Is the Arens's space actually sequential??

Question

The Arens's space is presented here (and in many questions on this web) as the canonical example of a sequential space that is not Frèchet(-Urysohn).

Let me recall how the space is constructed. We take as the total space $X=\mathbb Z^+\cup\{\ast\}\cup(\mathbb Z^+\times\mathbb Z^+)$, where $\{\ast\}$ denotes a point that is not an integer (I could wirte $\infty$ as in the link). In the link, the topology is given by the fundamental system of neighbourhoods of each point:

• Points $(n,m)$ are isolated.

• For positive integers, neighbourhoods of a point $n$ are of the form $B_m(n)=\{n\}\cup\{(n,j),j\geq m\}$, for $m\in\mathbb Z^+$.

• For $\ast$, the neighbourhoods are obtain removing from $X$ finitely many $B_1(n)$ and then removing finitely many $(m,n)$ of the resulting set.

Now, let me explain my doubt. I focus on the point $\ast$. Notice that every neighbourhood of $\ast$ contains $\mathbb Z^+$. Hence, the closure $\bar{\{\ast\}}$ is the set $\mathbb Z^+\cup\{\ast\}$. Then the set $\{\ast\}$ is not closed. On the otehr hand, consider a sequence $\{x_n\}$ on $\{\ast\}$. Clearly, $x_n=\ast$ for all $n\in\mathbb N$. Moreover, the unique limit for this sequence is $\ast$. Hence $\{\ast\}$ is a sequentially closed set. If $X$ is sequential, then $\{\ast\}$ should also be closed.

I'm wrong, of course. But where is my mistake??

Thanks in advance.

Answer 1

$\{\ast\}$ is certainly closed, Arens space is Hausdorff so $T_1$ in particular. In fact it's easy to check that all points of the space unequal to $\ast$ have the property that all its basic neighbourhoods do not contain $\ast$ at all, making $X \setminus \{\ast\}$ clearly open.

Every neighbourhood of $\ast$ intersects both $\mathbb{Z}^+$ and $\mathbb{Z}^+ \times \mathbb{Z}^+$, but does not contain that whole set, because we can omit points from it.

The sequential closure of $A = \mathbb{Z} \times \mathbb{Z}$ is $A \cup \mathbb{Z}^+$ and the sequential closure of that subset is $X$. So the space is sequential (the closure is the iteration of the sequential closure) but not Fréchet-Urysohn (this means that closure = sequential closure right away). Your link shows an alternate way of showing sequentiality by writing $X$ as a quotient of a metric space.

Answer 2

Notice that every neighborhood of $*$ contains $\mathbb{Z}^+$.

For any $n\in \mathbb{Z}^+$, $X\setminus B_1(n)$ is a neighborhood of $*$ with does not contain $n$.

Hence, the closure $\overline{\{*\}}$ is the set $\mathbb{ℤ}^+\cup \{*\}$.

Even if your previous assertion were correct, this would not follow. The closure of $\{*\}$ is the intersection of all closed sets containing $*$. Equivalently, it is the set of all $x\in X$ such that every neighborhood of $x$ contains $*$. In fact, no point in $X$ other than $*$ has any basic open neighborhood which contains $*$. So $\{*\}$ is obviously closed.