# Arens-Fort Space is not First-Countable

## Theorem

Let $T = \left({S, \tau}\right)$ be the Arens-Fort space.

Then $T$ is not a first-countable space.

## Proof

Assume that $T$ is first-countable.

Then there exists a countable local base $B_0 = \left\{{U_i}\right\}_{i=1}^\infty$ for $\left({0, 0}\right)$.

Suppose there does not exist a point $\left({n_i, m_i}\right) \in U_i$ such that $n_i > i$ and $m_i > i$.

Then:

- $\forall \left({n, m}\right) \in U_i: n \le i$ or $m \le i$

Now suppose that there does not exist some $\left({n, m}\right) \in U_i$ with $n > i$.

Let $S_k$ be defined as:

- $S_k = \left\{{n: \left({k, n}\right) \notin U_i}\right\}$

The definition of the Arens-Fort space clearly stipulates that $S_k$ can only be a infinite set for a finite number of $k \in \Z_{\ge 0}$.

But:

- $k > i \implies S_k = \Z_{\ge 0}$

So we have shown that:

- $\exists \left({n, m}\right) \in U_i: n > i$

So, let $n > i$, and let $m \le i$.

Then $\left\{{m: m > i}\right\} \subseteq S_n$.

Thus $S_n = \left\{{m: \left({n, m}\right) \notin U_i} \right\}$ is infinite for all $n > i$.

Again, this contradicts the definition of the Arens-Fort space.

So, from the definition of the neighborhood of $\left({0, 0}\right)$, for each $U_i \in B_0$ there is a point $\left({n_i, m_i}\right) \in U_i$ such that both $n_i > i$ and $m_i > i$.

Now let the set $E$ be constructed as:

- $E := X \setminus \left\{{\left({n_i, m_i}\right)}\right\}_{i=1}^\infty$

We now prove that $E$ is a neighborhood of $\left({0, 0}\right)$.

By the definition of Arens-Fort space, it is enough to show that:

- $\forall i \in \N: S_i$ is finite.

We have that:

\(\displaystyle S_i\) | \(=\) | \(\displaystyle \left\{ {m: \left({i, m}\right) \notin E}\right\}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left\{ {m: \left({i, m}\right) \in \left\{ {\left({n_j, m_j}\right)}\right\}_{j=1}^\infty}\right\}\) |

But we defined $n_j > j$.

So if $j > i \implies n_j > j > i$.

Thus $S_i \subseteq \left\{{m_j}\right\}_{j \mathop = 1}^i$ is finite.

However, $\left({n_i, m_i}\right) \notin E$.

Therefore there exists no $U_i \subseteq E$.

From this contradiction it follows that $T$ cannot be first-countable.

$\blacksquare$

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{II}: \ 26: \ 3$