Showing that a space is normal and not locally compact Let $E$ be the set of all ordered pairs $(m,n)$ of non-negative integers. Topologize $E$ as follows:


*

*For a point $(m,n)\neq (0,0)$, any set containing $(m,n)$ is a neighbourhood of $(m,n)$. 

*A set $U$ containing $(0,0)$ is a neighborhood of $(0,0)$ if and only if for all except a finite number of $m$s the set $\lbrace n: (m,n)\notin U \rbrace$ is finite.





*

*Show that $E$ is not locally compact.

*Show that $E$ is normal.
 A: I assume that you assume a neighbourhood to be open, otherwise i am not sure if your definition determines a topology.
$E$ is not locally compact: First observe that for each $(m,n) \neq (0,0)$ the set $\{(m,n)\}$ is open, since by definition of your topology $\{(m,n)\}$ is a neighbourhood of $(m,n)$.
Now consider an arbitrary neighbourhood $U$ of $\{(0,0)\}$. From the definition of the topology there exists an infinite family $\{m_i\}_{i \in I}$ such that all $m_i$'s are nonzero and for each $m_i$ the set $\{ n \vert (m_i,n) \in U\}$ is infinite. For each $i$ choose $n_i$ such that $(m_i,n_i) \in U$. Set $V := U \setminus \bigcup_i \{(m_i,n_i)\}$. Then $V$ is still open, as it can be seen easily from the construction. Therefore we obtain an open cover $U = V \cup \bigcup_i \{(m_i,n_i)\}$. Since all this sets are disjoint there does not exist a finite subcover. Therefore, $U$ cannot be compact.
$E$ is regular: Let $A$ be any closed subset and $(m,n) \notin A$. We need to seperate $A$ and $(m,n)$ by open neighbourhoods; 
Case 1: $(0,0) \in A$: Let $U$ be any open neighbourhood of $A$. Then $U$ is also an open neighbourhood of $(0,0)$. From the topologies definition it follows easily that also $V := U \setminus \{(m,n)\}$ is an open neighbourhood of $(0,0)$ and clearly $A \subset V$. As above $\{(m,n)\}$ is open and therefore we can seperate $A$ and $(m,n)$ by the disjoint open sets $V$ and $\{(m,n)\}$.
Case 2: $(m,n) = (0,0)$: $A$ is a closed set not containing $(0,0)$. Thus $A^c$ is an open neighbourhood of $(0,0)$. Moreover $A$ is also open since all points different from $(0,0)$ are open. So we can seperate $(0,0)$ and $A$ by the disjoint open sets $A^c$ and $A$.
Case 3: $(m,n) \neq (0,0)$ and $(0,0) \notin A$. In this case the sets $\{(m,n)\}$ and $A$ are both open and we are done.
A: The space you are concerned with is called the Arens–Fort space. For these questions, the important facts about this space are the following:


*

*Any subset of $E$ which does not contain $(0,0)$ is open.

*Any neighbourhood of $(0,0)$ is actually clopen (= closed and open).

*If $N$ is any neighbourhood of $(0,0)$ and $m > 0$, then $N \setminus ( \{ m \} \times \mathbb{N} )$ is also a neighbourhood of $(0,0)$.



It should be clear that the only point at which $E$ may not be locally compact is $(0,0)$. So let $N$ be any neighbourhood of $(0,0)$, and we'll show that it is not compact: exhibit an open cover with no finite subcover. There must be an $m>0$ such that the set $A = \{ n : (m,n) \in N \}$ is infinite (even cofinite). Now consider the following open cover of $N$: $$\{ N \setminus ( \{ m \} \times \mathbb{N} ) \} \cup \{ \{ (m,n) \} : n \in A \}.$$ It is easy to see that the sets in this cover are pairwise disjoint, and so it has no proper subcover (let alone a finite subcover).

For normality, suppose that $A \subseteq E$ is closed, and $U \subseteq E$ is open with $A \subseteq U$.  There are two cases.  If $(0,0) \in U$, then as above $U$ is clopen, so taking $V = U$ we have $A \subseteq V = \overline{V} \subseteq U$.  If $(0,0) \notin U$, then $(0,0) \notin A$, and so $A$ is clopen. Taking $V = A$ we have $A \subseteq V = \overline{V} \subseteq U$.
