S. D. V. Stephens

notes / Studies in Real and Complex Analysis / chapter 2: Connectedness and Compactness

4 Infinite Products and Connectedness

· Denis Auroux

Original Professor Auroux’s handwritten notes (PDF), behind a password for now:

Recall: in a general topological space, limits of sequences are not necessarily unique!

There’s a whole hierarchy of properties one can require of a topological space; a useful one is:

  • is Hausdorff if open sets , such that .
Two points x and y with disjoint open neighbourhoods U containing x and V containing y.
  • Metric space Hausdorff.
  • There exist Hausdorff spaces that are not metrizable (i.e. whose topology doesn’t come from a metric).

Theorem 4.1. If is Hausdorff then every sequence in converges to at most one limit.

Proof. Assume converges to , and let . Choose disjoint neighborhoods , . Since , such that , . Hence for , so the sequence doesn’t converge to . ∎

Metric spaces are nice, but they can still be pretty nasty. (We’ll see conditions such as local connectedness, local compactness etc. come up.) Algebraic topologists like to focus on even nicer spaces. For example:

Definition 4.2. An -dimensional topological manifold is a topological space such that every point has a neighborhood homeomorphic to (or equivalently, to an open ball in ).

Example 4.3. is a 1-dimensional topological manifold; the sphere and the torus in are 2-dimensional topological manifolds.

A circle S^1 in the plane, and a sphere and a torus in three-dimensional space, as examples of topological manifolds.

Example 4.4.

A Y-shaped graph made of three segments meeting at a single vertex, which is not a manifold because of the vertex.

This Y-shaped graph isn’t a topological manifold (the vertex looks wrong), but it is part of a more general class of spaces called CW complexes, built by attaching “cells” (closed balls of dimension ) onto each other inductively.

We’ll see more on this later when we get to algebraic topology. In decreasing order of generality:

Topologies on infinite products

Given topological spaces , (an index set), what is the natural topology on

First idea:

Definition 4.5. The box topology on has basis .

(This is a basis: box box = box, since .)

This is actually too fine for most purposes.

Example 4.6. Consider the diagonal map (), . Giving the box topology, is not continuous!! (Unlike the case of finite products.)

Indeed, let , which is open in the box topology. Then which is not open in .

Better:

Definition 4.7. The product topology on has basis

(This is the same as the box topology if is finite; for infinite it is coarser.) Unless otherwise specified, the product topology is the one we’ll use on .

Theorem 4.8. (with the product topology), , is continuous each component is continuous.

Proof.

  • The projection to the -th factor is continuous ( open, is open in the product topology). Hence, if is continuous, so is .
  • Conversely, assume all are continuous, and consider a basis element , where for all but finitely many . Then Each is open, and all but finitely many are , so they can be omitted from the intersection. So is the intersection of finitely many open sets in , hence open.

∎

Example 4.9. This now implies the diagonal map is continuous, since each = identity.

Similarly:

Exercise 4.10. A sequence , converges to , the sequence converges to .

Example 4.11. Given a set and a topological space , let with the product topology. Then a sequence converges to iff , in .

So: the product topology is the topology of pointwise convergence.

The uniform topology

On products of metric spaces, there is another natural topology, finer than the product topology but coarser than the box topology: the uniform topology. This works similarly to the construction of on , but for an infinite product the sup might be infinite. So:

  • First step: we can replace the metric on by . This is still a metric (check!) and induces the same topology as (same balls of radius !).
  • Now, given metric spaces , replace each by the bounded metric , and define a metric on ( if it’s , else ).

This is called the uniform metric, and it induces the uniform topology.

Example 4.12. On (with the usual distance on ), this is So : uniform convergence!

Remark. The ball of radius around is contained in , but not equal to it (unless is finite)!

Indeed, only implies ! The ball only contains those for which the sup is . In fact: and is not open for !

Theorem 4.13. The uniform topology on is finer than the product topology, and coarser than the box topology (strictly if is infinite).

Proof.

  1. Let , and a basis element in the product topology. Then such that . Without loss of generality we can assume , and for all but finitely many (whenever ). So , and So is open in the uniform topology: .
  2. , and each is open in the box topology balls of the uniform topology are open in the box topology, so .

∎

Notes not reached in this lecture, covered at the start of Lecture 5

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 (also Munkres, end of §21).

Example 4.14. In , converges pointwise (i.e. in the product topology) to but not uniformly ().

(constant functions) uniformly (and pointwise), but not in the box topology: is never in , 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 4.15. Given a topological space , a metric space , and a sequence of functions : if is continuous and uniformly, then is continuous.

Proof. Let be open and . such that . Let be such that . Let be open such that (continuity of ). Then, using the triangle inequality, , So . ∎

Corollary 4.16. is a closed subspace of .

Connected spaces

(Munkres §23–24.)

Definition 4.17. 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 .)

Two disjoint hatched regions U and V whose union is a space that is not connected.

Proposition 4.18. (standard topology) is connected.

Proof. Assume is a separation. Without loss of generality, . Let

  • , open for some , so .
  • We can’t have : since is open this would imply for some , hence for , hence , contradiction. So .
  • But if : open, such that , and by definition of , such that . Hence , contradicting the definition of .
  • Hence , and since is open, such that ; and by definition of , such that . Hence , and . Contradiction.

∎

Example 4.19. is not connected, since and are open in the subspace topology and provide a separation. More generally, if with and , then is disconnected.

Theorem 4.20. 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 .) ∎

Corollary 4.21 (Intermediate value theorem). Let be a connected topological space and continuous. If and lies between and , then such that .

Proof. Since is connected, so is . If , then and give a separation of (one contains and the other contains ): contradiction. So . ∎