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My text (Stoll, Introduction to Real Analysis, 2nd Ed) defined that $K$, a subset of $\mathbb R$, is compact if every open cover of $K$ has a finite subcover of $K$. Then, it proceeded to prove that every closed interval is compact. To my best knowledge, the proof is standard, but it feels like that it glosses over something:

Let $\{U_\alpha\}_{\alpha \in A}$ be an open cover of $[a, b]$,
$E = \{r \in [a, b] : [a, r]$ is covered by a finite number of $U_\alpha\}$

$E$ is bounded above by $b$, and is nonempty because $a \in E$. So, sup$E$ exists. Let $c =$ sup$E$. Furthermore, since $b$ is an upper bound, $c \le b$. Show that $c \in E$. Since $c \in [a, b]$, $c \in U_\beta$ for some $\beta \in A$. $U_\beta$ is open; there is a $\delta$ such that $N_\delta(c) \subset U_\beta$. Then, since $c - \delta$ is not an upper bound, there is a $r$ such that $c - \delta \lt r \le c$. $[a, r]$ is covered by a finite number of $U_\alpha$, say $\{U_{\alpha_1}, U_{\alpha_2}, ... U_{\alpha_n}\}$, and $[a, c]$ is covered by $\{U_{\alpha_1}, U_{\alpha_2}, ... U_{\alpha_n}, U_\beta\}$. Therefore, $c \in E$.

I will skip the part showing that $c = b$. My question has to do with the set $E$. At first, I find it circular. We are asked to prove that $[a, b]$ is compact, but from the start, we are already saying that $[a, r] \subset \bigcup^n_{j = 1} U_{\alpha_j}$. Here, I take $\{U_\alpha\}_{\alpha \in A}$ as arbitrary but fixed. Intuitively, it is possible that $[a, r] \subset \bigcup^n_{j = 1} U_{\alpha_j}$ because $r$ is a limit point of $[a, r]$, so every point "close enough" to $r$ is covered by the same $U_r$ that covers $r$. However, I don't think that alone justifies that $[a, r]$ is covered by a finite subcover. What am I missing?

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  • $\begingroup$ By definition, $r \in E$... $\endgroup$ – zcn Dec 31 '13 at 16:14
  • $\begingroup$ In the proof the sentence "Then, since $c-\delta$ is not an upper bound ..." should say " Since $c=\sup E$ there is $r\in (c-\delta,c]\cap E."$ ......In your last paragraph are you talking about any $r\in E$ or about the $r\in(c=\delta,c]$ that the text forgot to mention was in $E$? $\endgroup$ – DanielWainfleet Jul 10 '16 at 5:26
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I think you have to forget for a moment about limit points and look at the proof this way: it's a sort of "induction" on $r$. First you prove that $E$ is non empty because for $r=a$ you have at least one open set $U_\alpha$ such that $a\in U_\alpha$. Right? Then you want to prove that $r$ can "reach" $b$. There is nothing circular in this.

EDIT. Of course $E$ has infinitely many elements: $E = [a,b]$. :-) And all the trick is as follows (your proof already says it and you too are saying it): once you have some $r\in E$, by definition, you've got some $U_r$ such that $r\in U_r$ and hence some $\delta > 0$ such that $(r, r+\delta) \subset U_r$. So you can pick $r'\in (r,r+\delta)$ and you'll have that:

  • $[a,r]$ is covered by a finite number of open sets $U_\alpha$,
  • $[r,r']$ is covered by $U_r$.

Hence, $[a,r'] = [a,r] \cup [r,r']$ is covered by a finite number of open sets $U_\alpha$ (the ones that covered $[a,r]$, plus the one that covers $[r,r']$).

This way, you can "push" that $r$ to the right indifinitively.

(Of course, what I've written is not a proof: those "moving" $r$ could stop before reaching $b$ in my reasoning. The correct way to do it is as in your book, taking $\mathrm{sup}$ every time you "push" $r$ to the right. But if now you've understood the trick without the $\mathrm{sup}$ thing, maybe you can read again your book with it.)

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  • $\begingroup$ In fact, $E$ has infinitely many elements. If $a \in U_a$, which is open, then there is a $\delta$ such that $[a, a+\delta) \subset U_a$. I still am not sure how you can "push" the $r \in [a, a+\delta)$ to $b$, because $b$ does not have to be in $U_a$. I think this is my question. $\endgroup$ – Andy Tam Jan 6 '14 at 0:21
  • $\begingroup$ I've edited my answer: see if it inspires you. $\endgroup$ – Agustí Roig Jan 6 '14 at 2:13
  • $\begingroup$ Maybe you would like to take a look at the answer to this question too: math.stackexchange.com/questions/629309/… $\endgroup$ – Agustí Roig Jan 6 '14 at 19:36
  • $\begingroup$ TY, actually your advice to look at it as an "induction" is helpful. $\endgroup$ – Andy Tam Aug 3 '15 at 14:51
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I think a simpler and more intuitive way to approach this problem is the following:

  1. A (subset of a) metric space is compact iff it is complete and totally bounded.
  2. A closed subset of a complete metric space is complete.
  3. A closed interval is a closed set, and $\mathbb R$ is complete, so a closed interval is complete.
  4. A closed interval is bounded, and any bounded set in $\mathbb R$ is totally bounded.
  5. Hence a closed interval is compact.

This is a more general approach; the only "special" property of $\mathbb R$ that is used is in 4. In general, total boundedness (given $\varepsilon>0$, a set can be covered by finitely many open balls of radius $\varepsilon$) is not equivalent to boundedness, but in $\mathbb R^n$, it is - and this is the crux of why closed and bounded is equivalent to compact in $\mathbb R^n$.

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