \(H_{a}^{+}\) to the complete topological \(T_{2}\)-ring \(R\). Since \(S\) is a dense subsemiring of \(B\), for every \(a \in {E^{+}{(S)}}\), \(H_{a}^{+}\) is a dense subring of \({\overset{\sim}{H}}_{a}^{+}\). In fact, for any non-empty open set \(U\) in \({\overset{\sim}{H}}_{a}^{+}\), we have \(U\) is open in \(B\) and hence \({U \cap S} \neq \varnothing\). If \(y \in {U \cap S}\), then \(y \in H_{y}^{+} \subseteq {\overset{\sim}{H}}_{a}^{+}\) and thus \(U\) intersects \(H_{a}^{+}\). For \(x \in {\overset{\sim}{H}}_{a}^{+}\), if \(\mathcal{B}_{x}\) is the filter of all neighborhoods of \(x\) in \(B\) and \(\mathcal{B}_{x}' = {\{{U_{x} \cap H_{a}^{+}}:{U_{x} \in \mathcal{B}_{x}}\}}\), then using Theorem 4.5, it follows that for every \(a \in {E^{+}{(S)}}\), there exists a continuous ring homomorphism \({\overset{\sim}{f}}_{a}:{{\overset{\sim}{H}}_{a}^{+}\longrightarrow R}\), extending \(f_{a}\), defined by \({{\overset{\sim}{f}}_{a}{(x)}} = {\lim{f_{a}{(\mathcal{B}_{x}')}}}\). As \(B\) is disjoint union of open sets \({\overset{\sim}{H}}_{a}^{+}\), \(a \in {E^{+}{(B)}} = {E^{+}{(S)}}\), the mapping \(\overset{\sim}{f}:{B\rightarrow R}\); defined by \({\overset{\sim}{f}{(x)}} = {{\overset{\sim}{f}}_{a}{(x)}}\), for all \(x \in B\) such that \(x \in {\overset{\sim}{H}}_{a}^{+}\); is a continuous mapping from \(B\) into \(R\) extending \(f\). To complete the prove, it remains to show that \(\overset{\sim}{f}\) is a semiring homomorphism. For this purpose, let \(x \in {\overset{\sim}{H}}_{a}^{+}\) and \(y \in {\overset{\sim}{H}}_{b}^{+}\), where \({a,b} \in {E{(B)}} = {E^{+}{(S)}}\). We show that \({\overset{\sim}{f}{({x + y})}} = {{\overset{\sim}{f}{(x)}} + {\overset{\sim}{f}{(x)}}}\) and \({\overset{\sim}{f}{({x \cdot y})}} = {{{\overset{\sim}{f}{(x)}} \cdot \overset{\sim}{f}}{(x)}}\).

Let \(V\) be an arbitrary neighborhood of \(\overset{\sim}{f}{({x + y})}\) . Then there exists a neighborhood \(V_{1}\) such that \({cl_{_{R}}V_{1}} \subseteq V\). Since \({\overset{\sim}{f}{({x + y})}} = {\overset{\sim}{f_{a+b}}{({x + y})}}\), there exists a neighborhood \(U\) of \(x + y\) such that \({f_{a + b}{({U \cap H_{a + b}^{+}})}} \subseteq V_{1}\). Since \(S\) is an idempotent semiring of topological rings, there exist neighborhoods \(U_{x},U_{y}\) of \(x\text{and}y\) respectively such that \({{({U_{x} \cap H_{a}^{+}})} + {({U_{y} \cap H_{b}^{+}})}} \subseteq {U \cap H_{a + b}^{+}}\). Then \({\overset{\sim}{f}{(x)}} = {{\overset{\sim}{f}}_{a}{(x)}} \in {cl_{_{R}}f_{a}{({U_{x} \cap H_{a}^{+}})}}\) and \({\overset{\sim}{f}{(y)}} = {{\overset{\sim}{f}}_{b}{(y)}} \in {cl_{_{R}}f_{b}{({U_{y} \cap H_{b}^{+}})}}\) implies \({{\overset{\sim}{f}{(x)}} + {\overset{\sim}{f}{(y)}}} \in {{cl_{_{R}}f_{a}{({U_{x} \cap H_{a}^{+}})}} + {cl_{_{R}}f_{b}{({U_{y} \cap H_{b}^{+}})}}} \subseteq {cl_{_{R}}{({{f_{a}{({U_{x} \cap H_{a}^{+}})}} + {f_{b}{({U_{y} \cap H_{b}^{+}})}}})}} =\)
\({cl_{_{R}}{({{f{({U_{x} \cap H_{a}^{+}})}} + {({U_{y} \cap H_{b}^{+}})}})}} \subseteq {cl_{_{R}}{({f_{a + b}{({U \cap H_{a + b}^{+}})}})}} \subseteq {cl_{_{R}}V_{1}} \subseteq V\). Therefore, \({{\overset{\sim}{f}{(x)}} + {\overset{\sim}{f}{(y)}}} = {\overset{\sim}{f}{({x + y})}}\).

