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I'm trying to prove the following problem: let $(X,\tau_X)$ be a topological space and define $\tau_{\infty}=\lbrace U\subset Y| U\in\tau_X \text{ or } Y-U\text{ is a compact and closed subset of X}\rbrace$ where $Y=X\cup\lbrace \infty\rbrace$. Let $\tau_Y$ be the relative topology over $X$ inherited as a subset of $Y$. Show that $\tau_X=\tau_Y$.

Ok, this is my proof attempt: the inclusion $\tau_X \subset \tau_{Y}$ is obvious since any $U\in\tau_X$ is an element of $\tau_{\infty}$ and hence we can write $U=U\cap Y$, so $U\in \tau_{Y}$. Now, let $U\in \tau_{Y}$. Then $U=U_{\infty}\cap Y$, where $U_{\infty}\in \tau_{\infty}$. If $U_{\infty}\in \tau_X$, then clearly $U\in\tau_X$. Otherwise, $Y-U_{\infty}$ is compact and closed in $X$. I'm stucked in here since I can't guarantee that $U_{\infty}\in \tau_X$ in this case. Can anyone give me an advice about how can I proceed further? I will appreciate any given help.

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  • $\begingroup$ The complement of $U$ in $X$ is closed in both cases. $\endgroup$ – Henno Brandsma May 9 at 9:32
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You’re almost done. Let $K=Y\setminus U_\infty$, so that $K$ is closed in $X$. (The fact that $K$ is compact is irrelevant here; it serves only to ensure that $\langle Y,\tau_Y\rangle$ is compact.) Then $U_\infty=Y\setminus K$, so

$$X\cap U_\infty\cap X=X\cap(Y\setminus K=X\setminus K\in\tau_X\,,$$

as desired.

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If $C := Y \setminus U$ is compact within $X$, then $X \setminus C$ is open as a subset of $X$, i.e, $X \setminus C \in \tau_X$. But $X \setminus C = U \cap X$ which belongs to the subspace topology by definition.

Alternatively if $U \subseteq X$ such that $U \in \tau_Y$ then there's a $U_\infty \in \tau_\infty$ such that $U = U_\infty \cap X$ and we may assume it's of the second type, that is, $Y \setminus U_\infty =: C$ is closed and compact in $\tau_X$. So $X \setminus C = U \in \tau_X$.

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    $\begingroup$ Your first sentence assumes that a compact subset of $X$ is closed; this is true in Hausdorff spaces but not in general. That’s why the definition of $\tau_\infty$ in the question required that $Y\setminus U$ be compact and closed in $X$. $\endgroup$ – Brian M. Scott May 9 at 3:58

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