Tychonoffication with an application


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 $C_p$-theory, namely: Given space $X$ there is a Tychonoff space $X'$ such that $C_p(X)$ and $C_p(X')$ are isomorphic (homeomorphic and holomorphic).

Recall that when $X$ is a topological space then $C_p(X) = (C(X), \tau_p)$ is the space of real continuous functions equipped with the topology of pointwise convergence $\tau_p$ which is generated by a basis formed by $$ \gates{n}{x}{U} \coloneqq \{\,f \in C(X) : f(x_i) \in U_i \text{ for } 1 \leq i \leq n \,\} , $$ where $n \in \N$, each $x_i \in X$ and $U_i$ is open in $\R$. Equivalently, we can view the set of all continuous real functions on $X$ as a subset of all real functions, in symbols $C(X) \subset \R^X$. When we equip $\R^X$ with the product topology, then $C_p(X)$ has simply the subspace topology.


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 $X$ into $X'$ which is Tychonoff and $C_p(X) \cong C_p(X')$. This greatly simplifies $C_p$-theory: It allows us to assume that $X$ is Tychnoff while working with $C_p(X)$. See it in action in the theorem below.

Let $X$ be Tychonoff. $C_p(X)$ is metrizable if and only if $X$ is coutable.

⇒ Assume there is a countable base at $f$. Derive at contradiction using a Cantor-diagonal-argument-like trick to create an open neighborhood of $f$ which is not a supset of any set from a local base. ⇐ Note that $C_p(X)$ is a subset of $\mathbb{R}^X$ with product topology, hence metrizable as a countable product of metrizable space.

⇒ We are proving the contrapositive. Assume that $X$ is uncountable, but $C_p(X)$ is metrizable. Metric spaces are first countable so we can pick a countable local base $\mathcal{A}$ at the zero function $\theta$. For $A \in \mathcal{A}$ take a basic set $B_A \subset A$ which contains $\theta$. Let $\text{support}[x_1, \dots, x_n; U_1, \dots, U_n] \coloneqq \{x_1, \dots, x_n\}$ and
$$ Y \coloneqq \bigcup_{A \in \mathcal{A}}\text{support}(B_A) \subset X. $$ Set $Y$ is countable as a countable union of finite sets. Because $X$ is uncountable there is $x \in X \setminus Y$. Since $[x;(-1,1)] \eqqcolon V_x$ is an open neighborhood of $\theta$ and $\mathcal{A}$ is a base at that point there is $A \in \mathcal{A}$ such that
$$V_x \supset A \supset B_A = [x_1, \dots, x_k; U_1, \dots, U_k].$$ Since $X$ is Tychonoff we find $f\in C(X)$ such that $f(x_i) \in U_i$ for $1 \leq i \leq k$ and $f(x) = 1 \not \in (-1, 1)$. The function $f$ is well defined because $x \not\in \text{support}(B_A)$. By construction $f \in B_A$ but $f \not \in V_x$. Contradiction with $V_x \supset B_A$.

⇐ Let $X$ be countable. Note that $C_p(X) \subset \mathbb{R}^X$. Since $\mathbb{R}^X$ is metrizable as a countable product of metrizable space, space $C_p(X)$ is metrizable as a subset of a metrizable space $\mathbb{R}^X$.

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 $X$ is not Kolmogorov (not $T0$) when there is a pair of distinct points $x$ and $y$ such that every open set containing one point, contains the other. Any trivial topology on a set with at least two points is not $T0$.

We can think of it as if topology could not detect a difference between $x$ and $y$. If that is the case, it is a straightforward exercise to show: $f \in C(X) \implies f(x) = f(y)$.

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 $X$ even more?

Consider the following example. Let $X$ be an $\mathbb{R}$ with the excluded point topology with $0$ (that is $U \neq X, \emptyset$ is open only when $0 \in U$). This space is $T0$ but the only real continuous functions are constant. Hence $C_p(X) \cong C_p(\{0\})$.

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


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 $$x \sim y \iff \forall f \in C(X) \text{ we have } f(x) = f(y).$$ Then topologize $X/\!\sim$ 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 $\mathcal{F}$ be a family of functions from a set $X$ to a topological space $Y$. The weak topology on $X$, denoted $\sigma(X, \mathcal{F})$ is the smallest topology which makes functions from $\mathcal F$ continuous. Its subbasis consists of sets of the form $f^{-1}(U)$, where $f \in \mathcal{F}$ and $U \subset Y$ is open.