To show \({\overset{\sim}{f}{({x \cdot y})}} = {{{\overset{\sim}{f}{(x)}} \cdot \overset{\sim}{f}}{(x)}}\), let \(W\) be an arbitrary neighborhood of \(\overset{\sim}{f}{({x \cdot y})}\). Then there exists a neighborhood \(W_{1}\) such that \({cl_{_{R}}W_{1}} \subseteq W\). Since \({\overset{\sim}{f}{({x \cdot y})}} = {\overset{\sim}{f_{a⁢b}}{({x \cdot y})}}\), there exists a neighborhood \(N\) of \(x \cdot y\) such that \({f_{ab}{({N \cap H_{ab}^{+}})}} \subseteq W_{1}\). Since \(S\) is an idempotent semiring of topological rings, there exist neighborhoods \(N_{x},N_{y}\) of \(x\text{and}y\) respectively such that \({{({N_{x} \cap H_{a}^{+}})} \cdot {({N_{y} \cap H_{b}^{+}})}} \subseteq {N \cap H_{ab}^{+}}\). Then \({\overset{\sim}{f}{(x)}} = {{\overset{\sim}{f}}_{a}{(x)}} \in {cl_{_{R}}f_{a}{({N_{x} \cap H_{a}^{+}})}}\) and \({\overset{\sim}{f}{(y)}} = {{\overset{\sim}{f}}_{b}{(y)}} \in {cl_{_{R}}f_{b}{({N_{y} \cap H_{b}^{+}})}}\) implies \(\overset{\sim}{f}{(x)}\overset{\sim}{f}{(y)} \in cl_{_{R}}f_{a}{(N_{x} \cap H_{a}^{+})} \cdot cl_{_{R}}f_{b}{(N_{y} \cap H_{b}^{+})} \subseteq cl_{_{R}}{(f_{a}{(N_{x} \cap H_{a}^{+})}f_{b}{(N_{y} \cap}}\)
\({H_{b}^{+})}) = cl_{_{R}}(f{({(N_{x} \cap H_{a}^{+})}{(N_{y} \cap H_{b}^{+})})})\) \(\subseteq {cl_{_{R}}{({f_{ab}{({N \cap H_{ab}^{+}})}})}} \subseteq {cl_{_{R}}W_{1}} \subseteq W\). Therefore, \({\overset{\sim}{f}{(x)}\overset{\sim}{f}{(y)}} = {\overset{\sim}{f}{({x \cdot y})}}\). Consequently, \(\overset{\sim}{f}\) is a semiring homomorphism.

Since \(T\) is an idempotent semiring of topological rings, each \(\mathcal{H}^{+}\)-class in \(T\) is a topological ring. Also, for every \(z \in T\), since \(H_{z}^{+}\) is a clopen subset of \(T\), it follows that \(H_{z}^{+}\) is a complete \(T_{2}\)-ring. Moreover, for every \(a \in {E^{+}{(S)}}\), it is clear that \({f{(H_{a}^{+})}} \subseteq H_{f{(a)}}^{+}\) and hence for every \(a \in {E^{+}{(S)}}\), the mapping \({f_{a} = {f|}_{H_{a}^{+}}}:{H_{a}^{+}\longrightarrow H_{f{(a)}}^{+}}\) is a continuous ring homomorphism, where \(H_{a}^{+}\) is a topological ring and \(H_{f{(a)}}^{+}\) is a complete \(T_{2}\)-ring. Then using Theorem 4.5, each \(f_{a}\) can be extended to a continuous ring homomorphism \({\overset{\sim}{f}}_{a}:{{\overset{\sim}{H}}_{a}^{+}\longrightarrow H_{f{(a)}}^{+}}\), where \({\overset{\sim}{H}}_{a}^{+}\) is the \(\mathcal{H}^{+}\)-class containing the element \(a \in {E^{+}{(S)}} = {E^{+}{(B)}}\) in \(B\), i.e., for each \(a \in {E^{+}{(B)}}\), there is a continuous semiring homomorphism \({\overset{\sim}{f}}_{a}:{{\overset{\sim}{H}}_{a}^{+}\longrightarrow T}\). As \(B\) is disjoint union of open sets \({\overset{\sim}{H}}_{a}^{+}\), \(a \in {E^{+}{(B)}} = {E^{+}{(S)}}\), the mapping \(\overset{\sim}{f}:{B\rightarrow T}\); defined by \({\overset{\sim}{f}{(x)}} = {{\overset{\sim}{f}}_{a}{(x)}}\), for all \(x \in B\) such that \(x \in {\overset{\sim}{H}}_{a}^{+}\); is a continuous mapping from \(B\) into \(T\) extending \(f\). Similar to the proof of Theorem 4.8, one can easily prove that \(\overset{\sim}{f}\) is a semiring homomorphism. Consequently, there exists a continuous semiring homomorphism \(\overset{\sim}{f}:{B\rightarrow T}\) which extends \(f\).

