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Homeclosed subsets of a compact set are compact

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# closed subsets of a compact set are compact

###### Theorem 1.

Suppose $X$ is a topological space. If $K$ is a compact subset of $X$, $C$ is a closed set in $X$, and $C\subseteq K$, then $C$ is a compact set in $X$.

The below proof follows e.g. [3]. A proof based on the finite intersection property is given in [4].

###### Proof.

Let $I$ be an indexing set and $F=\{V_{\alpha}\mid\alpha\in I\}$ be an arbitrary open cover for $C$. Since $X\setminus C$ is open, it follows that $F$ together with $X\setminus C$ is an open cover for $K$. Thus, $K$ can be covered by a finite number of sets, say, $V_{1},\ldots,V_{N}$ from $F$ together with possibly $X\setminus C$. Since $C\subset K$, $V_{1},\ldots,V_{N}$ cover $C$, and it follows that $C$ is compact. ∎

The following proof uses the finite intersection property.

###### Proof.

Let $I$ be an indexing set and $\{A_{{\alpha}}\}_{{\alpha\in I}}$ be a collection of $X$-closed sets contained in $C$ such that, for any finite $J\subseteq I$, $\displaystyle\bigcap_{{\alpha\in J}}A_{{\alpha}}$ is not empty. Recall that, for every $\alpha\in I$, $A_{{\alpha}}\subseteq C\subseteq K$. Thus, for every $\alpha\in I$, $A_{{\alpha}}=K\cap A_{{\alpha}}$. Therefore, $\{A_{{\alpha}}\}_{{\alpha\in I}}$ are $K$-closed subsets of $K$ (see this page) such that, for any finite $J\subseteq I$, $\displaystyle\bigcap_{{\alpha\in J}}A_{{\alpha}}$ is not empty. As $K$ is compact, $\displaystyle\bigcap_{{\alpha\in I}}A_{{\alpha}}$ is not empty (again, by this result). This proves the claim. ∎

# References

- 1
J.L. Kelley,
*General Topology*, D. van Nostrand Company, Inc., 1955. - 2
S. Lang,
*Analysis II*, Addison-Wesley Publishing Company Inc., 1969. - 3
G.J. Jameson,
*Topology and Normed Spaces*, Chapman and Hall, 1974. - 4
I.M. Singer, J.A. Thorpe,
*Lecture Notes on Elementary Topology and Geometry*, Springer-Verlag, 1967.

## Mathematics Subject Classification

54D30*no label found*

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