Question

Hi, I have recently encountered these two definitions of a sequential space and a space of countable tightness. And I seem to have difficulty understanding what is the difference between these two definitions. For example, I know that the space of ultrafilters over ,say, R or N is not weakly Frechet Urysohn so it should not be sequential. But how can one show it directly from the definition? Also, Does these spaces have countble tightness? Thanks!

Answer 1

Just a partial answer.

For $\beta \mathbb{N}$ (the set of all ultrafilters on $\mathbb{N}$ with the Stone topology) it is not hard to see that a sequence converges iff it is eventually constant. Hence any subset of $\beta \mathbb{N}$ is sequentially open -- and of course, $\beta \mathbb{N}$ is not discrete, so it cannot be sequential. Similarly, for the ultrafilters on $\mathbb{R}$.

If I recall correctly, this 'trivial sequential convergence' holds in all extremally disconnected spaces -- this should be an exercise in the book 'Rings of continuous functions' by Gillman and Jerison (there is also a PDF/TeX-file with all exercise solutions freely available on the web somewhere).

Also, I could be wrong, but I think $\beta \mathbb{N}$ is not countably tight since its remainder is not (since there exist weak P-points). Maybe somebody else can confirm or reject.

Answer 2

All three notions, "countably tight," "sequential," and "Frechet-Urysohn," say that each point $p$ in the closure of a set $A$ can be "approached in some countable way" by points from $A$. The difference is in the "countable ways." The strongest of the three, Frechet-Urysohn, requires $p$ to be the limit of a sequence of points in $A$. "Sequential" allows iteration of this: Take the set of limits of sequences of points in $A$; then take limits of sequences of such points; then take limits ... ; eventually you get $p$. (More formally, let $A_0=A$, let $A_{\alpha+1}$ be the set of limits of sequences from $A_\alpha$, and for limit ordinals let $A_\lambda$ be the union of all $A_\alpha$'s for $\alpha<\lambda$. Then $p$ should be in $A_\alpha$ for some $\alpha$. The $\alpha$ here can always be taken to be countable, but that's the best bound you can get.) Finally, "countably tight" only requires $p$ to be in the closure of some countable subset of $A$; that can happen even if there are no convergent sequences of points from $A$ (except of course the eventually constant sequences).

Answer 3

Some examples to expand Andreas' answer might be of interest: (it's too much to fit in a comment so I'm adding it as an answer, though I think Andreas' response is great)

• any metric space will be Frechet-Urysohn (choose $x_n$ in $A$ within $1/n$ of $p$); (more generally, any first-countable space is F-U: just choose $x_n$ in the intersection of $A$ and $U_n$ where $\{U_n\}_n$ is a countable base at the limit point);

• a sequential but not Frechet-Urysohn space is given by taking $((\omega+1)\times\omega)\cup\{*\}$ where each copy of $\omega+1$ has the usual topology and a base for $*$ consists of sets $A_{m,n}=\{(m,n)|m>M,n>N_m\}$ for $M,N_m\in\omega$ (ie cofinitely many elements of cofinitely many fibers) - then $*$ is in the closure of $\omega\times\omega$ but is not the limit of any sequence of points in $\omega\times\omega$; however, it is the limit of the sequence $x_n=(\omega,n)$ and each $x_n$ is the limit of a sequence of points from $\omega\times\omega$;

• a countably-tight but not sequential space could be given by taking $(\omega\times\omega)\cup\{*\}$ where all points $(m,n)$ are open and a base for $*$ consist of sets $A_{M,N}=\{(m,n)|m>M, n>N_m\}$ for $M,N_m\in\omega$ - this space is trivially countably tight (it's countable) but is not sequential: $*$ is not the limit of any sequence in $\omega\times\omega$ (since we can always exclude any putative sequence converging to $*$);

• finally a non-countably-tight space is given by $\omega_1+1$ with the usual topology: $\omega_1$ (as a point) is in the closure of $\omega_1$ (as a set) but any countable subset of $\omega_1$ has bounded (countable) closure.

Answer 4

Does anyone here know how to show that sequential implies pytkeev? or, where can one find a proof of it?