Why is the sequential closure not sequentially closed?

Question

In my understanding, a sequentially closed subset $A\subset X$ of a topological space $X$ is one that contains every sequential limit point of itself, whereas the sequential closure of $A$ is defined as $[A]_{seq}=\{x\in X: \exists (x_n)\in A^{\mathbb{N}}: x_n\to x\}$.

I think that the sequential closure need not be sequentially closed, because assuming to the contrary that every sequential closure was sequentially closed, I can construct the contradiction that every sequential space is a Frechet-Urysohn space as follows:

Let $X$ be a sequential space. We must show $\overline{A}=[A]_{seq}$.

"$\overline{A}\subset[A]_{seq}$": $[A]_{seq}$ is sequentially closed, thus closed since $X$ is sequential. Also $A\subset[A]_{seq}$ trivially. Hence $\overline{A}=\bigcap_{A\subset F\subset X closed}F \subset [A]_{seq}$.

"$\overline{A}\supset[A]_{seq}$": This is true for every topological space (every limit of a sequence is part of the closure)

Am I right? If yes, isn't the notation exceptionally inconvenient?

Brian M. Scott · Answer 1 · 2013-10-12T19:01:51.833

7

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.

edited Oct 12 '13 at 19:01

answered Aug 02 '13 at 23:12

Brian M. Scott

616,228

Thanks for the elaborate answer. Andreas was faster though. I chose "exceptionally inconvenient" because I was slightly upset about how long it took me to figure that out ;) – Bananach Aug 02 '13 at 23:44
@Bananach: Andreas was faster because I had to hunt up the link. :-) (Not a problem.) You’re welcome. – Brian M. Scott Aug 02 '13 at 23:47
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@StefanH: Uryson is not a typo: it’s simply a different (and preferable) transliteration of the Cyrillic Урысон. – Brian M. Scott Oct 12 '13 at 19:03
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Ok. I really thought it was a typo, as I only ever encountered it as Urysohn. – Stefan Hamcke Oct 12 '13 at 19:09
@StefanH: No problem: I was just explaining. Thanks for catching the real typo in the formula. – Brian M. Scott Oct 12 '13 at 19:11

score 5 · Accepted Answer · answered Aug 02 '13 at 23:10

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.

Why is the sequential closure not sequentially closed?

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