The difference between a sequential space and a space with countable tightness - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:05:46Z http://mathoverflow.net/feeds/question/35348 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35348/the-difference-between-a-sequential-space-and-a-space-with-countable-tightness The difference between a sequential space and a space with countable tightness tali11 2010-08-12T12:52:35Z 2012-08-26T16:32:25Z <p>Hi, I have recently encountered these two definitions of a <a href="http://en.wikipedia.org/wiki/Sequential_space" rel="nofollow">sequential space</a> and a <a href="http://en.wikipedia.org/wiki/Countably_generated_space" rel="nofollow">space of countable tightness</a>. 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!</p> http://mathoverflow.net/questions/35348/the-difference-between-a-sequential-space-and-a-space-with-countable-tightness/35374#35374 Answer by Peter Krautzberger for The difference between a sequential space and a space with countable tightness Peter Krautzberger 2010-08-12T17:07:09Z 2010-08-12T17:07:09Z <p>Just a partial answer.</p> <p>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 $\mathbb{R}$.</p> <p>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).</p> <p>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.</p> http://mathoverflow.net/questions/35348/the-difference-between-a-sequential-space-and-a-space-with-countable-tightness/35376#35376 Answer by Andreas Blass for The difference between a sequential space and a space with countable tightness Andreas Blass 2010-08-12T18:02:23Z 2010-08-12T18:02:23Z <p>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 <code>$A_0=A$</code>, let <code>$A_{\alpha+1}$</code> be the set of limits of sequences from <code>$A_\alpha$</code>, and for limit ordinals let <code>$A_\lambda$</code> be the union of all <code>$A_\alpha$</code>'s for $\alpha&lt;\lambda$. Then $p$ should be in <code>$A_\alpha$</code> 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). </p> http://mathoverflow.net/questions/35348/the-difference-between-a-sequential-space-and-a-space-with-countable-tightness/54148#54148 Answer by Apollo for The difference between a sequential space and a space with countable tightness Apollo 2011-02-02T23:20:37Z 2011-02-04T19:49:23Z <p>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)</p> <ul> <li><p>any metric space will be Frechet-Urysohn (choose <code>$x_n$</code> in <code>$A$</code> within <code>$1/n$</code> of <code>$p$</code>); (more generally, any first-countable space is F-U: just choose $x_n$ in the intersection of <code>$A$</code> and <code>$U_n$</code> where <code>$\{U_n\}_n$</code> is a countable base at the limit point);</p></li> <li><p>a sequential but not Frechet-Urysohn space is given by taking <code>$((\omega+1)\times\omega)\cup\{*\}$</code> where each copy of $\omega+1$ has the usual topology and a base for <code>$*$</code> consists of sets <code>$A_{m,n}=\{(m,n)|m&gt;M,n&gt;N_m\}$</code> for <code>$M,N_m\in\omega$</code> (ie cofinitely many elements of cofinitely many fibers) - then <code>$*$</code> is in the closure of <code>$\omega\times\omega$</code> but is not the limit of any sequence of points in <code>$\omega\times\omega$</code>; however, it is the limit of the sequence <code>$x_n=(\omega,n)$</code> and each <code>$x_n$</code> is the limit of a sequence of points from <code>$\omega\times\omega$</code>;</p></li> <li><p>a countably-tight but not sequential space could be given by taking <code>$(\omega\times\omega)\cup\{*\}$</code> where all points <code>$(m,n)$</code> are open and a base for <code>$*$</code> consist of sets <code>$A_{M,N}=\{(m,n)|m&gt;M, n&gt;N_m\}$</code> for <code>$M,N_m\in\omega$</code> - this space is trivially countably tight (it's countable) but is not sequential: <code>$*$</code> is not the limit of any sequence in <code>$\omega\times\omega$</code> (since we can always exclude any putative sequence converging to <code>$*$</code>);</p></li> <li><p>finally a non-countably-tight space is given by <code>$\omega_1+1$</code> with the usual topology: <code>$\omega_1$</code> (as a point) is in the closure of <code>$\omega_1$</code> (as a set) but any countable subset of <code>$\omega_1$</code> has bounded (countable) closure.</p></li> </ul> http://mathoverflow.net/questions/35348/the-difference-between-a-sequential-space-and-a-space-with-countable-tightness/105551#105551 Answer by tali1 for The difference between a sequential space and a space with countable tightness tali1 2012-08-26T16:32:25Z 2012-08-26T16:32:25Z <p>Does anyone here know how to show that sequential implies pytkeev? or, where can one find a proof of it?</p>