The two definitions of a compact set

Question

1. In general, $A$ is compact if every open cover of $A$ contains a finite subcover of $A$.
2. In $R$, $A$ is compact if it is closed and bounded.

The second is very easy to understand because I can easily come up with an example like $[0,1]$ which is both closed and bounded so it's compact.

However, I am very confused at definition (1) because I don't really understand what is meant by a cover and I don't understand how this is really related to a set being closed and bounded?

Could someone please explain what is the relationship between (1) and (2)?

Thank you.

Answer 1

Let $X \subset R$

1) Compact => bounded.

I find it easy to just do this. For every $x \in X$ let $V_x = (x-1/2, x + 1/2)$. $V_x$ is open and $X \subset of \cup V_x$. So {$V_x$} is an open cover. So it has a finite subcover. So there is a lowest interval and there is a greatest interval in the finite subcollection of intervals and X is bounded between them.

2) Compact => closed

Let X not be closed. Then there is a limit point,y, of X that is not in X. Let's let $V_n$ = {$x \in \mathbb R| |x - y| > 1/n$}. As this covers all $\mathbb R$ except $y$ and $y \not \in X$ it covers X. Take any finite subcover the is a maximum value of $n$ so $(y - 1/n, y + 1/n)$ is not covered by the finite subcover. As $y$ was a limit point, $(y - 1/n, y + 1/n)$ contains points of X. So the subcover doesn't cover X. So X is not compact.

Unfortunately Closed and Bounded => compact is much harder.

But I hope I gave you a sense of the flavor of compact sets.

Answer 2

The definitions of open cover and finite subcover can be stated simply. An open cover of $A$ is a collection of open sets in the underlying topology whose union includes all of $A$. This definition of compactess says that if you have any open cover of any set $A$, you should be able to find a finite collection of sets in that cover that also cover $A$. As an example, let $A=[0,1]$ and my collection of sets be $(\frac 1{2n-1},2), n \in \Bbb N \cup (\frac {-1}{1000},\frac 1{1000})$. Every point in $[0,1]$ is in at least one of these sets, so it is a cover. We can find a finite subcollection by requiring that $n \lt 1000$ in the first piece that also covers $[0,1]$. To prove that $[0,1]$ is compact under this definition you would have to show that finding such a finite subcover was possible for any cover. On the other hand, let $B=\Bbb R$. We can cover $B$ with $(2n, 2n+2), n \in \Bbb Z$ but no finite subcollection covers $\Bbb R$, so $\Bbb R$ is not compact. The reason we specify an open cover is that we could cover $[0,1]$ with $[\frac 1{n+1}, \frac 1n] \cup \{0\}$ and not have a finite subcover. We want $[0,1]$ to be compact, so need to choose a definition that makes it so.

Answer 3

The relationship between 1 and 2 is due to the fact that R has the least upper bound property (aka R is complete). The proof of the equivalency can be found pretty easily online. Baby Rudin contains the proof in the 2nd chapter when Cauchy sequences are introduced.

Answer 4

The relation between the two exists in R^n only. The equivalence of the two conditions is given by Heine–Borel theorem. The proof is given nicely in S kumarsen's book named Topology of Metric Spaces. IN general topological space the 2nd definition is meaningless as there is no notion of boundedness.