Definition of compactness unnecessarily verbose?

Question

The definition of a compact set is given as a set, $X$, for which all open covers have a finite subcover. This seems unnecessarily verbose to me. Wouldn't it be sufficient to simply say that $X$ has a finite, open cover?

Answer 1

As @BrianM.Scott noted, every space $X$ has a finite open cover, namely $\{X\}$. So this tells us nothing about $X$.

One way to motivate the definition is to see that if $X = \mathbb{R}^d$ with the standard topology, it is satisfied by precisely the subsets which are closed and bounded.

To illustrate this, I'll show this equivalence in one direction: if $A$ is compact according to the open-cover definition, this forces $A$ to be closed and bounded. It's useful to look carefully at this case, because it shows exactly why the ability to reduce any open cover to a finite subcover is important.

So, suppose that $A \subseteq \mathbb{R}^d$ is compact according to the open-cover definition.

We first show that $A$ is bounded. Consider the open cover $\{B_n\}$ where $B_n$ is the open ball of radius $n$ centered at the origin. Since $A$ is compact, this cover has a finite subcover. Since the sets $B_n$ are nested, the finite subcover must be of the form $B_n$ for some particular $n$. So $A \subseteq B_n$ for some $n$, hence $A$ is bounded.

Next, we show that $A$ is closed. Let $y$ be any point in $A^c$. For each $x \in A$, we can find two disjoint open sets $U_x$ and $V_x$, such that $x \in U_x$ and $y \in V_x$. Then $\{U_x\}$ is an open cover of $A$. Since $A$ is compact, it has a finite subcover, so $A$ is contained in the union of finitely many of the $U$'s. Then $y$ is contained in the intersection $I$ of the corresponding $V$'s, and since $I$ is an intersection of finitely many open sets, $I$ is open. (Note: this would not generally true if the intersection were not finite!) Also, by construction, $I$ and $A$ are disjoint, so $I \subseteq A^c$. This shows that $y$ is an interior point of $A^c$. Since $y$ is an arbitrary element of $A^c$, this means that $A^c$ is open, hence $A$ is closed.

Answer 2

Wouldn't it be sufficient to simply say that X has a finite, open cover?

No. $\Bbb R$ has the finite open cover $\{(-\infty,1),(0,,\infty)\}$ but it is not compact.

Answer 3

It's not enough to demand that a space has a finite cover by open sets - every space has a finite cover by open sets, namely $\{$the whole space$\}$. This is a consequence of the fact that the definition of a topological space requires the whole space to be open.

The idea behind compactness is that the open covers of the space are, in a sense, the only covers we ever need to consider: any cover of a compact space is a finite cover, plus maybe a bunch of redundant open sets.