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I'm having a difficulty understanding how to work with the following definition of a compact set:

$$ \text{A set } A\subseteq \mathbb{R}^n \text{ is said to be}\, \textit{compact}\, \text{ if any finite open cover of } A \text{ contains a finite subcover of } A. $$

Now, I know that this is equivalent to the following characterization:

$$ \text{A set } A \text{ is}\, \textit{compact}\, \text{ if and only if it is closed and bounded.} $$

Now, by the second statement, $\bigl[0,1\bigr]$ would be a compact and something like $\bigl(0,1\bigr)$ and $\bigl(-\infty,1\bigr]$ wouldn't. Here comes my question: how would I show that $\bigl(0,1\bigr)$ and $\bigl(-\infty,1\bigr]$ aren't compact, and that $\bigl[0,1\bigr]$ is compact using the definition?

One further thing that's boggling my mind is that, to me, it looks like, at least intuitively, for every open cover of $\bigl(0,1\bigr)$, you should be able to produce a finite subcover. So here is where the definition is counter-intuitive to me as well. Why can't you?

Plungerdz
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    You seem to have mistranscribed the definition of compactness. It is “any open cover contains a finite subcover”. – Malady Feb 10 '25 at 17:32
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    The cover $(0,1-\frac1n)$ with $n \in \mathbb{N}$ and $n \geq 2$ is an open cover of $(0,1)$ but no finite subcover will work. You can use a similar idea for $(-\infty , 1]$. – CyclotomicField Feb 10 '25 at 17:33
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    To "know" that the definition with open covers is equivalent to being closed and bounded, you must "know" a proof of it. Scrutinize it with your examples in hand (including $[0,1]$). – Anne Bauval Feb 10 '25 at 17:36
  • As for why $[0,1]$ is compact, given an open cover of $[0,1]$, look at the set of $0<a<1$ such that $[0,a]$ has a finite subcover. This set is nonempty, (it contains $0$). Can you show that this set of $a$ is an interval? Then, can you show its supremum is $1$? – Malady Feb 10 '25 at 17:36
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    Note that your equivalent definition holds only valid in finite dimensions, e.g. $A \subseteq \mathbb{R}^n$. – epartow Feb 10 '25 at 17:45
  • https://math.stackexchange.com/questions/188996/showing-that-0-1-is-compact as well as https://math.stackexchange.com/questions/722506/why-isnt-0-1-compact should answer your question. This should be closed as a duplicate. – Malady Feb 10 '25 at 18:57
  • This question is similar to: [Why isn't $(0,1]$ compact?](https://math.stackexchange.com/questions/722506/why-isnt-0-1-compact). If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Malady Feb 10 '25 at 18:58
  • @CyclotomicField your idea was very helpful. Thanks a lot for your contribution! – Plungerdz Feb 11 '25 at 19:15
  • @Malady you're right about me mistranscribing the definition. Will take a look at your links. Really sorry if this has been answered before, but I searched for it and I couldn't find it. – Plungerdz Feb 11 '25 at 19:15

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It may be helpful to know that a closed set is defined to be a set that contains all of its limit points. So, consider a set $A$. Loosely, a limit point of $A$ would be a point $x$ where any epsilon neighborhood containing $x$ would also contain another element of $A$. I.e. if you imagine $A$ to be a circle, a limit point of $A$ would be any point in which no matter how small you could draw a circle (the circle being the epsilon neighborhood) around that point, it would still contain another element of $A$.

So, now lets consider our set $(0,1)$ and lets say $Z=(0,1)$. If we consider the set $Z$ on a number line, it can be seen that $0,1$ are both limit points of $Z$. If you were to try and draw a circle around $1$ without also circling an element of $Z$, you would have a very hard time. We also know by our definition of $Z$ that neither $0$ or $1$ are contained in $Z$, so $Z$ cannot be a closed set.

Using this same idea, finding limit points and showing they are either in the set or not, you can determine whether $(-\infty,1]$ or $[0,1]$ are closed. Additionally, a set $A$ is bounded if $\exists M>0$ s.t. $|a|\leq M \forall a\in A$. Finding a constant $M>\infty$ may prove tricky, so keep that in mind.

Magnus
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  • Sorry @Malady, was trying to speak very informally for the sake of understanding, but I see how that is confusing. Yes, "epsilon neighborhood" is what I was trying to express casually. – Magnus Feb 10 '25 at 19:32