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Problem Of The Week # 336 - Jan 06, 2019

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Euge

MHB Global Moderator
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Jun 20, 2014
1,925
Here is this week's POTW:

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Suppose $X$ is a compact Hausdorff space. Let $S$ be a closed subspace of $X$. Show that the one-point compactification of $X - S$ is homeomorphic to the quotient space $X/S$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,925
No one answered this week's problem. You can read my solution below.


Define a surjective map $f : X \to (X - S) \cup \{\infty \}$ by setting $$f(x) = \begin{cases}x&\text{if $x\in X - S$}\\\infty&\text{if $x\in S$}\end{cases}$$ If $V$ is an open subset of $X - S$, then $V$ is open in $X$ (since $X - S$ is open in $X$) and $f^{-1}(V) = V$. On the other hand, if $V$ is a neighborhood of $\infty$, set $U = V - \{\infty\}$. Then $(X - S) - U$ is a compact subset of $X$, i.e., $X - (S \cup U)$ is a compact subset of $X$. The Hausdorff property of $X$ implies $X - (S \cup U)$ is closed, and consequently, $S \cup U$ is open. Furthermore, $f^{-1}(V) = S \cup U$. This shows that $f$ is continuous. As $f(S) = \{\infty\}$, $f$ induces a bijective continuous map $\tilde{f}: X/S \to (X - S) \cup \{\infty\}$. Since $X$ and $(X - S) \cup \{\infty\}$ are compact Hausdorff spaces, it follows that $\tilde{f}$ is a homeomorphism.
 
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