What makes an interval compact or non-compact?

Question

This question has been asked here several times but I still have some gaps in my knowledge. So $(0,1)$ is non-compact while $[0,1]$ is. The explanations usually involve creating a set $(\frac{1}{n},1)$. Here is what I do not understand.

Why can't we just construct a set like $\{(-1,2)\}$? It is an open set. It is a finite collection. I know Heine-Borel theorem and that it implies that a closed and bounded set is compact. So somehow, I figure $[0,1]$ is compact because it a $1$-cell.

The book I am using (baby Rudin, I understand, it is called) has not discussed any sequences as such. So what am I doing wrong by choosing a set like $(-2,2)$ or some such.

Answer 1

Compactness means: every open cover has a finite subcover. It does not suffice to check that there is one finite subcover! You have to check them all.

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$(-1,2)$ has an open cover without finite subcover:

$$\bigcup_{n=1}^\infty (-1+1/n, 2-1/n) = (-1,2)$$

Thus $(-1,2)$ is not compact.

Alternatively,$(-1,2)$ is not closed (thus it can't be compact).

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The interval $[0,1]$ is compact because it is closed and bounded (Heine-Borel).

Answer 2

If I said "All odd numbers are prime" and to prove it I produced the number $5$, you wouldn't accept that.

If I said "All open covers contain a finite subcover" and produce only the open cover $(-1,2)$, why would you accept that?

Answer 3

Yes, $(0,1)\subset(-1,2)$, and yes, $(-1,2)$ is open and, yes, $\bigl\{(-1,2)\bigr\}$ is finite. However, asserting that $(0,1)$ is compact means that for every set oven open subsets of $\mathbb R$ whose union contains $(0,1)$, there is a finite subse whose union still contains $(0,1)$.

And now consider $\Lambda=\left\{\left(\frac1n,1-\frac1n\right)\,\middle|\,n\in\{3,4,5,\ldots\}\right\}$ has the property that the union of its elements conains $(0,1)$. However, this doesn't happen for any finite subset of $\Lambda$.

Answer 4

A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded $\textbf{or}$ if and only if every sequence in that set has a subsequence that converges to a limit point contained in the set $\textbf{or}$ if and only if every cover of the set consisting only of open sets has a finite subcover (i.e. you can somehow choose finitely many of those open sets such that they still cover the whole set). $(-2,2)$ isn't compact because it is not closed. Alternatively, you can pick a sequence that converges to $-2$ or $2$, e.g. $x_n = 2-\frac{1}{n}$. Then, $(x_n) \in (-2,2)^\mathbb{N}$ but $\lim_{n\rightarrow\infty} x_n \not\in (-2,2)$. Alternatively, you can find an open cover of $(-2,2)$ that does not have a finite subcover (what about $A_n = (-2+\frac{1}{n},2-\frac{1}{n})$?).