Let $A$ be a nontrivial closed set. Pick $x$ in $A^c$ which is open. Hence there exists a basic \nei $$ x \in \bigcap_{i=1}^n f_i^{-1}(U_i) \subset A^c, $$ where $f_i \in \mathcal{F}$ and $U_i \subset \R$ is open for each $1 \leq i \leq n$. Since $\R$ is completely regular and point $f_i(x)$ is not in the closed $U_i^c$, there is a continous function $g_i\colon \R \to [0,1]$ such that $g_i(f_i(x)) = 1$ and $g_i(U_i^c) = \{0\}$. Let $$ h \coloneqq (g_1 \circ f_1) \cdots (g_n \circ f_n). $$ The function $h\colon X \to [0,1]$ is continuous as a finite product of continuous functions. We have $h(x) = 1 \cdots 1 = 1$. Take $a \in A$. There has to be an index $i$ for which $f_i(a) \not\in U_i$, since otherwise $a \in \bigcap_{i=1}^n f_i^{-1}(U_i) \subset A^c$ which is a contradiction sine $a$ is in $A$. Thus $g_i(f_i(a)) = 0$ and $h(a) = 0$. Hence $h(A) = \{0\}$.

Now assume that $\mathcal{F}$ separates points in $X$. To show that $X$ is Tychonoff it remains to show that $X$ is Hausdorff. Take different points $x,y \in X$. By assumption, there exists $f \in \mathcal{F}$ such that $f(x) \neq f(y)$. Since $\R$ is Hausdorff there exist disjoint open neighborhoods $U, V$ of $f(x), f(y)$ respectively. Sets $f^{-1}(U) \ni x$ and $f^{-1}(V) \ni y$ are weakly open sets separating $x$ and $y$.


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

Step I. Define an equivalence relation: $$x \sim y \iff \forall f\in C : f(x) = f(y).$$

Step II. For $f\in C(X)$ define $$\phi_f = \{ ([x], f(y)) : [x] \in \quot{X}{\sim} \text{ and } y \in [x] \text{ is arbitrary}\}.$$ It’s straightfoward to see that $\phi_f$ is a function $\phi_f \colon \quot{X}{\sim} \to \R$.

Step III. Construct the Tychonoff topology on $\quot{X}{\sim}$. Let $\mathcal{F} = \{\, \phi_f : f \in C(X)\,\}$. Note that this family of functions separates points in $\quot{X}{\sim}$. By lemma the weak topology $\sigma(\quot{X}{\sim}, \mathcal{F}) \eqqcolon \tau$ is Tychonoff.

And that’s all!

Remark. The weak topology on $\quot X \sim$ from the above theorem may not be the same as a quotient topology $\tau_q$. Since projection $\pi$ is weakly continuous and $\tau_q$ is the finest topology that makes $\pi$ continuous, we have $\tau \subset \tau_q$. This inclusion might be strict as we show in the following example.

Example Take the $K$-topology on $\R$. A basis of this topology is formed by open intervals $(a, b)$ and $(a, b) - K$, where $K = \{\, 1/n : n \in \Z \,\}$. Because $K$-topology is strictly finer than $\tau_\R$, the identity function $f \colon (\R, \tau_K) \to (\R, \tau_\R)$ is continuous. This means that $x \sim y \iff x = y$. So each equivalence class is a singleton. Thus $(\R, \tau_K) \cong (\quot \R \sim, \tau_q)$. But $K$-topology is not regular. And in particular, it’s not Tychonoff. On the other hand $\quot{\R}{\sim}$ 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 $X$ be a space and $\quot{X}{\sim}$ its Tychonoffication. For any Tychonoff space $Z$ and continuous transformation $f\colon X \to Z$ there is a \emph{unique} and continuous $g \colon \quot{X}{\sim} \to Z$ such that $f = g \circ \pi$. In other words, the following diagram commutes: \begin{center} \begin{tikzcd} X && {\quot{X}{\sim}}\\ \\ && Z \arrow["\pi", from=1-1, to=1-3] \arrow["g", dashed, from=1-3, to=3-3] \arrow["f"', from=1-1, to=3-3] \end{tikzcd} \end{center}

Tychonoffication and Cp-theory

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 $\pi \colon X \to X/\!\sim$ is continuous.

Define $$ \phi \colon C_p(X) \to C_p(X/\!\sim) $$ $$ f \mapsto \phi_f $$ Show that $\phi$ is a bijection.

Moreover, $\phi$ is a homomorphism. For any $f,g \in C(X)$ and $[x]$ we have $$ \phi(f+g)([x]) = (f+g)(x) = f(x) + g(x) = \phi(f)([x]) + \phi(g)([x]). $$

Lastly, $\phi$ is a homeomorphism. We will only show that $\phi$ is continuous. Let $[[x]; U]$ be an arbitrary subbasic element of $C_p(\quot{X}{\sim})$ (it is enough to check continuity on subbasic sets). For any $x \in [x]$ the set $$ \phi^{-1}(U) = \{\, f \in C(X) : f(x) \in U \,\} $$ is open in $C_p(X)$ as standard subbasic element. Hence $\phi$ 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