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Let $X$ be a topological space and $q \in X$.

$X$ is strictly Frechet at $q$, if, for all $A_n \subset X, q \in \bigcap_{n \in \omega} \overline {A_n}$ implies the existence of a sequence $q_n \in A_n$ with $\lim_{n \rightarrow \infty} q_n = q$.

We say that $X$ is Strictly-Frechet, if it is strictly Frechet at any point $q \in X$.

$X$ is Frechet at $q \in X$, if, $A \subset X$, $q \in \overline A$, implies the existence of a sequence $\{ q_n \} \subset A$ such that $\lim q_n = q$.

We say that $X$ is Frechet, if it is Frechet at any point $q \in X$.

I have been trying for a while to think of a space which is Frechet but not Strictly-Frechet. Any ideas, directions or examples?

Thank you!

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2 Answers 2

up vote 4 down vote accepted

An example is the quotient space $\mathbb{R} / \mathbb{N}$ (all natural numbers are collapsed to a point, which I'll denote by $\mathord{*}$).

Claim. This space is Fréchet(–Urysohn).

sketch. Every point except $*$ has a countable neighbourhood base, so the only possible problem is at $\mathord{*}$. So let $A \subseteq \mathbb{R} / \mathbb{N}$, and suppose that $\mathord{*} \in \overline{A}$. Without loss of generality we may assume that $\mathord{*} \notin A$, so that $A \subseteq \mathbb{R} \setminus \mathbb{N}$. It is not too difficult to show that there must be some $n \in \mathbb{N}$ such that $n \in \mathrm{cl}_{\mathbb{R}} ( A )$, taking any sequence in $A$ converging (in the real line) to $n$ we get a sequence in $A$ which converges (in $\mathbb{R} / \mathbb{N}$) to $\mathord{*}$.

Claim. The space is not strictly Fréchet(–Urysohn).

sketch. For each $n \in \mathbb{N}$ let $A_n$ be the open interval $( n , n + \frac{1}{2} )$. Clearly $\mathord{*} \in \overline{A_n}$ for all $n$. However, if you choose a sequence $\langle a_n \rangle_{n \in \mathbb{N}}$ such that $a_n \in A_n$ for each $n$, then you can construct an open neighbourhood of $\mathord{*}$ which contains no element of this sequence. (In fact, $(\mathbb{R} / \mathbb{N}) \setminus \{ a_n : n \in \mathbb{N} \}$ will be such an open neighbourhood of $\mathord{*}$.)

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OK, understood. Thank you for the simple and readable example! –  Shir Sivroni Apr 16 at 13:26

According to this paper by Gruenhage $S_\omega$, the countable sequential fan, is an example of a Fréchet but not strictly Fréchet space. (Note that the paper calls this property $\alpha_4$-FU, countably bisequential, or strongly Fréchet; this might help finding references). $S_\omega$ is the quotient space that results when we take countably many disjoint convergent sequences, where we identify the limits to one point.

It's not too hard to see that $S_\omega$ works: take $A_n$ be the $n$-th sequence (without the limit), then when we take $q$ as the identified limit, then $q \in \cap_n \overline{A_n}$, but taking only one point $q_n$ from each $A_n$ the set $\{q\} \cup \cup_n (A_n \setminus \{q_n\})$ is an open neighbourhood of $q$ that misses all point of the sequence, so $(q_n)$ does not converge to $q$.

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OK, I will go over this article. Thank you! –  Shir Sivroni Apr 16 at 13:25
I think that the property of being strictly Fréchet is actually $α_2$-FU, which is stronger than $α_4$-FU, which corresponds to the property of being strongly Fréchet. Of course this doesn't matter for the counterexample. –  user87690 Dec 7 at 14:12
@user87690 it seems you're right. –  Henno Brandsma Dec 7 at 14:15

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