# Pete L. Clark's Convergence Notes

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 $A = \overline{A}$.

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Maybe the man will give you an answer here! I always love seeing (moreso on MO, if memory serves) authors' input on questions. – The Chaz 2.0 May 1 '12 at 19:09
Agreed! I was quite eager to read them in particular when I saw the author's name. As he has given many great insightful answers to my questions here already. :) – Kyle May 1 '12 at 19:11
Pete, you repeated "a mapping" in the first paragraph. (consider it for the next edition) – Peter Tamaroff May 1 '12 at 19:12

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.

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 I haven't read through this yet, but I skimmed it and believe it. Thanks! I'll go through it in more detail now. – Kyle May 1 '12 at 19:54 OK. Nice argument! Thanks very much! – Kyle May 1 '12 at 20:00

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. $\dashv$

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 This is very helpful, I will think about this for a while. Thanks! – Kyle May 1 '12 at 19:55 As for the topology generated by $D$, do you just mean the metric topology? Also, this may not be related, but this reminds me of a problem I worked on in my first course in topology which dealt with the order topology applied to $\mathbb{R}^{2}$, where the order was the dictionary order: $(a,b)\leq (c,d)$ if $a\leq b$ or $a = b$ and $c\leq d$...... or perhaps I am prematurely losing my mind. Thanks again! – Kyle May 1 '12 at 20:04 Every point of $\Bbb R^2$ is the intersection of some two crosses, so the topology generated by this set of crosses is the discrete topology, which is even first countable. The term cross topology usually refers to the topology in which a set is open iff it intersects each vertical and horizontal line in an open set. – Brian M. Scott May 1 '12 at 20:10 @Kyle this is one of the things I don't remember well. I believe it is correct to say that a set $\mathcal{O}$ is open in the cross topology iff it is contained in an "interval cross" within the set $\mathcal{O}$. – rschwieb May 1 '12 at 20:10 @Brian M. Scott the one that you describe as "usual" is the one I intended. Please help make sure this is how it is written. – rschwieb May 1 '12 at 20:15