# Baby Rudin 2.33: How can we be sure there is an open cover for K relative to Y?

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?

• 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 $K$. – not an analyst Oct 24 '20 at 17:45
• I don't understand your last paragraph. Why are you trying to find such a $q$? – Thorgott Oct 24 '20 at 17:45

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 $$d(p,q): the definition just says that if $$q\in Y$$ is such that $$d(p,q), then $$q\in V_\alpha$$. It is entirely possible that the only point $$q\in Y$$ that satisfies $$d(p,q) is $$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 $$Y$$.

• Thank you. That concrete example cleared up a lot of confusion for me. – laichzeit0 Oct 25 '20 at 6:18

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.

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 $$Y$$.

• "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? – laichzeit0 Oct 25 '20 at 9:47
• Not necessarily. The important thing is that the neighborhoods are open relative to $Y$ and that $K \subseteq \cup_{q \in Y}N_1(q)$. – user800827 Oct 25 '20 at 13:40