Since the mapping \(\gamma:\underset{x\mapsto x^{\prime}}{\overset{S\longrightarrow S}{}}\) is continuous, it is clear that \(U' \in \mathcal{U}\) for each \(U \in \mathcal{U}\). Let \(\mathcal{B}_{L} = {\{ U_{L}:{U \in \mathcal{U}}\}}\). Since the mapping \(\phi:\underset{{(x,y)}\mapsto{x^{\prime}+y}}{\overset{{S\times S}\longrightarrow S}{}}\) is continuous, \(U_{L} = {{\phi^{- 1}{(U)}} \cap {({\bigcup\limits_{a \in {E^{+}{(S)}}}{H_{a}^{+} \times H_{a}^{+}}})}}\) is open in \(S \times S\). Clearly, \({\bigtriangleup S} = {\{{(x,x)}:{x \in S}\}} \subseteq U_{L}\) for each \(U \in \mathcal{U}\). Let \(N \in \mathcal{B}_{L}\). Then \(N = U_{L}\), for some \(U \in \mathcal{U}\). Now, \({(x,y)} \in {(U_{L})}^{- 1}\) if and only if \({(y,x)} \in U_{L}\) if and only if \({y' + x} \in U\) and \(0_{x} = 0_{y}\) if and only if \({x' + y} \in U'\) and \(0_{x} = 0_{y}\) if and only if \({(x,y)} \in U_{L}'\) implies that \(N^{- 1} = U_{L}' \in \mathcal{B}_{L}\). Let \({M_{1},M_{2}} \in \mathcal{B}_{L}\). Then \(M_{1} = U_{1L}^{}\), \(M_{2} = U_{2L}^{}\), for some \({U_{1},U_{2}} \in \mathcal{U}\). Then \({M_{1} \cap M_{2}} = {({U_{1} \cap U_{2}})}_{L}\), infact \({(x,y)} \in {M_{1} \cap M_{2}}\) if and only if \({x' + y} \in {U_{1} \cap U_{2}}\) and \(0_{x} = 0_{y}\) if and only if \({(x,y)} \in {({U_{1} \cap U_{2}})}_{L}\). It follows that \({M_{1} \cap M_{2}} \in \mathcal{B}_{L}\). Let \(M \in \mathcal{B}_{L}\). Then \(M = W_{L}\), for some \(W \in \mathcal{U}\). Since \({{E^{+}{(S)}} + {E^{+}{(S)}}} \subseteq {E^{+}{(S)}} \subseteq W\), by [7, Theorem 1.1], there exist \({V_{1},V_{2}} \in \mathcal{U}\) such that \({V_{1} + V_{2}} \subseteq W\). Suppose \(P = {({V_{1} \cap V_{2}})}_{L}\). We show that \({P \circ P} \subseteq M\). Indeed, if \({(x,z)} \in {P \circ P}\) then there exists \(y \in S\) such that \({{(x,y)},{(y,z)}} \in P\). This implies that \({{x' + y},{y' + z}} \in {V_{1} \cap V_{2}}\) and \(0_{x} = 0_{y} = 0_{z}\). Then \({x' + z} = {x' + 0_{z} + z} = {x' + 0_{y} + z} = {x' + y + y' + z} \in {V_{1} + V_{2}} \subseteq W\) and \(0_{x} = 0_{z}\). Hence \(\mathcal{B}_{L}\) is a base for some uniformity \(\mathcal{L}\) on \(S\). Let \(\tau_{_{\mathcal{L}}}\) be the corresponding topology on \(S\). We show that \(\tau_{_{\mathcal{L}}} = \tau\). For each \(x \in S\) and \(M \in \mathcal{B}_{L}\), \({M{(x)}} = {\{{y \in S}:{{(x,y)} \in M}\}}\) is an open set in \((S,\tau_{_{\mathcal{L}}})\). For each \(x \in S\) and \(M \in \mathcal{B}_{L}\), \({M{(x)}} = {\{{y \in S}:{{{x' + y} \in U},{0_{x} = 0_{y}}}\}} = {\{{y \in S}:{{{x + y'} \in U'},{0_{x} = 0_{y}}}\}} = {({x + U'})}^{\ast}\), where \(M = U_{L}\), for some \(U \in \mathcal{U}\). Since \(S\) is a b-lattice of topological rings, for each \(x \in S\) and \(U \in \mathcal{U}\), \({M{(x)}} = {({x + U'})}^{\ast}\) is open in \((S,\tau)\). Therefore, \(\tau_{_{\mathcal{L}}} \subseteq \tau\). Suppose \(G\) be an open set in \((S,\tau)\) and \(x \in G\). Then \(0_{x} \in {({x' + G})}^{\ast}\). Let \(V = {{({x' + G})}^{\ast} \cup {\{ H_{_{y}}^{+}:{y \notin H_{_{x}}^{+}}\}}}\). Then \(V \in \mathcal{U}\) which implies \(Q = V_{L} \in \mathcal{B}_{L}\) and \(x \in {Q{(x)}} \subseteq G\). Indeed, if \(z \in {Q{(x)}}\), then \({(x,z)} \in Q\) which implies \({x' + z} \in V\) and \(0_{x} = 0_{z}\). Also, \({x' + z} \notin {\cup {\{ H_{_{y}}^{+}:{y \notin H_{_{x}}^{+}}\}}}\). Because if \({x' + z} \in H_{_{y}}^{+}\), for some \(y \notin H_{_{x}}^{+}\), then \(0_{x} = 0_{y}\) which implies that \(y \in H_{_{x}}^{+}\), a contradiction. Hence \({x' + z} \in {({x' + G})}^{\ast}\) and this implies that \(z \in G\). It follows that \(\tau \subseteq \tau_{_{\mathcal{L}}}\). Thus, we have \(\tau_{_{\mathcal{L}}} = \tau\) and so \(S\) is a uniform space.

Since \(S\) is a b-lattice of topological rings, by Theorem 5.2 there is a uniformity \(\mathcal{L}\) on \(S\). Suppose \(U \in \mathcal{U}\), where \(\mathcal{U}\) is defined as in Theorem 5.2. If \(V = U\), then for \({x,y} \in S\) with \({(x,y)} \in V_{L}\), we have \({{({\gamma{(x)}})}' + {\gamma{(y)}}} = {{(x')}' + y'} = {x + y'} \in U\) and \(0_{\gamma{(x)}} = 0_{\gamma{(y)}}\) implies \({({\gamma{(x)}},{\gamma{(y)}})} \in U_{L}\). Therefore, \(\gamma\) is uniformly continuous.

