notes / Math 55b / chapter 2: Connectedness and Compactness
5 Uniform Convergence and Connectedness
· Denis Auroux
Last time
Topologies on products , e.g. :
: basis , open, for all but a finite number of .
(for metric spaces): if this is , else .
: basis , open.
For :
in the product topology, for all :pointwise convergence;
in the uniform topology, :uniform convergence.
Remark. On the product topology is actually metrizable, using a clever modification of (see Munkres Thm. 20.5), while the box topology isn’t metrizable (Munkres, end of §21). On uncountable products, neither box nor product are metrizable (Munkres, end of §21).
Example 5.1. In , converges pointwise (i.e. in the product topology) to
but not uniformly ().
(constant functions) uniformly (and pointwise), but not in the box topology: never belongs to , a box neighborhood of . In fact it’s almost impossible for a nonconstant sequence of functions to converge in the box topology.
The notion of uniform convergence is important in real analysis because it is well behaved with respect to continuity and differentiability. For example:
Theorem 5.2. Given a topological space , a metric space , and a sequence of functions : if is continuous for all and uniformly, then is continuous.
Proof. Let be open and . There is such that . Let be such that . Let be open such that (continuity of ). Then, using the triangle inequality, for all ,
So .∎
Corollary 5.3. is a closed subspace of .
Connected spaces (Munkres §23–24)
Definition 5.4. A topological space is connected if it cannot be written as where are disjoint nonempty open sets. (Such a decomposition is called a separation of .)
Note. This is equivalent to: , disjoint nonempty, ,. The latter is easier to check for the subspace topology, but equivalent: since and vice versa, the separation condition implies and are closed in , hence also open (,).
Proposition 5.5. is connected (in the subspace topology) iff for all ,.
I.e. connected subspaces of = intervals (all kinds: closed/open/half-open, semi-infinite, all of , …) (“no gaps”).
Proof.
If there are with ,, then
gives a separation, so is not connected.
Assume ,, and assume isn’t connected, i.e. we have a separation , open in the subspace topology, nonempty. Let ,; without loss of generality . By assumption . Let (, so or ).
First assume : then open in gives such that . But then contains no element of , contradicting .
So ; necessarily (since ), and (since ). But open in gives such that , hence , giving a contradiction.
∎
Theorem 5.6. continuous, connected is connected.
Proof. If is a separation of , then is a separation of , contradiction! (Subspace topology: with open in is open in ; and .)∎
This implies the intermediate value theorem:
Theorem 5.7 (Intermediate value theorem). Let be a connected topological space and continuous. If and lies between and , then there is such that .
Proof. Since is connected, so is . The result then follows from the above characterization of connected subspaces of (“no gaps”). ∎
Remark (Fact). connected (for the subspace topology) connected. Example:
But things are better for unions of connected sets, provided they overlap.
Theorem 5.8. Let be connected subspaces, all containing some point (i.e. ). Then is connected.
Proof. Assume , disjoint, open in . Without loss of generality, . Then and are disjoint and open in . Since is connected and , we must have for all . Hence (and ). So is connected. ∎
Theorem 5.9. connected is connected.
Proof. Fix . Then for all , is connected by the previous theorem (both pieces contain ), and now (all containing ), so is connected.
∎
In fact, more is true (won’t prove):
Theorem 5.10., connected with the product topology is connected.
(This is false for the uniform and box topologies: e.g. for infinite . Say is bounded if is a bounded subset. Then is a separation of in the uniform topology.)
Notes not reached in this lecture, covered at the start of Lecture 6
Path-connectedness
Definition 5.11. 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 5.12. 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 5.13. 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 5.14 (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 .
On the other hand, is not path-connected: there’s no path connecting to . (Proof later, using compactness: the image of such a path is a closed subset of and must contain all of , but that set isn’t closed: is a limit point of it not in .)
However, for nice enough spaces the two notions are equivalent.
Theorem 5.15. open is connected iff is path-connected.
Proof. Already seen: path-connected connected. We show: not path-connected not connected.
Assume open in : then the path components of 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. ∎
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 5.16. An open cover of a topological space is a collection of open subsets such that .
Definition 5.17. is compact if every open cover of admits a finite subcover, i.e. there are such that .