Arens-Fort Space is not First-Countable

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Theorem

Let $T = \struct {S, \tau}$ be the Arens-Fort space.


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


Proof

Aiming for a contradiction, suppose $T$ is first-countable.

Then there exists a countable local base $B_0 = \sequence {U_i}_{i \mathop = 1}^\infty$ for $\tuple {0, 0}$.


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

Then:

$\forall \tuple {n, m} \in U_i: n \le i$ or $m \le i$

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

Let $S_k$ be defined as:

$S_k = \set {n: \tuple {k, n} \notin U_i}$

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:

$\exists \tuple {n, m} \in U_i: n > i$

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

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

Thus $S_n = \set {m: \tuple {n, m} \notin U_i}$ is infinite for all $n > i$.

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


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


Now let the set $E$ be constructed as:

$E := X \setminus \sequence {\tuple {n_i, m_i} }_{i \mathop = 1}^\infty$

We now prove that $E$ is a neighborhood of $\tuple {0, 0}$.

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

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

We have that:

\(\ds S_i\) \(=\) \(\ds \set {m: \tuple {i, m} \notin E}\)
\(\ds \) \(=\) \(\ds \set {m: \tuple {i, m} \in \sequence {\tuple {n_j, m_j} }_{j \mathop = 1}^\infty}\)

But we defined $n_j > j$.

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

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

However, $\tuple {n_i, m_i} \notin E$.

Therefore there exists no $U_i \subseteq E$.

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

$\blacksquare$


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