Uniform continuity of \(f\) implies the continuity of \(f\). For the converse part, using Theorem 5.2, let \(\mathcal{U}\) and \(\mathcal{U}'\) be the family of open sets containing \(E^{+}{(S)}\) and \(E^{+}{(T)}\) defining the uniformities \(\mathcal{L}_{S}\) and \(\mathcal{L}_{T}\) on \(S\) and \(T\) respectively. Suppose \(U_{L} \in \mathcal{L}_{T}\) for some \(U \in \mathcal{U}'\). Then \(f^{- 1}{(U)}\) is an open set containing \(E^{+}{(S)}\) in \(S\). Then there exists \(V \in \mathcal{U}\) such that \({E^{+}{(S)}} \subseteq V \subseteq {f^{- 1}{(U)}}\). Now, if \({x,y} \in S\) with \({(x,y)} \in V_{L}\), then \({x' + y} \in V\) and \(0_{x} = 0_{y}\) implies \(0_{f{(x)}} = 0_{f{(y)}}\) and \({{({f{(x)}})}' + {f{(y)}}} = {f{({x' + y})}} \in U\), which implies that \({({f{(x)}},{f{(y)}})} \in U_{L}\). Therefore, \(f\) is uniformly continuous.

Since \(S\) is a b-lattice of topological rings, so using Theorem 5.2, we conclude that there is a uniformity \(\mathcal{L}\) on \(S\) defined by \(\{ U_{L}:{U \in \mathcal{U}}\}\), where \(\mathcal{U} = {\{{U \in \tau}:{{E^{+}{(S)}} \subseteq U}\}}\). Suppose \(S\) is Hausdorff. Then \({\bigcap\limits_{M \in \mathcal{L}}M} = {\bigtriangleup S}\). Let \(x \in {\bigcap\limits_{U \in \mathcal{U}}U}\). Then \(x \in U\), for all \(U \in \mathcal{U}\). This implies that \({(0_{x},x)} \in U_{L}\), for all \(U \in \mathcal{U}\). So \({(0_{x},x)} \in {\bigcap\limits_{M \in \mathcal{L}}M} = {\bigtriangleup S}\) which implies that \(x = 0_{x}\) and so \({\bigcap\limits_{U \in \mathcal{U}}U} \subseteq {E^{+}{(S)}}\). Therefore, \({\bigcap\limits_{U \in \mathcal{U}}U} = {E^{+}{(S)}}\).

Conversely, let \({\bigcap\limits_{U \in \mathcal{U}}U} = {E^{+}{(S)}}\). Suppose \({(a,b)} \in {\bigcap\limits_{M \in \mathcal{L}}M}\). Then \({(a,b)} \in U_{L}\) for every \(U \in \mathcal{U}\), which implies \({a' + b} \in U\) and \(0_{a} = 0_{b}\), for all \(U \in \mathcal{U}\). So \({a' + b} \in {\bigcap\limits_{U \in \mathcal{U}}U} = {E^{+}{(S)}}\), which implies that \(a = b\) and so \({(a,b)} \in {\bigtriangleup S}\). Hence \({\bigcap\limits_{M \in \mathcal{L}}M} = {\bigtriangleup S}\) which implies that \(S\) is Hausdorff.

Let \({E^{+}{(S)}} = {\{ e_{1},e_{2},\ldots,e_{n}\}}\) and let \(U\) be an open set containing \(E^{+}{(S)}\) in \(\hat{S}\). Now, \(E^{+}{(S)}\) being compact, there exists an open set \(W\) in \(\hat{S}\) such that \({E^{+}{(S)}} \subseteq W \subseteq {cl_{\hat{S}}W} \subseteq U\). Then \(W \cap S\) is open in \(S\). Since \(S\) is a regular topological space, for each \(i = {1,2,3,\ldots,n}\), there exists an open set \(V_{i}\) containing \(e_{i}\) in \(S\) such that \(e_{i} \in {cl_{S}V_{i}} \subset {W \cap S}\). Then \({E^{+}{(S)}} \subset {\bigcup\limits_{i = 1}^{n}{cl_{S}V_{i}}} \subset {W \cap S}\). This implies that \({E^{+}{(S)}} \subset {cl_{S}{({\bigcup\limits_{i = 1}^{n}V_{i}})}} \subset {W \cap S}\). Suppose \(V = {\bigcup\limits_{i = 1}^{n}V_{i}}\). Then \(V\) is open in \(S\) and \({cl_{S}V} \subset {W \cap S}\). Now, \({cl_{S}V} = {{cl_{\hat{S}}V} \cap S}\) and this implies that \({cl_{\hat{S}}V} \subseteq {cl_{\hat{S}}W} \subseteq U\).

