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Let $X$ be a compact Hausdorff space and $A \subseteq X$ be closed. Then show that one point compactification of $X \setminus A$ is homeomorphic to $X / A.$

$\textbf {My attempt} :$

What we know is that one point compactification of $X \setminus A$ is nothing but the space $Y = (X \setminus A) \cup \{\infty\}$ equipped with the topology $\tau_Y = \tau_1 \cup \tau_2,$ where $\tau_1$ is the topology on $X \setminus A$ and $\tau_2$ is given by $$\tau_2 : = \{Y \setminus C\ |\ C \subseteq X \setminus A\ \text {is compact} \}.$$ It is clear that as a set $X / A$ and $Y$ are in a bijective correspondence. There are obvious natural embeddings of $X \setminus A$ into the spaces $X / A$ and $Y$ and it is clear that both the spaces can be obtained by adjoining one single point to $X \setminus A$ (when viewed in terms of the embeddings). So in order to prove $Y$ is homeomorphic to $X/A$ it is enough to show that $X/A$ is compact and Hausdorff. Compactness of $X/A$ is clear as it is the image of the compact set $X$ under the quotient map $p : X \longrightarrow X/A.$ Also since $X$ is compact and Hausdorff it is normal and hence regular in particular and since $A \subseteq X$ is closed it follows that $X / A$ is Hausdorff. This proves the result.

Would anybody please have a look at my solution and check whether it holds good or not?

Thanks for reading.

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  • $\begingroup$ If $\phi: X\rightarrow X/A$ is the quotient map, why don't you just show directly that a set $U \subseteq X/A$ is open if and only if one of the two disjoint cases hold either 1) the point $A/A$ is not a member of $U$ and $\phi^{-1}(U)$ is an open subset of $X\setminus A$ or 2) the point $A/A$ is a member of $U$ and $\phi^{-1}(U)$ is the complement of a compact subset of $X$ disjoint from $A$. That is the definition of the Alxandroff topology, and it's a lot more direct than relying on all those results like normality and universality of the Alexandroff construction. $\endgroup$ Commented Jun 4, 2021 at 17:04
  • $\begingroup$ That's not what you have to show. The topology of $X/A$ is the collection of all subsets $U$ such that $q^{-1}(U)$ is open in $X$. So $U$ with $[A]\in U$ is open iff $q^{-1}(U)$ is open in $X$ and $A\subseteq q^{-1}(U)$ which is iff the complement of $q^{-1}(U)$ is a compact subset of $X$ disjoint from $A$. $\endgroup$ Commented Jun 6, 2021 at 18:43
  • $\begingroup$ @O. Peters Here $X/A$ is Hausdorff since $X$ is regular and $A \subseteq X$ is closed. So singletons are closed there. So if $U$ is an open subset of $X/A$ containing $[A]$ then $U \setminus \{[A]\}$ is open in $(X / A) \setminus \{[A]\}.$ But then $q^{-1} (U \setminus \{[A]\})$ is an open subset of $X$ disjoint from $A.$ Let $q^{-1} (U \setminus \{[A]\}) = V.$ Then clearly $q^{-1} (U) = V \cup A.$ So we need to show that $V \cup A$ is open in $X$ and complement of $V \cup A$ is a compact subset of $X$ disjoint from $A.$ I don't have any idea how to show that. $\endgroup$ Commented Jun 7, 2021 at 4:10
  • $\begingroup$ Dear @O. Peters I have added an answer to my question. Kindly look into it when you will find some time and feel free to give any kind of suggestion if some improvement is required. $\endgroup$ Commented Jun 7, 2021 at 5:28

2 Answers 2

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We need to also assume that $A$ is not open. Let $x\in X\setminus A$, then since $X$ is compact Hausdorff, and $x\notin A$, there is a neighbourhood $U$ of $x$ whose closure is disjoint from $A$ and is compact (follows from the regularity of $X$). So, $X\setminus A$ is locally compact, Haudorff.

The restriction of the quotient map $q\colon X\to X/A$ to $X\setminus A$ is an injective continuous map whose image misses the point corresponding to $A$. Moreover, the closure of this image is $X/A$ because $A$ is not open (else $X\setminus A$ is already compact).

Being the continuous image of a compact set, $X/A$ is compact and by regularity of $X$, it is Hausdorff. By the uniqueness of one point compactification, $X/A$ is the one point compactification of $X\setminus A$.

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  • $\begingroup$ It also works for open $A$. You didn't use that $A$ is not open. $\endgroup$ Commented Jun 10, 2021 at 19:27
  • $\begingroup$ @PaulFrost I did. When $A$ is open, the image of $X\setminus A$ is not dense in $X/A$. $\endgroup$ Commented Jun 11, 2021 at 6:42
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    $\begingroup$ It also works for open $A$. Note that the one-point compactification $Y^+ = Y \cup \{\infty\}$ is defined for each locally compact Hausdorff $Y$. If $Y$ is compact, then $Y^+$ is the disjoint union of $Y$ and a a single point space. For eaxmple, $[0,1]^+ \approx [0,1] \cup \{2\}$. But in fact you have to modify your proof a little. $\endgroup$ Commented Jun 11, 2021 at 10:22
  • $\begingroup$ @PaulFrost Oh, ok, difference in terminology. I considered one point compactifications for locally compact Hausdorff spaces that aren't already compact, i.e, an embedding as a dense subset in a compact set with one extra point. Yes, otherwise it works when $A$ is open as well. $\endgroup$ Commented Jun 12, 2021 at 3:44
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Let us denote the one point compactification of $X \setminus A$ by $(X \setminus A)^{+}.$ Now consider the map $f : X \longrightarrow (X \setminus A)^{+}$ defined by $$f(x) = \begin {cases} x\ & \text {if}\ x \in X \setminus A \\ \infty\ & \text {if}\ x \in A \end {cases}$$ Claim $:$ $f$ is continuous.

Case 1 $:$ Let $U \subseteq_{\text {open}} (X \setminus A)^{+}$ such that $\infty \notin U.$ Then $U \subseteq_{\text {open}} X \setminus A.$ Since $U$ is open in $X \setminus A$ and $X \setminus A$ is open in $X$ (because $A$ is closed in $X$) we have that $U$ is open in $X.$ Also since $f \big \rvert_{X \setminus A} = \text {Id}_{X \setminus A}$ it follows that $f^{-1} (U) = U.$ So we are through in this case.

Case 2 $:$ Let $U \subseteq_{\text {open}} (X \setminus A)^{+}$ such that $\infty \in U.$ Then $(X \setminus A)^{+} \setminus U$ is a compact subset of $X \setminus A$ and hence a compact subset of $X$ and thus it is closed in $X$ (because $X$ is Hausdorff). But on the other hand since $(X \setminus A)^{+} \setminus U \subseteq X \setminus A$ and $f \big \rvert_{X \setminus A} = \text {Id}_{X \setminus A}$ it follows that $$(X \setminus A)^{+} \setminus U = f^{-1} ((X \setminus A)^{+} \setminus U) = X \setminus f^{-1} (U).$$ Thus we have $X \setminus f^{-1} (U)$ is closed in $X$ or in other words $f^{-1} (U)$ is open in $X.$

This proves our claim.

Now by universal property of quotient topology $f$ induces a bijective continuous map $\widetilde {f} : X/A \longrightarrow (X \setminus A)^{+}.$ But since $X/A$ is compact and $(X \setminus A)^{+}$ is Hausdorff (by definition) it follows that $\widetilde {f}$ is a homeomorphism.

This completes the proof.

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