Problem Of The Week # 336 - Jan 06, 2019

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Euge

MHB Global Moderator
Staff member
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!

Euge

MHB Global Moderator
Staff member
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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