Clearly, \(Z = {S \times T}\) is a b-lattice of topological rings and \({E^{+}{(Z)}} = {{{E^{+}{(S)}} \times E^{+}}{(T)}}\) which is a finite set. Let \(\mathcal{U} = {\{{U \in \tau}:{{E^{+}{(S)}} \subseteq U}\}}\) and \(\mathcal{V} = {\{{V \in \sigma}:{{E^{+}{(T)}} \subseteq U}\}}\) as defined in Theorem 5.2 generate the uniformities \(\mathcal{F}_{_{\mathcal{L}}}^{S}\) and \(\mathcal{F}_{_{\mathcal{L}}}^{S}\) respectively. Then \(\{{U \times V}:{{U \in \mathcal{U}},{V \in \mathcal{V}}}\}\) constitutes a base of open neighborhoods of \(E^{+}{(Z)}\). Now, \(Z\) has a uniformity \(\mathcal{F}_{_{\mathcal{L}}}^{Z}\) generated by \(\mathcal{G}_{U,V} = {\{{{({(x,y)},{(x_{1},y_{1})})} \in {Z \times Z}}:{{{{x' + x_{1}} \in U},{{y' + y_{1}} \in {V\text{with}\,\, 0_{x}} = 0_{x_{1}}}},{0_{y} = 0_{y_{1}}}}\}}\), where \(U \in \mathcal{U}\) and \(V \in \mathcal{V}\). For any \(U \in \mathcal{U}\) and \(V \in \mathcal{V}\), set \(W_{_{U_{L},V_{L}}} = {\{{({{(x,y)}{(x_{1},y_{1})}})}:{{(x,x_{1})} \in {U_{L}\text{and}{(y,y_{1})}} \in V_{L}}\}}\). Then \(\{ W_{_{U_{L},V_{L}}}:{{U \in \mathcal{U}},{V \in \mathcal{V}}}\}\) forms a base for the product uniformity \(\mathcal{F}_{_{\mathcal{L}}}^{S} \times \mathcal{F}_{_{\mathcal{L}}}^{T}\) on \(S \times T\), where \(\mathcal{F}_{_{\mathcal{L}}}^{S}\) and \(\mathcal{F}_{_{\mathcal{L}}}^{S}\) are uniformities on \(S\) and \(T\) respectively. Clearly, for any \(U \in \mathcal{U}\) and \(V \in \mathcal{V}\), \(\mathcal{G}_{U,V} = W_{_{{L{(U)}},{L{(V)}}}}\). Therefore, it follows that \(\mathcal{F}_{_{\mathcal{L}}}^{Z} = {\mathcal{F}_{_{\mathcal{L}}}^{S} \times \mathcal{F}_{_{\mathcal{L}}}^{T}}\).

Let \(x \in S\). Then \(H_{x}^{+}\) is a topological ring and so it has a uniformity say, \(\mathcal{L}_{_{x}}\). Now, we show that \(\{{U \cap H_{x}^{+}}:{U \in \mathcal{U}}\}\) is a fundamental set of neighborhoods of \(0_{x}\), where \(\mathcal{U} = {\{{U \in \tau}:{{E^{+}{(S)}} \subseteq U}\}}\). Let \(W\) be an open set containing \(0_{x}\) in \(H_{x}^{+}\). Then \(W\) is open in \(S\) and therefore, \(W \cup {({S \smallsetminus H_{x}^{+}})}\) is an open set containing \(E^{+}{(S)}\) in \(S\). Now, \(\mathcal{U}\) being a base at \(E^{+}{(S)}\) in \(S\), there exists \(U \in \mathcal{U}\) such that \(U \subset {W \cup {({S \smallsetminus H_{x}^{+}})}}\). Then \({U \cap H_{x}^{+}} \subseteq W\) and thus it follows that \(\{{U \cap H_{x}^{+}}:{U \in \mathcal{U}}\}\) is a fundamental set of neighborhood of \(0_{x}\). Therefore, \(\{{({U \cap H_{x}^{+}})}_{L}:{U \in \mathcal{U}}\}\) is a basis for the uniformity \(\mathcal{L}_{_{x}}\), where \({({U \cap H_{x}^{+}})}_{L} = {\{{{(a,b)} \in {H_{x}^{+} \times H_{x}^{+}}}:{{a' + b} \in {U \cap H_{x}^{+}}}\}}\). Since \(S\) is a uniform space and \(H_{x}^{+}\) is a subset of \(S\), \(\mathcal{L}_{_{x}}'\) be the uniformity induced on \(H_{x}^{+}\) by the uniformity of \(S\). Then \(\{{U_{L} \cap {({H_{x}^{+} \times H_{x}^{+}})}}:{U \in \mathcal{U}}\}\) is a basis for the uniformity \(\mathcal{L}_{_{x}}'\) on \(H_{x}^{+}\). Then \({\{{U_{L} \cap {({H_{x}^{+} \times H_{x}^{+}})}}:{U \in \mathcal{U}}\}} = {\{{({U \cap H_{x}^{+}})}_{L}:{U \in \mathcal{U}}\}}\) indeed, for any \(U \in \mathcal{U}\), \({U_{L} \cap {({H_{x}^{+} \times H_{x}^{+}})}} = {L{({U \cap H_{x}^{+}})}}\). Therefore, \(\mathcal{L}_{_{x}} = \mathcal{L}_{_{x}}'\). Also, the restriction of the translation \(\lambda_{_{x}}:{S\longrightarrow S}\) on \(H_{x}^{+}\) is a translation on \(H_{x}^{+}\) and therefore, \({\lambda_{_{x}}|}_{_{H_{x}^{+}}}:{H_{x}^{+}\longrightarrow H_{x}^{+}}\) is an isomorphism in the sense of uniformity. Now, \(H_{x}^{+}\) being a closed subset of \(S\), \(V \cap H_{x}^{+}\) is a closed subset of the complete set \(V\) and so \(V \cap H_{x}^{+}\) is complete. Since \({{\lambda_{_{x}}|}_{_{H_{x}^{+}}}{({V \cap H_{x}^{+}})}} = {x + {({V \cap H_{x}^{+}})}}\) and \({({x + V})}^{\ast} = {x + {({V \cap H_{x}^{+}})}}\), it follows that \({({x + V})}^{\ast}\) is complete.

Let \(V\) be complete with respect to the uniformity, and let \(\mathcal{F}\) be a Cauchy filter on \(S_{s}\). Then there exists an element \(A\) in \(\mathcal{F}\) such that \({A \times A} \subseteq V_{L}\) and this implies that for any \({a,x} \in A\), \({{x' + a} \in V},{0_{x} = 0_{a}}\). So, for any \(x \in A\), \(a \in {({x + V})}^{\ast}\), for all \(a \in A\). This implies that the trace of \(\mathcal{F}\) on the complete subspace \({({x + V})}^{\ast}\) of \(S_{s}\) is a Cauchy filter which converges to a point \(x'\). Since \(x'\) is a cluster point of \(\mathcal{F}\), by [4, Chapter II, §3, No. 2, Corollary 2], \(\mathcal{F}\) converges to \(x'\).

