Relative toplogy inherited as subset of compactification

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.

• The complement of $U$ in $X$ is closed in both cases. – Henno Brandsma May 9 at 9:32

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
$$X\cap U_\infty\cap X=X\cap(Y\setminus K=X\setminus K\in\tau_X\,,$$
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 $$X \setminus C = U \in \tau_X$$.
• 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 $X$. – Brian M. Scott May 9 at 3:58