Timeline for Punctured space of compact Hausdorff space and one-point compactification
Current License: CC BY-SA 3.0
6 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Aug 18, 2016 at 17:41 | comment | added | Brian M. Scott | @Nesta: You’re welcome! | |
| Aug 18, 2016 at 17:34 | comment | added | Nesta | Ah of course, I was confused for a second. Thanks again! | |
| Aug 18, 2016 at 17:30 | comment | added | Brian M. Scott | @Nesta: You’re right, but you don’t have to work that hard: $V\cap Y$ is the intersection of two open sets in $X$, so it’s automatically open in $X$. | |
| Aug 18, 2016 at 17:23 | comment | added | Nesta | Thank you Professor Scott! Just to make sure I get the details: since $X$ is Hausdorff, the singleton $\lbrace x\rbrace$ is closed. And so, $Y=X\setminus\lbrace x\rbrace$ is open in $X$. Since $U$ is open in $Y$ we can find a set $V$ which is open in $X$ such that $U=V\cap Y$. Then we easily find that $X\setminus U = (X\setminus V)\cup (X\setminus Y)$, which is closed in $X$, and so $U$ is open. Together with what I did in the OP this shows that $f$ is continuous on all of $X$, am I right? I will check out your linked answer as well! | |
| Aug 18, 2016 at 16:24 | history | answered | Brian M. Scott | CC BY-SA 3.0 |