Let \(U\) be an open set containing \(E^{+}{(S)}\) and \({E^{+}{(S)}} = {\{ e_{1},e_{2},\ldots,e_{n}\}}\). Since \(S\) is locally compact, for each \(i = {1,2,3,\ldots,n}\), there exists an open set \(V_{i}\) containing \(e_{i}\) such that \(e_{i} \in {cl_{S}V_{i}} \subseteq U\), where each \(cl_{S}V_{i}\) is compact. Now, \({E^{+}{(S)}} \subseteq {\bigcup\limits_{i = 1}^{n}{cl_{S}V_{i}}} \subseteq U\), which implies that \({E^{+}{(S)}} \subseteq {cl_{S}{\bigcup\limits_{i = 1}^{n}V_{i}}} \subseteq U\). Set \(V = {\bigcup\limits_{i = 1}^{n}V_{i}}\). Then \(V\) is an open neighbourhood of \(E^{+}{(S)}\) in \(S\) and \(cl_{S}V\) is compact with \({cl_{S}V} \subseteq U\). Now, every compact space being complete with respect to its unique uniformity, so by Proposition 5.18, it follows that \(S\) is complete.

Let \(W\) be an open set containing \(E^{+}{(S)}\) in \(S\). Since \(E^{+}{(S)}\) is compact, by Wallace theorem, there exist open set \(V\) containing \(E^{+}{(S)}\) in \(S\) such that \({V + V + V} \subseteq W\). Now \(\mathcal{G}\) being a Cauchy filter, there is an element \(B \in \mathcal{G}\) such that \({B \times B} \subseteq V_{L}\). Let \(b \in B\). Then \(U = {({b + V + b'})}^{\#} = {\{{x \in S}:{{b' + x + b} \in V}\}}\) is an open set containing \(E^{+}{(S)}\). Now, \(\mathcal{F}\) being a Cauchy filter, there is an element \(A \in \mathcal{F}\) such that \({A \times A} \subseteq U_{L}\). Then \({A + B} \in {\mathcal{F} + \mathcal{G}}\) and \({{({A + B})} \times {({A + B})}} \subseteq W_{L}\), indeed, for any \({a_{1},a_{2}} \in A\) and \({b_{1},b_{2}} \in B\), \({{(b_{1},b)},{(b,b_{2})}} \in {B \times B} \subseteq V_{L}\) impliy that \({{b_{1}' + b},{b' + b_{2}}} \in V\) and \(0_{b_{1}} = 0_{b} = 0_{b_{2}}\). Also, \({(a_{1},a_{2})} \in {A \times A} \subseteq U_{L}\) implies that \({a_{1}' + a_{2}} \in U = {({b + V + b'})}^{\#}\) and \(0_{a_{1}} = 0_{a_{2}}\). Then \({b' + a_{1}' + a_{2} + b} \in V\) which implies \({{({a_{1} + b_{1}})}' + {({a_{2} + b_{2}})}} = {b_{1}' + a_{1}' + a_{2} + b_{2}} = {{({b_{1}' + b})} + {({b' + a_{1}' + a_{2} + b})} + {({b' + b_{2}})}} \in {V + V + V} \subseteq W\) and \(0_{a_{1} + b_{1}} = {0_{a_{1}} + 0_{b_{1}}} = {0_{a_{2}} + 0_{b_{2}}} = 0_{a_{2} + b_{2}}\).

Since \(E^{+}{(S)}\) is compact, by Wallace theorem, there exist open set \(W \in \mathcal{B}\) such that \({W + {W \cdot W}} \subseteq V\). Now, \(\mathcal{F}\) being a Cauchy filter, there is an element \(F \in \mathcal{F}\) such that \({F + F'} \subseteq W\). Let \(a \in F\). Then using the continuity of the mappings \(R_{m}:\underset{x\mapsto{a\cdot x}}{\overset{S\longrightarrow S}{}}\) and \(L_{m}:\underset{x\mapsto{x\cdot a}}{\overset{S\longrightarrow S}{}}\) on \(E^{+}{(S)}\), there is an open set \(U\) containing \(E^{+}{(S)}\) such that \(U \subseteq W\), \({a \cdot U} \subseteq W\) and \({U \cdot a} \subseteq W\). Now, \(F \subseteq {a + W}\) implies \({F \cdot U} \subseteq {{a \cdot U} + {W \cdot U}} \subseteq {W + {W \cdot W}} \subseteq V\). Similarly, we can show that \({U \cdot F} \subseteq V\).

Let \(W\) be any neighbourhood of \(E^{+}{(S)}\) in \(S\). Since \(E^{+}{(S)}\) is compact, by Wallace theorem, there exist open set \(V\) containing \(E^{+}{(S)}\) in \(S\) such that \({V + V + {V \cdot V}} \subseteq W\). By lemma 5.24 there is an element \(U \in \mathcal{B}\), \(F_{1} \in \mathcal{F}\), \(G_{1} \in \mathcal{G}\) such that \(U \subseteq V\), \({F_{1} \cdot U} \subseteq V\) and \({U \cdot G_{1}} \subseteq V\). Now \(\mathcal{F}\) and \(\mathcal{G}\) being Cauchy filters, there exist \(F_{2} \in \mathcal{F}\), \(G_{2} \in \mathcal{G}\) such that \(F_{2} \subseteq F_{1}\) and \(G_{2} \subseteq G_{1}\) and \({F_{2} + F_{2}'} \subseteq V\), \({G_{2} + G_{2}'} \subseteq V\). Let \({a,a_{1}} \in F_{2}\) and \({b,b_{1}} \in G_{2}\). Then \(a = {a_{1} + w_{1}}\), \(b = {b_{1} + w_{2}}\) for some \({w_{1},w_{2}} \in W\). Then \({a \cdot b} = {{a_{1} \cdot b_{1}} + {a_{1} \cdot w_{2}} + {w_{1} \cdot b_{1}} + {w_{1} \cdot w_{2}}} \in {{a_{1} \cdot b_{1}} + V + V + V^{2}} \subseteq {{a_{1} \cdot b_{1}} + W}\), which implies that \({{F_{2} \cdot G_{2}} + {({F_{2} \cdot G_{2}})}'} \subseteq W\).

