Why is the sequential closure not sequentially closed?

Solution 1:

Yes, you're right: Fréchet-Uryson spaces are precisely the spaces in which the sequential closure is the same as the ordinary closure. For a specific example you can use the Arens space, which is discussed in Dan Ma’s Topology Blog: the sequential closure of the set of isolated points of the Arens space is not closed.

It's true that the sequential closure operator is not a true closure operator in the usual sense of the term; this is a minor nuisance, and one might wish for a better term, but I hardly think that it qualifies as exceptionally inconvenient.

Added: To turn the sequential closure operator into a true closure operator, you have to iterate it, possibly transfinitely. That is, if $\operatorname{scl}$ denotes the sequential closure, define

$$\operatorname{scl}^\eta A=\begin{cases} A,&\text{if }\eta=0\\\\ \operatorname{scl}\bigcup_{\xi<\eta}\operatorname{scl}^\xi A,&\text{if }\eta>0\;. \end{cases}$$

There is always an ordinal $\eta$ such that $\operatorname{scl}^\xi A=\operatorname{scl}^\eta A$ for all $\xi\ge\eta$; if we define $\operatorname{scl}^*A=\operatorname{scl}^\eta A,$ then $\operatorname{scl}^*$ is a true closure operator, and a space $X$ is sequential iff $\operatorname{scl}_X^*$ is identical to the ordinary closure.

Solution 2:

It is indeed the case that the sequential closure $[A]_{\text{seq}}$ of a set $A$ need not be sequentially closed. Even though $[A]_{\text{seq}}$ contains limits of all convergent sequences from $A$, it need not contain the limits of convergent sequences from $[A]_{\text{seq}}-A$. I'm inclined to agree with you that the terminology "sequential closure" is not very good.