# In the definition of 'Compact set', why do we require the existence of finite subcover for any open cover? Why not countable subcover? [duplicate]

For a given topological space, in the definition of 'Compact set', why do we require the existence of finite subcover for any open cover? Why not countable subcover? In particular, what is the importance of having finite subcover for an arbitrary space?

Any help in this regard would be appreciated.

## marked as duplicate by Chill2Macht, Daniel W. Farlow, Alex Provost, Behrouz Maleki, PragabhavaAug 14 '16 at 18:41

• math.stackexchange.com/questions/485822/… – Asaf Karagila Aug 13 '16 at 17:49
• The finiteness condition makes compact sets behave like finite sets in many important respects. If you weaken to a countable cover, you lose all the major theorems. Consider that, according to your definition, the set of rationals is a compact subset of the set of reals (with the standard topology). – symplectomorphic Aug 13 '16 at 17:49
• Countable subcover leads to a Lindelöf topological space. – Santiago Aug 13 '16 at 17:50
• Furthermore, on $\mathbf R$ as the topology has a countable basis, any subset of $\mathbf R$ would be compact. – Bernard Aug 13 '16 at 17:51

I think the best way to convince yourself of the value of the definition of compactness is to look at some theorems which require it as a hypothesis.

For example, consider the following: if a topological space $X$ is compact, then any continuous map from $X$ to $\mathbb{R}$ is bounded (this is a slightly more abstract version of the Extreme Value Theorem from calculus). If we weaken the hypothesis to Lindelof (= every cover has a countable subcover), then the conclusion need not hold: $\mathbb{R}$ with the usual topology is Lindelof (exercise), but there are unbounded maps $\mathbb{R}\rightarrow\mathbb{R}$ (e.g. the identity map).

There are lots of other examples, which you can find if you search around.

Note: in my example above, the converse is not true: there are non-compact spaces which satisfy the extreme value property. I'm not giving an equivalent form of compactness, just a useful consequence of compactness for which its variants (e.g. Lindelof-ness) are not sufficient.

How I always think of it is the following:

the infimum of a finite set of strictly positive numbers has to be strictly positive, but the infimum of an infinite set of strictly positive numbers can be zero.

This fact leads to examples where the finiteness of the subcover has practical importance.

Lebesgue numbers are used to prove that every continuous function from a compact space is uniformly continuous. (See for example here).

If the set is not compact, then we cannot necessarily define a Lebesgue number (see here).

With compactness, if we have a positive number associated to every open set in the open cover, then we can always take a finite subcover and thus find a "smallest" number which is still greater than zero.

If we didn't have compactness, then the "smallest" number might be zero, and then the proof would no longer work.