Pete L. Clark's Convergence Notes

Question

I had initially sought out a better understanding of filters and nets, and a few quick google searches showed this document as highly recommended. (And they are excellent!)

I'm having a bit of trouble verifying one of the facts which is stated on page 6: fact 2. Link: http://math.uga.edu/~pete/convergence.pdf

It says that $X$ is Frechet if and only if every subspace $Y$ of $X$ is sequential.

It is a few lines to show the $(\Rightarrow)$ direction, but I'm really stuck on the $(\Leftarrow)$ direction. In fact, I'm actually not able to find the difference between a sequential space (one where sequentially closed subsets are closed) and a Frechet space (one where the sequential closure of subsets coincides with the topological closure).

It seems that the definitions are the same if $sc(sc(A)) = sc(A)$ for all subsets $A$ of some topological space. I can't prove this fact, but cannot come up with a counter-example either.

Of course Frechet implies sequential: If $A\subset X$ is sequentially closed, then $sc(A) = A$. Since $X$ is Frechet, $sc(A) = \overline{A}$. So $A = \overline{A}$.

Answer 1

Let $X$ be a space.

1. $X$ is Fréchet if the following is true: if $A\subseteq X$, and $x\in\operatorname{cl}A$, then there is a sequence of points of $A$ converging to $x$.

2. $X$ is sequential if every sequentially closed subset of $X$ is closed. A set $A\subseteq X$ is sequentially closed iff the following is true: if $x\in X$ is the limit of a sequence of points of $A$, then $x\in A$.

Suppose that $X$ is not Fréchet; then there are a subset $A\subseteq X$ and a point $x\in\operatorname{cl}A$ such that no sequence of points of $A$ converges to $X$. Let $Y=\{x\}\cup A$; I claim that $A$ is a sequentially closed subset of $Y$ that is not closed in $Y$. Clearly $A$ is not closed in $Y$, since $x\in\operatorname{cl}_YA\setminus A$. To see that $A$ is sequentially closed in $Y$, just note that if $A$ has any convergent sequences at all, their limits must already be in $A$, since $x$ is not the limit of any convergent sequence in $A$. Thus, $Y$ is a non-sequential subspace of $X$. This proves the ($\Leftarrow$) direction.

This post from Dan Ma's Topology Blog contains a description of the Arens space, which is the classical example of a sequential space that is not Fréchet. You'll probably also want to look at some of his earlier posts on sequential and Fréchet spaces; there are links at the end of the post.

Answer 2

For the plane $\mathbb{R}^2$ the cross topology is the topology in which a set $\mathcal{O}$ is open iff each point $x\in\mathcal{O}$ is contained in a "cross" formed by a horizontal open interval and vertical open interval within $\mathcal{O}$. More precisely, for $r>0$ and $p=\langle x,y\rangle\in\Bbb R^2$ let $$C(p,r)=\Big((x-r,x+r)\times\{y\}\Big)\cup\Big(\{x\}\times(y-r,y+r)\Big)\;;$$ $\mathcal{O}$ is open iff for each $p\in\mathcal{O}$ there is an $r_p>0$ such that $C(p,r_p)\subseteq\mathcal{O}$.

Let $A$ be the open first quadrant; clearly the origin is in the closure of $A$, but it's not in the sequential closure of $A$, so the plane with the cross topology is not Fréchet.

To see this, suppose that $\sigma=\langle p_n:n\in\Bbb N\rangle$ is a sequence in $A$, where $p_n=\langle x_n,y_n\rangle$. The cross topology is finer than the Euclidean topology, so $\sigma$ cannot converge to the origin in the cross topology unless it does so in the Euclidean topology. Thus, we may assume that $\langle x_n:n\in\Bbb N\rangle\to 0$ and $\langle y_n:n\in\Bbb N\rangle\to 0$. By passing to a subsequence, if necessary, we may further assume that $\langle x_n:n\in\Bbb N\rangle$ and $\langle y_n:n\in\Bbb N\rangle$ are strictly decreasing. For $n\in\Bbb N$ let $S_n$ be the segment $\overline{p_np_{n+1}}$, and let $S=\bigcup_{n\in\Bbb N}S_n$; $S$ is the graph of a piecewise linear function bijection of $(0,x_0]$ onto $(0,y_0]$ (with countably infinitely many pieces). Let $U=\Bbb R^2\setminus S$; then $U$ is an open neighborhood of the origin in the cross topology that contains no term of $\sigma$, so $\sigma$ does not converge to the origin in the cross topology. $\dashv$

Essentially the same argument shows that a sequence converges to a point $p$ in the cross topology iff it is eventually in every $C(p,r)$. We can use this to show that the plane is sequential in the cross topology.

To see this, suppose that $S\subseteq\Bbb R^2$ is sequentially closed in the cross topology. If $p=\langle x,y\rangle\in\Bbb R^2\setminus S$, no sequence in $S$ converges to $p$, so there is some $r>0$ such that $C(p,r)\cap S=\varnothing$; let $V_0=C(p,r)$. No sequence in $S$ converges to any point of $V_0$, so for each $q\in V_0$ there is an $r_q>0$ such that $C(q,r_q)\cap S=\varnothing$; let $V_1=\bigcup\{C(q,r_q):q\in V_0\}$. In general, if $V_n$ is disjoint from $S$, then for each $q\in V_n$ there is an $r_q>0$ such that $C(q,r_q)\cap S=\varnothing$, and we let $V_{n+1}=\bigcup\{C(q,r_q):q\in V_n\}$. Finally, let $V=\bigcup\{V_n:n\in\omega\}$; the construction ensures that $V$ contains a cross centred at each of its points, so in the cross topology $V$ is an open nbhd of $p$ disjoint from $S$. Thus, $S$ is closed in the cross topology, which is therefore sequential. $\dashv$