# Weierstrass and Heine-Borel theorems in Rudin's 3rd

This is a question about the logic of Theorems 2.41, 2.42 in Rudin's 3rd Ed, which deal with the Heine-Borel and Weierstrass properties of sets of $R^k$, respectively.

A quick version of my question is: doesn't 2.41 (together with 2.40, on which it depends) moot 2.42?

2.41. (Heine-Borel+) If a set E in $R^k$ has one of these three properties, it has the other two: (a) E is closed and bounded; (b) E is compact; (c) Every infinite subset of E has a limit point in E.

2.42. (Weierstrass) Every bounded infinite subset of $R^k$ has a limit point in $R^k$.

The text explains that 2.41(a) implies (b) implies (c) implies (a). So the only step to be supplied for 2.42 is that a bounded subset of $R^k$ is compact (and therefore closed, to bring it into the ambit of 2.41).

But in both theorems Rudin resorts to an extrinsic proof (his 2.40) to show that that "k-cells" are compact.

I think (am not sure) there is a theorem stating that Heine-Borel implies Weierstrass (and conversely) but H-B consists of 2.41(a) and (b), according to Rudin's note preceding 2.41. I wonder if, with the addition of 2.41(c), he needs to prove Weierstrass separately?

This is a sort of fussy question but an answer might help me understand the relationship between these ideas better. Thanks. EDITED: so the quick version of the question includes the ref. to 2.40.

-
Condition (c) is very close related to Weierstress theorem but the result does not directly applies since there are bounded no closed subsets. –  azarel Mar 11 '12 at 20:11
I see your point. If $T$ is infinite and bounded, get a closed ball $E$ that contains it. The ball is closed and bounded so use $(a)\to (c)$ from $2.41$. –  Patrick Mar 11 '12 at 20:32

2.41 does not trivialize 2.42, as it is not necessarily true that a bounded subset of $\mathbb{R}^k$ is compact. Why? Consider the following set:

$$\left\{\frac{1}{n} | n\in \mathbb{N}\right\}\subset \mathbb{R}$$

This is bounded but not compact. Why? $$\bigcup_{i\in\mathbb{N}} \left(\frac{1}{i+1/2},\frac{1}{i-1/2}\right)$$ is an open cover with no finite subcover. So you really do need both closed and bounded for compact. As a result, you have to supply an extra argument to get 2.42 from 2.41- an outline of such an argument is as follows:

Every bounded infinite subset $A\subset\mathbb{R}^k$ has a closure $A\cup \{\mathrm{limit points of A}\}=\overline{A}\subset\mathbb{R}^k$, which is also a bounded infinite subset of $\mathbb{R}^k$. But $\overline{A}$ is closed, so since it is also bounded, it is compact by 2.41 (a)$\Rightarrow$(b). Then we can apply 2.41 (b)$\Rightarrow$(c) to see that such a set has a limit point in $\overline{A}\subset\mathbb{R}^k$ and thus every bounded infinite subset of $\mathbb{R}^k$ has a limit point in $\mathbb{R}^k$.

-
I'm sorry, I don't have a copy of Rudin with me. Could you perhaps post the statement of 2.40? –  KReiser Mar 11 '12 at 20:24
I believe my answer still holds even if we know 2.40. There are lots of subsets of $\mathbb{R}^k$ which are infinite and bounded but not $k$-cells. You still do need something extra. –  KReiser Mar 11 '12 at 20:37
Yes, and this is the assertion that makes the whole thing work, more or less. I use this in the first sentence of my last paragraph- explicitly, if $A$ is bounded, then $A$ is a subset of some $k$-cell, and thus the closure of $A$ is a subset of this same $k$-cell. –  KReiser Mar 11 '12 at 21:22
Everything you have said, to include your answer above, is correct. But 2.40 is precisely the "other argument" you present in the last paragraph--it gets us compactness for 2.42. But it was used to prove compactness in 2.41(b). I am upvoting this answer (which is essentially azarel's) because it is correct, but I will wait to accept it, because...well for the reason I just gave. 2.40 is the extra argument. –  daniel Mar 12 '12 at 20:50

The connection between the topics is strong, but not as strong as you currently believe. The problem is that a bounded set of $\mathbb{R}^k$ is not necessarily compact: take the open unitl ball in $\mathbb{R}^k$, or more generally any bounded set in $\mathbb{R}^k$ that is not closed. The k-cells are invoked to say that our closed and bounded set $E$ is a subset of some k-cell $I$, as it is bounded. It would be enough to take $E$'s closure, but this approach also works: as k-cells are compact, 2.41 tells us that every infinite subset of $I$ (which includes the infinite subsets of $E$) has a limit point in $I \subset \mathbb{R}^k$ (and not necessarily $E$).

Further examining Rudin's setup, both the given proofs for Heine-Borel and Weierstrass use the general observations that infinite subsets of compact sets have a limit point in that compact set, as well as the compactness of k-cells and that closed sets of compact sets are compact.

-
In fairness, I would upvote this answer if you remove the phrase "extremely related" from the second paragraph. –  daniel Mar 13 '12 at 9:41
too pedantic? In any case, it's gone. –  JakeR Mar 13 '12 at 14:23