If $A\subset K$ and $K$ compact, does $\bar A$ compact?

Question

Let $(X,\mathcal T)$ a topological space. Let $K\subset X$ a compact set and $A\subset K$. Is $\bar A$ compact ? (where $\bar A$ denote the closure).

---

This result is true if $\mathcal T$ is metrizable. But if $\mathcal T$ is not metrizable, is it true ? I tried as follow. Let $\mathcal U$ an open cover of $\bar A$. Let $\mathcal V$ an open cover of $K$. In particular, $\mathcal V\cup\mathcal U$ cover $K$ and thus there is a finite subcovering of $K$ of $\mathcal U\cup \mathcal V$, let say $\bigcup_{i=1}^n U_i\cup V_i$ In particular it cover $\bar A$.

Question : How can I be sure that $\bigcup_{i=1}^nU_i$ cover $\bar A$ ?

Answer 1

If $X$ is Hausdorff, then yes. This follows from the fact that in a Hausdorff space, compact subsets are closed. Thus, $K$ is closed, and thus, since $A\subseteq K$, we have $\overline A\subseteq \overline K=K$. Since closed subsets of compact spaces are compact, we are done.

If $X$ is not Hausdorff, then no. For example, if $A=K$ is compact and dense in $X\neq K$ which is not compact, then $\overline A=X$ is not compact.

For a concrete example, let $X=\mathbf N$ be equipped with topology such that the for each $n\in \mathbf N$, the neighbourhoods of $n$ are exactly the sets containing $\{0,n\}$. Then $A=K=\{0\}$ is compact and dense in $X$, so $\overline A=X$, but $X$ is obviously not compact.

Answer 2

You can take $P=\{A_i,i\in N\}$ as an open cover of $\bar{A}$. now $\bar{A}^{\complement}$ is an open set. so $\bar{A}^{\complement} \cup P $ covers $K$. you now use compactness ok K, you will find your solution.