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!

Does anyone here know how to show that sequential implies pytkeev? or, where can one find a proof of it? – user25968 Aug 26 '12 at 16:32

Wikipedia claims to give a proof at en.wikipedia.org/wiki/Pytkeev_space but it involves terminology that I'm not familiar with ($\pi$net) and too busy to look up just now, so I don't guarantee that the proof is correct. – Andreas Blass Aug 26 '12 at 22:20
All three notions, "countably tight," "sequential," and "FrechetUrysohn," 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, FrechetUrysohn, 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).
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 FrechetUrysohn (choose $x_n$ in $A$ within $1/n$ of $p$); (more generally, any firstcountable space is FU: 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 FrechetUrysohn 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 countablytight 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 noncountablytight 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.
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/TeXfile 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 Ppoints). Maybe somebody else can confirm or reject.