Baby Rudin 2.33: Relative Compactness

Question

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!

Answer 1

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 $X$.

Answer 2

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.