On Pseudo-finite topological spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:11:17Z http://mathoverflow.net/feeds/question/99169 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99169/on-pseudo-finite-topological-spaces On Pseudo-finite topological spaces AliReza Olfati 2012-06-09T10:56:36Z 2012-06-12T05:23:29Z <p>We recall that a topological space $(X,\tau)$ is Pseudo-finite, if each compact subset of $X$ is finite.</p> <p>One of the classical example of Pseudo-finite topological spaces can be considered as an uncountable set $X$ with the co-countable topology.(i.e.each subset with countable complement is open)</p> <p>The above topology has no isolated point but it fails to be at least Hausdorff. On the base of my Knowledge there are two <strong>Tychonoff</strong> Pseudo-finite topological spaces as follows:</p> <p>A. All discrete spaces are trivial examples of these spaces.</p> <p>B. Consider the set $\Sigma=\mathbb{N}$$\cup {p}, and topologize it as follows:</p> <ul> <li>Consider a free ultrafilter \mathcal{U} on \mathbb{N}.</li> <li>All points of \mathbb{N} are isolated.</li> <li>The Neighborhoods of p are of the form: U$$\cup$ {$p$}, where $U \in \mathcal{U}$.</li> </ul> <p>We must recall that Case "B" is a special Example of maximal Hausdorff topologies on a set.</p> <p>But I think there is no example of a Pseudo-finite Tychonoff space without isolated point !. and I guess the following statement:</p> <p><strong>Statement</strong>:Every Pseudo-finite Tychonoff space has an isolated point.</p> <p>Is there a counterexample of the above statement?</p> <hr> http://mathoverflow.net/questions/99169/on-pseudo-finite-topological-spaces/99184#99184 Answer by David Feldman for On Pseudo-finite topological spaces David Feldman 2012-06-09T17:02:23Z 2012-06-09T17:02:23Z <p>I believe you can grow your example B into a counterexample. </p> <p>Stage 0 is $\emptyset$</p> <p>Stage 1 is <code>$\{p\}$</code>.</p> <p>Stage 2 is your $B$: you've added a copy of ${\Bbb N}$ for each point newly added in the previous stage (all one of them) with an ultrafilter to define neighborhoods of the old point(s).</p> <p>And the recipe for Stage $n+1$ adds a copy of ${\Bbb N}$ for each point newly added in Stage $n$ with an ultrafilter to define neighborhoods of the old points.</p> <p>Natural numbers suffice to index the stages (no transfinite induction necessary).</p> <p>Clearly we kill all the isolated points in the limit. I believe you get pseudo-finiteness much as you get it for $B$, but there are details.</p> http://mathoverflow.net/questions/99169/on-pseudo-finite-topological-spaces/99257#99257 Answer by Warren McG for On Pseudo-finite topological spaces Warren McG 2012-06-10T20:22:32Z 2012-06-10T20:22:32Z <p>The rationals are pseudo-finite.</p> http://mathoverflow.net/questions/99169/on-pseudo-finite-topological-spaces/99265#99265 Answer by Joseph Van Name for On Pseudo-finite topological spaces Joseph Van Name 2012-06-10T23:18:24Z 2012-06-10T23:18:24Z <p>Every $P$-space (a $P$-space is a completely regular space where every G_{ \delta}-set is open) is pseudofinite since one can easily show that every subspace of a $P$-space is a $P$-space and every compact $P$-space is finite. However there are many $P$-spaces with no isolated points. For instance, take the $P$-space coreflection of an uncountable product of the space {0,1} and you will get a pseudofinite space with no isolated points.</p> http://mathoverflow.net/questions/99169/on-pseudo-finite-topological-spaces/99333#99333 Answer by David Feldman for On Pseudo-finite topological spaces David Feldman 2012-06-12T05:23:29Z 2012-06-12T05:23:29Z <p>The counterexamples so far depend on AC, but one can have such spaces just from ZF.</p> <p>In particular, instead of an ultrafilter on ${\Bbb N}$ as in your example $B$, one can use the filter of subsets with asymptotic density 1. The rest follows along the lines of my previous answer (and David Milovich's details).</p>