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Oct 23, 2012 at 23:25 comment added Camilo Arosemena Serrato sorry, I hadn't seen your "moreover"
Oct 23, 2012 at 21:57 comment added Brian M. Scott @Camilo: My statement that $\tau^*$ and $\tau'$ have the same nbhds of $p$ means exactly the same as your version. And yes, we do necessarily have $\tau^*\subseteq\tau'$; what my example shows is that we don’t necessarily have $\tau'\subseteq\tau^*$.
Oct 23, 2012 at 21:47 comment added Camilo Arosemena Serrato and also think that what you wanted to show in the first part was that $p$ has the same neighborhoods in $\left<{X,\tau^´}\right>$ and $\left<{X,\tau^*}\right>$; note that not necessarly we have $\tau^*\subseteq{\tau^´}$, because of your counterexamnple; note that the space of your counterexample is locally compact.
Oct 23, 2012 at 21:40 vote accept Camilo Arosemena Serrato
Oct 23, 2012 at 21:39 comment added Brian M. Scott @Camilo: I did indeed; thanks for catching it.
Oct 23, 2012 at 21:39 history edited Brian M. Scott CC BY-SA 3.0
Fixed a typo.
Oct 23, 2012 at 21:34 comment added Camilo Arosemena Serrato I think what you meant was $\left\{{0}\right\}\cup{\left\{{\displaystyle\frac{1}{n}:n\in{\mathbb{Z}^+}}\right\}}$, correcting this your proof, is correct Thank you very much!!!!!!!!
Oct 23, 2012 at 7:04 history answered Brian M. Scott CC BY-SA 3.0