[Math] baby rudin 2.33, relative compactness

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my question is relative to baby rudin theorem 2.33 which states; $$ \ suppose \ K \subset Y \subset X. \ then\ K \ is\ compact\ relative \ to\ X \ iff\ K\ is\ compact\ relative \ to \ Y.$$

honestly, i think i only have maybe a superficial understanding of what Rudin is even saying here. however, i become more uncertain in his proof.
i would say that i feel i have a pretty good understanding of theorem 2.30, the preceding theorem, which says

$$ suppose \ Y \subset X. a \ subset \ E \ of \ Y \ is \ open \ relative \ to \ Y \ iff \ \ E = Y \bigcap G \ for \ some \ open \ subset \ G \ of \ X.$$

which, as i understand the idea of openness as $E \subset Y$ may be open in $Y$ but may not be open in $X$ where $E \subset Y \subset X$.

also i feel pretty comfortable with the idea of a compact set as being one where it is a subset of a finite union of a family of sets, the finite subcover. compared to a general open cover, which is just a union of any family of open sets, which is a superset of some other set which it is the open cover for.

now that i have explained the relative parts of what i do (think) i understand, let me clarify what about theorem 2.33 i am uncomfortable with;

i really am not sure what it even means for sets to be compact relative to another set. in the topological sense, is compactness not a invariant property of a topological space?

Rudin proceeds on with the proof as follows;

suppose $K$ is compactive relative to $X$, and let $\{V_{\alpha}\}$ be a collection of sets, open relative to $Y$, such that $K\subset \bigcup_{\alpha} V_{\alpha}$.

this is the first part of the proof i am confused by. $K$ is assumed to be compact relative to $X$ but Rudin describes $K$ as being covered by $V_{\alpha}$, where $\{V_{\alpha}\}$ is an open subset of $Y$. would not $K$ being covered by a family of sets, subsets of $X$, follow immediately from the fact that $K$ is compact relative to $X$? i dont understand the motivation for this part.

carrying on for the moment. By theorem 2.30 there are sets $G_{\alpha}$, open relative to $X$, such that $V_{\alpha}=Y \bigcap G_{\alpha}$, for all $\alpha$; and since $K$ is compact relative to $X$ we have $$(22) \ K \subset G_{\alpha_1} \bigcup ….. \bigcup G_{\alpha_n}$$ for finitely many indices $\alpha_1,…\alpha_n$ which i dont argue with any of.

since $K \subset Y$, (22) implies $$ (23) \ K \subset V_{\alpha_1} \bigcup … \bigcup V_{\alpha_n} $$. and this proves $K$ is compact relative to $Y$. this is the last part i dont understand, how does $K$ being a subset of $Y$ force (22) to imply (23)?

of course this is only one direction in the bijection, but i was so bothered by the theorem/proof i havent even gotten to the second part of the bijection.

Best Answer

I think that you need first to understand the idea behind "open relative to". Rudin clarifies this concept in 2.29, so I recommend you read it. If you see the definition of an open set $A$ in a metric space $X$ you could change such $X$ by a $Y\subset X$ an that's all the deal of "open relative to".

$A$ is an open set if for every $p\in A$ there's a neighborhood $G_r(p)$ with center $p$ and radius $r>0$ such that $G_r(p)\subset A$. And a neighborhood $G_r(p)$ is a subset of $X$ with elements $q$ which satisfies $d(p,q)<r$.

$A$ is open relative to $Y$ if for every $p\in A$ there's $r>0$ such that every $q\in Y$ which satisfies $d(p,q)<r$ also it's in $A$. This is equivalent to say there's a neighborhood $G_r(p)$ such that $G_r(p)\cap Y \subset A$.

Now, why Rudin starts off with

suppose $K$ is compactive relative to $X$, and let $\{Vα\}$ be a collection of sets, open relative to $Y$, such that $K⊂⋃_αV_α$.

I think that if you need to prove that a set is compact relative to $Y$ then you need to show that any cover relative to $Y$ have finitely many sets open relative to $Y$.

And why (22) implies (23)?

Well, it's clear that :

  1. if $A\subset B$ then $A\cup C\subset B\cup C $
  2. $\{G_{\alpha_1} \bigcup ..... \bigcup G_{\alpha_n}\}\bigcap Y = \{G_{\alpha_1}\cap Y\} \bigcup ..... \bigcup \{G_{\alpha_n}\cap Y \} =V_{\alpha_1} \bigcup ... \bigcup V_{\alpha_n}$ by 2.30.

so if $K\subset G_{\alpha_1} \bigcup ..... \bigcup G_{\alpha_n}$ then $K\cap Y \subset \{G_{\alpha_1} \bigcup ..... \bigcup G_{\alpha_n}\}\bigcap Y$ and (23) is the simplified form.

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