Rudin Theorem 2.35: Why does Rudin suppose F is closed relative to X ？

analysis

I have a question regarding the following proof：

2.35 Theorem: Closed subsets of compact sets are compact.

Suppose $F\subset K\subset X$, $\mathit F\,$ is closed (relative to $\mathit X$), and $\mathit K$ is compact. Let $\mathit \{V_\alpha\}\,$ be an open cover of $\mathit F$. If $\mathit F^{\,c}$ is adjoined to $\mathit \{V_\alpha\}$, we obtain an open cover $ Ω $ of $\mathit K $. Since $\mathit K \,$ is compact, there is a finite subcollection $\, Φ $ of $ Ω \,$ which covers $\,\mathit K $, and hence $\,\mathit F$. If $\,\mathit F^{\,c}$ is a member of $\, Φ $, we may remove it from $\, Φ \,$ and still retain an open cover of $\,\mathit F$. We have thus shown that a finite subcollection of $\mathit \{V_\alpha\}$ covers $\mathit F$.

My question is the following:

In the theorem，$\mathit F\,$ is a closed subset of the compact set $\mathit K$，but in the proof Rudin supposes $\mathit F\,$ is closed relative to $\mathit X$ .

Best Answer

Rudin's approach is a bit unusual. Given a subset $Y \subset X$ of a metric space, he does not state explicitly that $Y$ inherits a metric from $X$ and is therefore itself a metric space. But without regarding $Y$ as a metric (sub-)space of $X$ the concept of an open / closed subset of $Y$ is undefined. He substitutes this by defining for subsets $E \subset Y$ the concept of being an open / closed subset relative to $Y$. See Remark 2.29. It is easy to check that $E \subset Y$ is an open / closed subset relative to $Y$ if and only if $E$ is an open / closed subset of the metric subspace $Y$ of $X$.

In the light of the above remarks we can say that the statement

$(*) \quad$ Closed subsets of compact sets are compact

is fairly vague. However, reading the proof it is quite obvious that Rudin's Theorem 2.35 has to be understood as follows:

  • Let $K \subset X$ be a compact subset of $X$ and $F \subset X$ be a closed subset of $X$ such that $F \subset K$. Then $K$ is a compact subset of $X$.

I do not think this is a particularly elegant formulation, and in fact Rudin expressed it more catchy. Another interpretation of $(*)$ is

  • Let $K \subset X$ be a compact subset of $X$ and $F \subset K$ be a subset which is relatively closed in $K$. Then $K$ is a compact subset of $X$.

Both variants are equivalent. To see this, we need the following

Lemma. Let $Y \subset X$ be closed and $E \subset Y$. Then $E$ is relatively closed in $Y$ if and only if $E$ is closed in $X$.

Rudin does not mention it explicitly, but the proof is very easy. Together with the fact that compact subsets are closed this shows the equivalence of both variants.

Related Question