Theorem 2.33 in Baby Rudin says: "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 ($\Rightarrow$), he starts off by writing, "Suppose $K$ is compact relative to $X$, and let {$V_\alpha$} be a collection of sets, open relative to $Y$, such that $K \subset \bigcup_\alpha V_\alpha$".
The part that I don't understand is how we are justified in supposing that such a collection {$V_\alpha$} of open relative sets exists?
In proving Theorem 2.30 Rudin uses the definition of open relative: "Suppose $E$ is open relative to $Y$. To each $p \in E$ there is a positive number $r_p$ such that the conditions $d(p,q) < r_p, q \in Y$ imply $q \in E$. Now going back to the initial argument, we can replace $E$ with $V_\alpha$ for some $\alpha$ and what we are saying is that for each $p \in {V_\alpha}$ there is a positive number $r_p$ such that the conditions $d(p,q) < r_p, q \in Y$ imply $q \in {V_\alpha}$.
My difficulty is in understanding how we can be assured that we can always find a $q \in Y$ for each $p \in V_\alpha$ with $d(p,q) < r_p$ and also having $q \in V_\alpha$ given that $r_p > 0$? The reason being that since $r_p > 0$ this implies we are looking for a $q \in Y$ such that $q \neq p$ (otherwise $r_p = 0)$. My feeling is that the only way this would be possible is if we assume that $X$ and $Y$ are open sets, so there are always neighbourhoods around $p$ that are subsets of $Y$, but the theorem doesn't make this assumption?
Best Answer
The problem is that you’ve misunderstood the definition. Given $p\in V_\alpha$, you don’t have to find a $q\in Y$ such that $d(p,q)<r_p$: the definition just says that if $q\in Y$ is such that $d(p,q)<r_p$, then $q\in V_\alpha$. It is entirely possible that the only point $q\in Y$ that satisfies $d(p,q)<r_p$ is $p$ itself.
There is an easier way to think about this. A set $V$ is open relative to $Y$ if and only if there is an open set $U$ in $X$ such that $V=V\cap Y$. This is equivalent to the characterization that you quote and a bit simpler, and it would be a good exercise to prove it; the proof isn’t hard.
Note that $Y$ is always open relative to itself, since it is equal to $X\cap Y$, where $X$ is certainly open in $X$. Thus, $\{Y\}$ is a cover of $K$ by sets that are open relative to $Y$. But in fact you can start with any family $\mathscr{U}$ of open sets in $X$ such that $K\subseteq\bigcup\mathscr{U}$ and let $\mathscr{V}=\{U\cap Y:U\in\mathscr{U}\}$: then $\mathscr{V}$ will be a cover of $K$ by sets that are open in $Y$.