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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?

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  • $\begingroup$ I think it is not needed to prove that a such collection of $V_\alpha$ exists. The $'\implies'$ part of the theorem says that if, given a collection of open sets in $X$ that covers $K$, you can always find a finite subcollection that covers $K$, then $\mathrm{given}$ an open (relative to $Y$) cover of K, you can always find a finite subcover of $K$. But the collection of open sets in $Y$ that covers $K$ is given, you just want to show that if it exists, it has a finite subcollection that covers $K$. $\endgroup$ – T_Kln Oct 24 at 17:45
  • $\begingroup$ I don't understand your last paragraph. Why are you trying to find such a $q$? $\endgroup$ – Thorgott Oct 24 at 17:45
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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.

Example: Let $X=\Bbb R$, $Y=(0,3)\cup\{5\}$, and $K=[1,2]\cup\{5\}$. Let $V=\big((1,3)\cup(4,6)\big)\cap K$. If $p=5$, let $r_p=1$; if $q\in Y$ and $d(5,q)<1$, then $q=5$, so it is true that $q\in V$. If we take $p=2$, we can again let $r_p=1$: if $q\in Y$ and $d(2,q)<1$, then $q\in(1,2]\subseteq K$.

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 $Y$.

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  • $\begingroup$ Thank you. That concrete example cleared up a lot of confusion for me. $\endgroup$ – laichzeit0 Oct 25 at 6:18
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The statement "Every relatively open cover of $K$ has a finite subcover" is effectively a conditional statement. It doesn't assert the existence of a relatively open cover of $K$. What it asserts is that "If a collection of sets is a relatively open cover of $K$, then there is a finite subcollection that also covers $K$".

To prove a conditional statement, you get to assume the hypothesis, hence to prove that $K$ is relatively compact, you get to assume the existence of a relatively open cover of $K$ (you don't have prove the existence), and the goal is to prove for that assumed relatively open cover, that there is a finite subcover.

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First of all, to answer the question in the title, we can be sure that if $K \subset Y$ then we can find an open cover for $K$ in $Y$, for example we can always choose the cover $\{Y\}$ which is an open cover of $K$ relative to $Y$, because $Y$ is open in $Y$. A second example of such cover would be $\{N_1(q)|q\in Y\}$, which is the set of all neighborhoods of radius 1, centered about a point of $Y$. So as you see there is no problem with assuming the existence of an open cover for $K$ relative to $Y$.

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  • $\begingroup$ "A second example of such cover would be {𝑁1(𝑞)|𝑞∈𝑌}, which is the set of all neighborhoods of radius 1, centered about a point of 𝑌". But for it to be a "relative open" cover for K, those points q of Y must also be elements of K, by the definition of relative openness? $\endgroup$ – laichzeit0 Oct 25 at 9:47
  • $\begingroup$ Not necessarily. The important thing is that the neighborhoods are open relative to $Y$ and that $K \subseteq \cup_{q \in Y}N_1(q)$. $\endgroup$ – user800827 Oct 25 at 13:40

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