Clearly, \(N\) and \(N'\) are b-lattices of topological rings such that \({E^{+}{(N)}} = {E^{+}{(S)}}\) and \({E^{+}{(N')}} = {E^{+}{(S')}}\). Since \(f\) is a continuous homomorphism from \(N\) into \(N' \subseteq S'\), by Proposition 5.6, \(f\) is uniformly continuous function from \(N\) into \(N'\). So, by [4, Chapter II, §3, no. 6, Theorem 2], \(f\) can be extended uniquely to a mapping \(\hat{f}\) of \(S\) into \(S'\) such that \(\hat{f}\) is uniformly continuous. The mappings \(\phi:\underset{{(x,y)}\mapsto{\hat{f}⁢{({x+y})}}}{\overset{{S\times S}\longrightarrow S^{\prime}}{}}\) and \(\psi:\underset{{(x,y)}\mapsto{{\hat{f}⁢{(x)}}+{\hat{f}⁢{(y)}}}}{\overset{{S\times S}\longrightarrow S^{\prime}}{}}\) are continuous and agree on the dense subset \(N\) of \(S\). Therefore \(\phi = \psi\) on \(S\) which implies \({\hat{f}{({x + y})}} = {{\hat{f}{(x)}} + {\hat{f}{(y)}}}\) for every \({x,y} \in S\). In a similar way we can show that \({\hat{f}{({x \cdot y})}} = {{{\hat{f}{(x)}} \cdot \hat{f}}{(y)}}\). So \(\hat{f}\) is a homomorphism.

To prove the second part, let \(S\) be Hausdorff and complete, and let \(f\) is an isomorphism from \(N\) onto \(N'\). Then there exist an isomorphism \(g:{N'\longrightarrow N}\) such that \(f^{- 1} = g\). By the first part, \(g\) has a continuous extension \(\hat{g}:{S'\longrightarrow S}\) which is a homomorphism also. Now, \({g \circ f} = {id_{_{N}}}\) and \({f \circ g} = {id_{_{N'}}}\). Since \(S\) is Hausdorff, \({\hat{g} \circ \hat{f}} = {id_{_{S}}}\). Analogously, \({\hat{f} \circ \hat{g}} = {id_{{}_{S}^{}'}}\). This implies that \(\hat{f}\) is bijective and consequently, \(\hat{f}:{S\longrightarrow S'}\) is an isomorphism.

\({(i)}\Longrightarrow{({ii})}\) : Since \(S\) is isomorphic to a dense subsemiring of a complete idempotent semiring of topological rings \(\hat{S}\), from Proposition 5.29 the inverse map \(\gamma:\underset{x\mapsto x^{\prime}}{\overset{S\longrightarrow S}{}}\) is uniformly continuous, which implies that the image of a Cauchy filter under \(\gamma\) with respect to the uniformity of \(S\) is a Cauchy filter with respect to this uniformity.

\({({ii})}\Longrightarrow{(i)}\) : By virtue of [12, Chapter II, §7, Theorem 7.4] and Proposition 5.22 and 5.26, it is clear that the extensions of the binary operations \(a:\underset{{(x,y)}\mapsto{x+y}}{\overset{{S\times S}\longrightarrow S}{}}\) and \(m:\underset{{(x,y)}\mapsto{x⁢y}}{\overset{{S\times S}\longrightarrow S}{}}\) on \(\hat{S}\) respectively, \(\hat{a}:\underset{{(x,y)}\mapsto{x+y}}{\overset{{\hat{S}\times\hat{S}}\longrightarrow\hat{S}}{}}\) and \(\hat{m}:\underset{{(x,y)}\mapsto{x⁢y}}{\overset{{\hat{S}\times\hat{S}}\longrightarrow\hat{S}}{}}\) are continuous. Now we show that \(\hat{S}\) is an idempotent semiring of topological rings. Consider the functions \({f_{\hat{a}},g_{\hat{a}}}:{{\hat{S} \times \hat{S} \times \hat{S}}\longrightarrow{\hat{S} \times \hat{S} \times \hat{S}}}\) by \({f_{\hat{a}}{(x,y,z)}} = {x + {({y + z})}}\) and \({g_{\hat{a}}{(x,y,z)}} = {{({x + y})} + z}\). Since \(f_{\hat{a}} = g_{\hat{a}}\) on the dense subspace \(S\), it follows that the law \({(x,y)}\longmapsto{x + y}\) is associative on \(\hat{S}\). By a similar way we can prove that the binary operation \({(x,y)}\longmapsto{x \cdot y}\) is associative on \(\hat{S}\). Therefore, \(\hat{S}\) is a topological semiring.

