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I have a question about Theorem 2.33 in Baby Rudin: "Suppose $K \subset Y \subset X$. Then $K$ is compact relative to $X$ if and only if $K$ is compact relative to $Y$."

To prove ($\Leftarrow$), he starts off by writing, "Suppose $K$ is compact relative to $Y$, let {$G_\alpha$} be a collection of open subsets of $X$ which covers $K$..."

I am a little confused about how I know the open cover {$G_\alpha$} exists. The argument I made to myself is:

  1. Since $K$ is compact relative to $Y$, there exists a collection {$V_\alpha$} of open subsets in $Y$ that cover $K$.
  2. From Theorem 2.30, if $V_i \subset Y \subset X$ and $V_i$ is open relative to $Y$, then there exists some open subset $G_i$ of $X$ such that $V_i = Y \cap G_i$.
  3. Therefore, the cover {$G_\alpha$} of open subsets of $X$ exists.

(Then he essentially follows these steps to show that $K$ is in fact compact relative to $X$.)

Is this the correct line of reasoning to simply show that the (not necessarily finite) open cover {$G_\alpha$} exists in the first place?

Thanks!

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  • $\begingroup$ You can cover any subset of any metric (more generally, topological) space by open subsets. The existence of open covers is completely unproblematic. The thing to show here is that every open cover has a finite subcover. $\endgroup$ – Daniel Fischer Jan 5 '14 at 19:24
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An open cover always exists, just take $\left\{X\right\}$. This is an open cover of $K$, and so there is no problem in taking an open cover of $K$ in $X$.

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Besides, with compactness, you are not arguing that a finite subcover always exists. You are only asking for a finite subcover to exist whenever an open cover exists.

You are allowed to assume that an open cover exists, in order to argue that this means a finite subcover also exists.

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