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Let $X$ a topological space and $\infty \in X$ with the topology given by $\tau=\lbrace A \subset X \mid \infty \not \in A, X \setminus A \text{ finite } \rbrace $ Interpret $X$ as a one point compactification.

Let $M=X \setminus \lbrace \infty \rbrace$ and $m\in M$. Equipped with the topology $\tau=\lbrace U \subset M \mid X \setminus U \, \, \text{Is finite}, m\not \in U\rbrace $.

I claim that $M$ is locally compact and Hausdorff.

In case of can prove it I like see that $\overline{M}=X$ and hence $X$ was the compactification of $M$.

I have a hour trying to prove that $M$ is locally compact, but I don´t get it. Otherwise is easy see that $M$ is Hausdorff.

Any advice of how to construct $M$ or what things I forget consider for makes $X$ the compactification of $M$.

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The definition of $\tau$ is very poorly written. I would normally interpret it to mean that $\tau$ is the family of all subsets of $X$ that have finite complements and do not contain $\infty$, but that is not a topology on $X$, since it does not include the set $X$. Apparently the comma is being (mis)used to stand for or, and what is actually intended is that

$$\tau=\{A\subseteq X:\infty\notin A\text{ or }X\setminus A\text{ is finite}\}\,.$$

Now let $M=X\setminus\{\infty\}$. Then every $A\subseteq M$ satisfies the condition $\infty\notin A$, so every subset of $M$ is open in $X$. Now you should have no trouble showing that $M$ with the relative topology that it inherits from $X$ is locally compact and Hausdorff.

Your $\{U\subseteq M:X\setminus U\text{ is finite and }m\notin U\}$ is not a topology, since it does not include $M$, and is not the set of $A\cap M$ such that $A\in\tau$ no matter how the comma in the original definition of $\tau$ is interpreted.

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  • $\begingroup$ I can see that for any $x,y\in M$ then $\lbrace x \rbrace$, $\lbrace y \rbrace$ are disjoint open sets that contains $x$ and $y$, then $M$ is hausdorff, for other hand I was trying for $x\in M$ find $U$ open in $M$ and $K$ compact in $M$ such that $x\in U \subset K$, but I don´t have idea of how to find $K$ $\endgroup$ Commented Mar 5, 2021 at 0:47
  • $\begingroup$ @JuanT: For local compactness you want to know that each $x\in M$ has a compact nbhd, and that’s easy: $\{x\}$ is open and finite, so it is a compact nbhd of $x$. (There are several definitions of local compactness, but in a Hausdorff space they’re all equivalent.) $\endgroup$ Commented Mar 5, 2021 at 1:58
  • $\begingroup$ Thanks a lot, I don´t know the other equivalences of compactness $\endgroup$ Commented Mar 5, 2021 at 2:01
  • $\begingroup$ @JuanT: You’re welcome. (I mentioned them because I wasn’t sure which definition you were using.) $\endgroup$ Commented Mar 5, 2021 at 2:44

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