Now the image under the inverse mapping \(x\longmapsto x'\) of a Cauchy filter with respect to the uniformity of \(S\) being a Cauchy filter with respect to this uniformity, it follows that the inverse mapping \(\gamma:\underset{x\mapsto x^{\prime}}{\overset{S\longrightarrow S}{}}\) has a continuous extension on \(\hat{S}\). Let \(\hat{\gamma}:{\hat{S}\longrightarrow\hat{S}}\) be the continuous extension of \(\gamma:\underset{x\mapsto x^{\prime}}{\overset{S\longrightarrow S}{}}\). Since \(S\) is dense in \(\hat{S}\) and \(\hat{S}\) is a Hausdorff uniform space, this extension \(\hat{\gamma}\) is unique. Now we show that \(\hat{S}\) is a completely regular semiring. Let \(a \in \hat{S}\). Since \(S\) is dense in \(\hat{S}\), there exists a net \((a_{_{\alpha}})\) in \(S\) such that \(a_{_{\alpha}}\longrightarrow a\). Now \(\hat{\gamma}\) being continuous, \({\hat{\gamma}{(a_{_{\alpha}})}}\longrightarrow{\hat{\gamma}{(a)}}\). Since \(\hat{\gamma}\) is an extension of \(\gamma\), it follows that \(a_{_{\alpha}}'\longrightarrow{\hat{\gamma}{(a)}}\). Now \(S\) being completely regular semiring, for each \(\alpha\), \({a_{_{\alpha}} + a_{\alpha}' + a_{_{\alpha}}} = a_{_{\alpha}}\), \({a_{_{\alpha}}' + a_{\alpha} + a_{_{\alpha}}'} = a_{_{\alpha}}'\) and \({a_{_{\alpha}} \cdot {({a_{_{\alpha}} + a_{\alpha}'})}} = {a_{_{\alpha}} + a_{\alpha}'}\). This implies that \({a + {\hat{\gamma}{(a)}} + a} = a\), \({{\hat{\gamma}{(a)}} + a + {\hat{\gamma}{(a)}}} = {\hat{\gamma}{(a)}}\) and \({a \cdot {({a + {\hat{\gamma}{(a)}}})}} = {a + {\hat{\gamma}{(a)}}}\). This implies that \(a\) is completely regular and so \(\hat{S}\) is a topological completely regular semiring. Now we show that \({E^{+}{(\hat{S})}} = {E^{+}{(S)}}\). Clearly, \({E^{+}{(S)}} \subseteq {E^{+}{(\hat{S})}}\). Let \(\hat{e} \in {E^{+}{(\hat{S})}}\). Clearly, \({\hat{\gamma}{(\hat{e})}} = \hat{e}\). Since \(\hat{e} \in \hat{S} = \overline{S}\), there is a net \(e_{_{\alpha}}\) in \(S\) such that \(e_{_{\alpha}}\longrightarrow\hat{e}\). This implies that \({e_{_{\alpha}} + e_{_{\alpha}}'}\longrightarrow{\hat{e} + {\hat{\gamma}{(\hat{e})}}} = \hat{e}\). Now for each \(\alpha\), \({e_{_{\alpha}} + e_{_{\alpha}}'} \in {E^{+}{(S)}}\) and \(E^{+}{(S)}\) is closed in \(S\). So \(\hat{e} \in {E^{+}{(S)}}\). Therefore, \({E^{+}{(\hat{S})}} = {E^{+}{(S)}}\). This implies that \(E^{+}{(\hat{S})}\) is compact in \(\hat{S}\). Now we show that each \(\mathcal{H}^{+}\)-class is open in \(\hat{S}\). Now from [10, Theorem 2.1] it follows that \(\varphi:\underset{x\mapsto\, 0_{x}}{\overset{\hat{S}\longrightarrow\hat{S}}{}}\) is a continuous function. We denote the \(\mathcal{H}^{+}\)-class containing an element \(a\) in \(\hat{S}\) by \({\hat{H^{+}}}_{_{a}}\). Now for any \(a \in \hat{S}\), \({\varphi^{- 1}{({\{ 0_{a}\}})}} = {\hat{H^{+}}}_{_{a}}\) and this implies that each \(\mathcal{H}^{+}\)-class is closed in \(\hat{S}\). Since \(E^{+}{(\hat{S})}\) is finite, each \(\mathcal{H}^{+}\)-class is open in \(\hat{S}\). Therefore, by [10, Corollary 2.16], \(\hat{S}\) is an idempotent semiring of topological rings with compact \(E^{+}{(\hat{S})}\).

Let \(\hat{\mathcal{U}}\) be the uniformity on \(\hat{S}\). We show that \((\hat{S},\hat{\mathcal{U}})\) is a complete uniform space. Let \(\mathcal{U}\) be the uniformity on \(\hat{S}\) obtained by completing the uniformity of \(S\). Then \(\hat{\mathcal{U}}\) and \(\mathcal{U}\) induce the same uniformity on \(S\) and this implies that every Cauchy filter base on \(S\) with respect to \(\hat{\mathcal{U}}\) is also a Cauchy filter base on \(S\) with respect to \(\mathcal{U}\). Let \(\mathcal{B}\) be a Cauchy filter base on \(S\) with respect to the uniformity \(\hat{\mathcal{U}}\). Then \(\mathcal{U}\) being a complete uniformity, \(\mathcal{B}\) converges in \(\hat{S}\). Now \(\hat{\mathcal{U}}\) and \(\mathcal{U}\) induce the same topology on \(\hat{S}\). So, by proposition 5.28, \(\hat{\mathcal{U}}\) is a complete uniformity. Hence \(\hat{S}\) is a complete uniform space. In particular, \(\mathcal{U}\) and \(\hat{\mathcal{U}}\) coincide.