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I am going through Baby Rudin and am having trouble comprehending how any closed and bounded Euclidean set is compact via the coverage definition:

If a set $E$ in $X$ is compact, then for any open cover of $E$ {consisting of subsets of $X$} there exists a finite subcover of $E$ consisting of a finite portion of the open cover.

Consider the interval $[a,b]\in\mathbb{R}$. Obviously open covers exist of it, e.g. $(a-1, b+1)$ is also a trivial subcover. But since there are an infinite number of real numbers between any two real numbers, if we take the open cover that is literally "every" real number between $a$ and $b$, how can there be a finite subcover?

Since all real numbers in $[a,b]$ are limit points, any open set in it is some neighborhood $(p-r, p+r)$ that contains an infinite number of points of $[a,b]$. So don't we still need an infinite number of these neighborhoods to cover $[a,b]$...?

Thanks!

user402857
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2 Answers2

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If I'm interpreting your cover correctly, you want to take the collection of single-element sets $\{r\}$, where $r$ ranges over all real numbers in $[a, b]$. But each $\{r\}$ is not open, so this isn't an open cover, and doesn't contradict compactness.

ckefa
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  • I think I was more asking about the open cover of (p-r, p+r) neighborhoods – user402857 Jan 09 '25 at 06:54
  • Oops I see how I did initially ask the question you answered. Now how about the (p-r, p+r) neighborhoods? – user402857 Jan 09 '25 at 06:59
  • @user402857 Can you elaborate on what the $(p-r, p+r)$ neighborhoods are? Does $p$ range over all points in $[a, b]$? Does $r$ range over all positive real numbers? – ckefa Jan 09 '25 at 12:35
  • Yes, p is over all points in [a,b], and each r is some very small real number. I guess my strugfles stems from the fact there are infinitely many reals between a and b, so why can a finite number of open neighborhoods cover the interval? Every p in [a,b] is a limit point - every neighborhood (however small) of p has infinitely many points in [a,b]. If a finite number of these neighborhoods cover the interval that means there is a finite number F for which infinity times F is infinity? – user402857 Jan 10 '25 at 23:27
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    @user402857 The point is that even a tiny interval $(p-r, p+r)$ contains infinitely many points. For an exaggerated example, let $[a, b] = [0, 1]$, and consider the open intervals $(-1, 1)$, $(0, 2)$. Clearly this gives an open cover for $[a, b]$, even though there are only two open sets here. (You could even take a single open set $(-1, 2)$ as a cover.) So it's entirely possible for a finite number of open sets to cover $[a, b]$; this is not an obstacle. – ckefa Jan 10 '25 at 23:52
  • Thanks I see how this is not an obstacle now. I am still having trouble directly seeing how a finite number of these neighborhoods can cover it. Why not an infinity of infinite sets? I feel like I am missing the key – user402857 Jan 12 '25 at 07:13
  • @user402857 What issue do you have with the example $(-1, 1) \cup (0, 2)$ covering $[0, 1]?$ These are two finite open sets whose union covers $[0, 1].$ – ckefa Jan 12 '25 at 07:43
  • Yes but the existence of a finite cover does not imply that every infinite open cover has a subcover. I am struggling with how every single infinite open cover has a finite subcover – user402857 Jan 14 '25 at 20:44
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There's a trick: use "proof by contradiction".

Let $\mathcal O$ be an open cover for $[a,b]$.

Take $\mathcal A$ the elements $x\in[a,b]$ such that the subinterval $[a,x]$ can be covered by finite member of $\mathcal O$.

Since $\mathcal O$ covers there exist some $U_0$ in this cover such that $a\in U_0$. Suppose that open set is $U_0=]a_0,b_0[$ then $[a,\frac{a+b_0}{2}]\subset U_0$, hence $\frac{a+b_0}{2}\in\mathcal A$, and so $\mathcal A$ is not empty.

Now take $\xi=\sup\mathcal A$ and suppose that $\xi<1$.

So there are finite open sets in that cover giving $$[a,\xi]\subset U_{\xi1}\cup U_{\xi2}\cup\cdots\cup U_{\xi n}.$$

Let $U_1=]a_1,b_1[$ in $\mathcal A$ be such that $\xi\in U_1$. Then $$[a,\frac{\xi+b_1}{2}]\subset U_{\xi1}\cup\cdots\cup U_{\xi n}\cup U_1.$$ If that is true then $\frac{\xi+b_1}{2}$ is in $\mathcal A$ and contradicts the purported essence of $\xi$.

This contradiction dissolves when $\xi=1$. This means that any cover of $[a,b]$ has a finite subcover.

janmarqz
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