Is Appert space Fréchet-Urysohn?

Question

I was asked if the set of positive natural numbers $\mathbb{N}$ with the Appert topology is a Fréchet-Urysohn space, that is, for every $A\subset \mathbb{N}$ and $p\in \bar{A}$, there is a sequence in $A$ converging to $p$. Could someone give me any ideas on how to solve this problem?

Recall that the Appert topology on $\mathbb{N}$ is defined as follows:

$$G\in \tau \iff 1\notin G \,\, or \,\, 1\in G \,\, and \,\,\lim\limits_{n\to\infty}\dfrac{|N(n,G)|}{n}=1.$$ where $N(n,G)=\{m\in G|m\leq n\}$.

Answer 1

Let $D=\Bbb N\setminus\{1\}$; clearly $1\in\operatorname{cl}D$. Let $\sigma=\langle n_k:k\in\Bbb N\rangle$ be any sequence in $D$, and let $S=\{n_k:k\in\Bbb N\}$, the set of distinct terms of $\sigma$.

• Show that if $S$ is finite (i.e., if $\sigma$ assumes only finitely many distinct values), then $\Bbb N\setminus S$ is an open nbhd of $1$.
• Show that if $S$ is infinite, then $\sigma$ has a strictly increasing subsequence $\langle n_{m_k}:k\in\Bbb N\rangle$, and verify that $n_{m_k}\ge k$ for each $k\in\Bbb N$.
• Let $A=\{n_{m_{2^k}}:k\in\Bbb N\}$, and show that $\Bbb N\setminus A$ is an open nbhd of $1$.
• Conclude that $\sigma$ does not converge to $1$ and hence that the space is not Fréchet-Uryson.