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:
- Since $K$ is compact relative to $Y$, there exists a collection {$V_\alpha$} of open subsets in $Y$ that cover $K$.
- 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 $V_i = Y \cap G_i$.
- 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!