# Summary

In this post, I show how to turn any topological space into a Tychonoff space (which meets a certain universal property). This construction has a neat application in the theory of spaces of real continuous functions equipped with the topology of pointwise convergence, known as -theory, namely: Given space there is a Tychonoff space such that and are isomorphic (homeomorphic and holomorphic).

Recall that when is a topological space then is the space of real continuous functions equipped with the topology of pointwise convergence which is generated by a basis formed by where , each and is open in . Equivalently, we can view the set of all continuous real functions on as a subset of all real functions, in symbols . When we equip with the product topology, then has simply the subspace topology.

# Context

Let’s look at the broader context first. I identified two motifs.

**Motif 1. X → X’**. Given a topological space and a
desired property, how to change it in some minimal or optimal way.
Compactification is a good example. In our case, the desired properties
are separation axioms.

**Motif 2: Interplay between X, Cp(X) and Cp(Y)**. This
is fun for the sake of exploring and finding counterexamples. For
example, check Drawing
Sorgenfrey continuous functions from Dan Ma’s Topology Blog.

Our goal is to turn into which is Tychonoff and . This greatly simplifies -theory: It allows us to assume that is Tychnoff while working with . See it in action in the theorem below.

Let be Tychonoff. is metrizable if and only if is coutable.

Let be Tychonoff. is metrizable if and only if is coutable.

⇒ Assume there is a countable base at . Derive at contradiction using a Cantor-diagonal-argument-like trick to create an open neighborhood of which is not a supset of any set from a local base. ⇐ Note that is a subset of with product topology, hence metrizable as a countable product of metrizable space.

⇒ We are proving the contrapositive. Assume that
is uncountable, but
is metrizable. Metric spaces are first countable so we can
pick a countable local base
at the zero function
. For
take a basic set
which contains
. Let
and

Set
is countable as a countable union of finite sets. Because
is uncountable there is
. Since
is an open neighborhood of
and
is a base at that point there is
such that

Since
is Tychonoff we find
such that
for
and
. The function
is well defined because
. By construction
but
. Contradiction with
.

⇐ Let be countable. Note that . Since is metrizable as a countable product of metrizable space, space is metrizable as a subset of a metrizable space .

# Hints that we can simplify space X to X’ so that CpX ≃ CpX’

Before we move on to the Tychonoffication let’s wonder a bit.

Recall that that a topological space
is *not Kolmogorov* (not
) when there is a pair of distinct points
and
such that every open set containing one point, contains
the other. Any trivial topology on a set with at least two points is not
.

We can think of it as if topology could not detect a difference between and . If that is the case, it is a straightforward exercise to show: .

This *hints to us that we can simplify X by ‘gluing’ points that
are topologically indistinguishable* and expect that spaces of
continuous functions will be isomorphic (this is Kolmogorov
quotient or Kolmogorofication if you like). Now the question is: Can
we simplify
even more?

Consider the following example. Let be an with the excluded point topology with (that is is open only when ). This space is but the only real continuous functions are constant. Hence .

Now we advise you to ponder: Which points in we can glue so that (where is a space with glued points)? The answer is in the next section.

# Tychonoffication

We will not glue the points which are topologically indistinguishable, but instead, those points which are indistinguishable by continuous real functions!

Formally we define an equivalence relation Then topologize with the weak topology on a certain family of functions — the quotient topology might be too fine for our purposes (we will provide an example later).

Before the construction, we need a definition and an important lemma.

**Definition.** Let
be a family of functions from a set
to a topological space
. The *weak topology* on
, denoted
is the smallest topology which makes functions from
continuous. Its subbasis consists of sets of the form
, where
and
is open.

Let be a set and . Then is a completely regular space. Moreover, if separates points in , then this space is Tychonoff.

Let be a nontrivial closed set. Pick in which is open. Hence there exists a basic where and is open for each . Since is completely regular and point is not in the closed , there is a continous function such that and . Let The function is continuous as a finite product of continuous functions. We have . Take . There has to be an index for which , since otherwise which is a contradiction sine is in . Thus and . Hence .

Now assume that separates points in . To show that is Tychonoff it remains to show that is Hausdorff. Take different points . By assumption, there exists such that . Since is Hausdorff there exist disjoint open neighborhoods of respectively. Sets and are weakly open sets separating and .

## Construction

We outline the construction of Tychonoffication given any topological space. It is straightforward to verify each step.

**Step I.** Define an equivalence relation:

**Step II.** For
define
It’s straightfoward to see that
is a function
.

**Step III.** Construct the Tychonoff topology on
. Let
. Note that this family of functions separates points in
. By lemma the weak topology
is Tychonoff.

And that’s all!

**Remark.** The weak topology on
from the above theorem may not be the same as a quotient
topology
. Since projection
is weakly continuous and
is the finest topology that makes
continuous, we have
. This inclusion might be strict as we show in the
following example.

**Example** Take the
-topology on
. A basis of this topology is formed by open intervals
and
, where
. Because
-topology is strictly finer than
, the identity function
is continuous. This means that
. So each equivalence class is a singleton. Thus
. But
-topology is not regular. And in particular, it’s not
Tychonoff. On the other hand
with the weak topology from the preceding theorem is
Tychonoff.

## Universal property

The Tychonofication process described is optimal in the sense that it enjoys the following universal property:

Let be a space and its Tychonoffication. For any Tychonoff space and continuous transformation there is a and continuous such that . In other words, the following diagram commutes:# Tychonoffication and Cp-theory

For any space let be its Tychoniffication. Then .

Everything that was built so far, is for this proof to be boring. That’s why I give only an outline.

Show that canonical projection is continuous.

Define Show that is a bijection.

Moreover, is a homomorphism. For any and we have

Lastly, is a homeomorphism. We will only show that is continuous. Let be an arbitrary subbasic element of (it is enough to check continuity on subbasic sets). For any the set is open in as standard subbasic element. Hence is continuous.

And we are done! In Tkachuk’s book *A Cp-Theory Problem Book:
Topological and Function Spaces* all of the exercises numbered
101–500 spaces are assumed to be Tychonoff thanks to this single
theorem.

# Notes and related concepts

- Most part of this post is written based on a solution to the 100th problem in Tkatchuk’s aforementioned book.
- Metrizatoin theorem might be proved in many other ways. For example
by noting that
is a convex space. See Theorem 4 in
*Topological Vector Spaces*by Robertsons. Or by noting that is a topological group and using the Birkfoff-Kakutani theorem. I really liked how it was proved in the*Lectures in Functional Analysis and Operator Theory*(Theorem 6.3) by Berberian. - Turing into Kolmogorov space is very easy, into Tychonoff is just a bit laborious. How about othe