notes / Math 55b / chapter 2: Connectedness and Compactness
6 Path-Connectedness and Compact Spaces
· Denis Auroux
Last time: is connected if there is no separation with disjoint nonempty open subsets.
Path-connectedness
Definition 6.1. Let be a topological space, . A path from to is a continuous map (where has the subspace topology of the standard ) such that and .
Definition 6.2. is path-connected if every pair of points in can be joined by a path.
Note. The relation and can be connected by a path is an equivalence relation, i.e.
(constant path );
(backwards path );
and (concatenate paths: ).
The equivalence classes are called the path components of . (Will return to these in algebraic topology!)
Theorem 6.3. If is path-connected then is connected.
Proof. Assume not, i.e. disjoint open, ,. Pick a path connecting to . Then are open subsets (containing and respectively). This contradicts the connectedness of .∎
The converse is false in general, but true for nice enough spaces, e.g. CW-complexes.
Example 6.4 (The “topologist’s sine curve”). Let
The “main” part is connected, since it’s the image of (connected) under the continuous map .
Hence is connected: if disjoint open, then is disjoint and open, so one of them (e.g. ) is empty. Then . But is not open in , so in fact .
But is not path-connected: there is no continuous
because the intermediate value property for gives a sequence such that and , so . (Cf. Munkres, end of §24.)
However, for nice enough spaces the two notions are equivalent. For example:
Theorem 6.5. open is connected iff is path-connected.
Proof. Assume open in : then the path components of (i.e. equivalence classes for the equivalence relation “can be joined by a continuous path in ”) are open. Indeed, if then there is such that , and any two points of can be connected inside by a straight line segment. So all of is in the same path component. Now: if is not path-connected then
gives a separation (while we’ve already seen that if is path-connected then is connected). ∎
This implies similar results for other classes of spaces, e.g. topological manifolds and CW-complexes.
For these kinds of spaces, path components are also connected components, i.e. they give a partition of into disjoint connected open (and closed) subsets. Such a partition only exists if is locally connected, i.e. the topology has a basis consisting of connected open subsets. (Counterexample: isn’t locally connected: each point of is its own path component, but these aren’t open.)
Compactness (Munkres §26–…)
Compactness is a “finiteness/boundedness” property of nice topological spaces such as closed bounded intervals , or more generally, closed bounded subsets of .
E.g.: any continuous map with compact achieves its maximum and minimum.
The definition isn’t very intuitive.
Definition 6.6. An open cover of a topological space is a collection of open subsets such that .
Notes not reached in this lecture, covered at the start of Lecture 7
Definition 6.7. is compact if every open cover of admits a finite subcover, i.e. there are such that .
Showing a space is not compact is much easier than showing it is!
Example 6.8. is not compact: the open cover has no finite subcover.
Neither is with the subspace topology: has no finite subcover.
Example 6.9. is compact: given any open cover , let be such that ; then also contains for all large , hence containing , together with , give a finite subcover.
Theorem 6.10. If is compact and is continuous, then (with the subspace topology) is compact.
(Remark: an open cover of with the subspace topology open, .)
Proof. Let be an open cover of . Then is an open cover of , hence there are such that . So for all ,, i.e. , a finite subcover. ∎
Once we know subsets of are compact iff closed and bounded, taking , this gives the extreme value theorem. To get started on this right away:
Theorem 6.11. (with the subspace topology ) is compact.
Proof. Let be an open cover of . Let
(it contains ). We want to show . Let .
First we show : there is such that ; since is open, there is such that . On the other hand , so there is with , and a finite subcover . Therefore , and .
Next, assume : since , there are such that , which is open, so there is such that . Hence covers for some (e.g. if this is , else ), which contradicts .
So : there is a finite subcover.
∎
Theorem 6.12. compact, closed is compact.
Proof. Given an open cover of , i.e. open with , let , which is open: then is an open cover of , hence there is a finite subcover. Discarding , this gives a finite subcover for .∎
The converse is true in Hausdorff spaces!
Theorem 6.13. Hausdorff, compact is closed in .
Proof. We show that is open. Let . Since is Hausdorff, for every there are disjoint open subsets ,.
Now is an open cover, so by compactness there are such that .
Let , which is open. Then , so .
Hence: for all , there is an open such that .
∎
(If we tried this for an arbitrary subset of , we’d find that isn’t a neighborhood of anymore. Compactness lets us reduce an infinite process to a finite one.)
Remark. We’ve actually shown more: Hausdorff, compact, there exist disjoint open subsets ,,. I.e.: we can separate points from compact subsets!
Example 6.14. When isn’t Hausdorff, compact closed in : e.g. with the finite complement topology: any subset is compact. Indeed, a nonempty open subset contains all but finitely many points, so given an open cover it is easy to find a finite subcover: take one nonempty , with finite complement , then take containing for .
Another instance of compactness allowing us to intersect infinitely many opens (or rather, reduce to a finite intersection) is the tube lemma:
Proposition 6.15 (Tube lemma). Let be a topological space, a compact topological space, . If is open and , then there exists a neighborhood of in such that .
Proof. For all , open, so there is a basis open set , a neighborhood of in , a neighborhood of in , such that .
Now is an open cover. (Remark: , but is not open!)
Since is compact, there are such that . Let . Then is a neighborhood of in , and .∎
Theorem 6.16. compact is compact.
Proof. Let be an open cover of . For any given , is compact, so there is a finite subcollection which suffices to cover .
is open, so by the tube lemma there is a neighborhood in such that . Now is compact, and form an open cover, so there are such that .
Now ,,, is a finite subcover for .∎
We now have all the ingredients needed to prove:
Theorem 6.17. is compact iff is closed and bounded.