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.

  • 1
    $\begingroup$ Who is baby Rudin? (I know of Walter Rudin.) $\endgroup$ – Sujaan Kunalan Oct 8 '13 at 1:56
  • $\begingroup$ 'baby rudin' is a reference to walter rudin's work "principles of mathematical analysis". $\endgroup$ – alienfetuseater Oct 8 '13 at 2:23
  • $\begingroup$ ‘Baby Rudin’ is the nickname for ‘Principles of Mathematical Analysis’ by Walter Rudin; ‘Papa Rudin’ is the nickname for ‘Real and Complex Analysis’ by Walter Rudin; ‘Grandpa Rudin’ is the nickname for ‘Functional Analysis’ by Walter Rudin; source: the Wikipedia article on Walter Rudin: en.wikipedia.org/wiki/Walter_Rudin $\endgroup$ – Mike Jones May 19 '17 at 9:03

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.


I see you are making a small mistake in your definition of compactness.

In my opinion, getting the definition is the most important part. I'll state it simply:

The set $K$ is compact iff the existence of an open cover implies the existence of a finite subcover.

Just because $K$ is compact relative to $X$ doesn't mean that there is an open cover for K.

$K$ being compact only tells us that a finite subcover exists whenever an open cover does.

The reason for choosing a set of $V_{\alpha}$ which is open relative to $Y$ is that we are assuming that an open cover for $K$ exists. We try to prove that, in turn, a finite subcover also exists in $Y$. Since you understand the rest of the logic well, and the justification of (22) implies (23) is given by Yesid already, it looks like you shouldn't have any problem with this proof.


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