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A metric space $X$ is called compact iff each of its open cover has a finite subcover. For example $[0,1)$ is not compact. I understand what open covers are; for example set $A = \{(-1,1)\}$ may be one of open coverings of $[0,1)$.

How do I check if set $A$ admits a finite sub covering? I understand what open covers are.

Attempt:

Let $O$ be the set of open intervals of the form $(-1/2, 1-1/t)$. $O$ is an open cover but it does not admit a finite subcover. My question is related with understanding of how can we say $O$ doesn't admit a finite subcover.

I find it difficult to apply the open covering definition of compactness. If someone can explain this definition with examples it would be good for me.

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Srijan
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  • I can't quite tell what you're asking. Some thoughts: The $A$ you give is already finite. Do you want an example of an open covering of $[0, 1)$ that has no finite subcover? – Dylan Moreland Jun 02 '12 at 04:17
  • @DylanMoreland I have edited my question. – Srijan Jun 02 '12 at 04:26
  • Cool. What is $t$ allowed to be? – Dylan Moreland Jun 02 '12 at 04:27
  • @DylanMoreland take largest t – Srijan Jun 02 '12 at 04:28
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    Exactly! Take a finite subcover $O_f$. Then there is a largest $m$ such that $(-\frac{1}{2},1-\frac{1}{m})$ is in $O_f$. But then the point $1-\frac{1}{2m}$ is not covered by any element of $O_f$. – André Nicolas Jun 02 '12 at 04:32
  • @AndréNicolas how can we say that 1-1/2m is not covered by $O_f$ . I am sorry for my stupidity. I have just started studying metric spaces. – Srijan Jun 02 '12 at 04:40
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    We are just on the line, nothing fancy. Think geometrically. The distance from $1$ of any point in $-1/2,1-1/t)$ is greater than $1/t$. Let $m$ be the largest number such that $(-1/2,1/m)$ is in our supposed finite subcover. The point $1-1/(2m)$ is distance $1/(2m)$ from $1$, which is closer than any point in the subcover. Actually, already $1-1/m$ is (barely) too close to $1$ to be covered. Sketch a picture, pick a finite number of $t$ like $5$, $8$, and $17$. Find a point not covered. – André Nicolas Jun 02 '12 at 05:14
  • @AndréNicolas Thank you very much. I understand now. – Srijan Jun 02 '12 at 05:29
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    Good. You shouldn't apologize, these things are hard at the beginning. But remember that there are simple pictures associated with most of these ideas. We use the pictures to write down a formal argument, because our mental picture could be wrong. There are for example some continuous functions that are weirder than anything one could imagine drawing. – André Nicolas Jun 02 '12 at 05:36
  • @AndréNicolas My sincere thanks to you. Your comments "The distance from 1 of any point in ...." was sufficient for me to understand things. Thanks again. – Srijan Jun 02 '12 at 05:46
  • @AndréNicolas I think all the points of [1-1/2m, 1) can not be covered by $O_f$. Am i right ? – Srijan Jun 02 '12 at 05:59
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    That's right, none of them can be covered. – André Nicolas Jun 02 '12 at 06:01
  • @AndréNicolas Wow I understand now.:) many thanks – Srijan Jun 02 '12 at 06:03

1 Answers1

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Here's the first task.

Let $U_n = (0, 1 - \frac1n)$ for $n \in \mathbf N$. Show that the open covering $\{U_n\}$ of $(0, 1)$ has no finite subcover.

Some steps you could follow:

  1. A finite subcover is of the form $\{U_n\}_{n \in S}$ for some finite subset $S$ of $\mathbf N$.
  2. If $S$ is non-empty then let $N$ be the largest element of $S$. Then $U_n \subset U_N$ for all $n \in S$, so it is enough to show that $U_N$ does not contain all of $(0, 1)$.
  3. $1 - \frac1N$ is an element of $(0, 1) \setminus U_N$.

I can't tell you how to understand the definition in a few words, but here are some thoughts. It's often enough to understand a special case of something, while keeping in mind that exceptions exist. Here, the familiar case is that of $\mathbf R$. More generally, in a metric space compactness is equivalent to sequential compactness. This might be a more visceral notion, and it might help to see the proof of equivalence in section 2 of this handout of Brian Conrad's.

For me, this justifies thinking of non-compact sets as those in which points can run off, either to a point outside of the set or "to infinity". See also this (closed) question at MO. Try thinking about why you can't construct an example like the one you gave for $[0, 1]$.

As opposed to applying the open covering definition directly, it's often enough to remember the fundamental fact that a subset of $\mathbf R^n$ is compact if and only if it is closed and bounded, together with standard theorems on compactness.

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Dylan Moreland
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  • Please check whether I am right or not. If yes then i got the concept. I took U = (0, 1- 1/2n) as open covering of (0,1). Now took a finite subcover say A ={(0, 1-1/n)} OF (0,1). But there is largest m such that point 1-1/2m is not covered by A. – Srijan Jun 02 '12 at 04:57
  • @srijan I can't quite follow the notation. What I wrote above makes some sense to me. Does it look alright to you? – Dylan Moreland Jun 02 '12 at 05:39