Let me first make sure I have the correct definitions because my question will be about the difference about the two and there may be some massive confusion on my part.

A topological space $X$ is said to be completely regular or Tychonoff when it is Hausdorff and satisfies the following equivalent conditions:

  • For every $x \in X$ and closed $F \subseteq X$ such that $x\not\in F$, there exists a continuous $f\colon X\to\mathbb{R}$, which we can assume to have values in $[0,1]$, such that $f(x) = 0$ and $f|_F = 1$.

  • The map $X \to [0,1]^{C(X,[0,1])}$ taking $x\in X$ to the family $(f(x))_{f\in C(X,[0,1])}$ of its images under every continuous $f\colon X\to[0,1]$ defines a homeomorphism of $X$ to its image.

  • The Stone-Čech compactification map $X \to \beta X$ defines a homeomorphism of $X$ to its image.

  • There exists a compact [Hausdorff] space $K$ such that $X$ is homeomorphic to a subspace of $K$.

On the other hand, a (necessarily Hausdorff) topological space $X$ is said to be functionally Hausdorff (or Urysohn, but some people use this to mean something different, so it's probably best to avoid this terminology) when it satisfies the following equivalent conditions:

  • For every $x,y \in X$ such that $x\neq y$, there exists a continuous $f\colon X\to\mathbb{R}$, which we can assume to have values in $[0,1]$, such that $f(x) = 0$ and $f(y) = 1$.

  • The map $X \to [0,1]^{C(X,[0,1])}$ taking $x\in X$ to the family $(f(x))_{f\in C(X,[0,1])}$ of its images under every continuous $f\colon X\to[0,1]$ is injective.

  • The Stone-Čech compactification map $X \to \beta X$ is injective.

  • There exists a continuous injective map $X \to K$ with $K$ a compact [Hausdorff] space.

I note that example 91 (the “deleted Tychonoff corkscrew”) in Steen & Seebach's Counterexamples in Topology gives an example of a functionally Hausdorff space which is not completely regular, showing that the two notions are not equivalent.

Since until recently I thought these two notions were equivalent (I somehow thought that $X \to \beta X$ was automatically an embedding when it is injective), my goal is essentially to dispel the confusion I had; I first have to ask:

Question 0a: Is the above account correct? (Are the properties I claim to be equivalent indeed equivalent, and equivalent to standard definitions for the terms they claim to define?)

Every topological space $X$ has a complete regularization or Tychonoff-ization, namely a continuous map $X \to X'$ with $X'$ a completely regular space, such that every continuous map $X \to Y$ with $Y$ completely regular uniquely factors as $X \to X'\to Y$. (I.e., the functor $X \mapsto X'$ is left adjoint to the inclusion functor from the full subcategory of completely regular spaces to that of topological spaces.) This $X'$ can be defined as the image of the Stone-Čech compactification map $X \to \beta X$ with the subspace topology; in particular, $X \to X'$ is always surjective.

Question 0b: Is this still correct?

Edit (2019-06-08): Essentially the above description of complete regularization functor is given, under the name “Tychonoff functor” in the paper “The Tychonoff Functor and Related Topics” by T. Ishii, chapter 6 (p.203–243) in K. Morita & J. Nagata (eds.), Topics in General Topology (1989). So it would appear that it is correct.

Now I thought $X'$ was a quotient space of $X$. This can't be the case because, if what I wrote above is correct, the equivalence relation (“having the same image in $X'$”) is simply “having the same image under every continuous function $X\to\mathbb{R}$ (or equivalently $X\to[0,1]$)”, which is trivial for a functionally Hausdorff space, yet the latter is not necessarily completely regular.

But this is problematic because in section d-2 (“Higher Separation Axioms”, p.158–159) of the Encyclopedia of General Topology (Hart, Nagata & Vaughan eds.) one reads:

“To every space $X$ one can associate a Tychonoff space $Y$ as follows. Two points $x$ and $y$ in $X$ are equivalent if $f(x) = f(y)$ for all continuous real-valued functions $f$ on $X$. The corresponding quotient space $Y$ is Tychonoff and the rings $C(X)$ and $C(Y)$ of real-valued continuous functions are isomorphic; the same holds for the rings $C^*(X)$ and $C^*(Y)$ of bounded real-valued continuous functions.”

It would seem to me that this assertion is contradicted by the existence of the aforementioned counterexample in Steen & Seebach.

Question 0c: Am I correct in believing that the above quote is in error? (Or did I miss some fine print or hidden assumption?)

Now assuming all of the above is correct, there are two natural questions which are left open:

Question 1a: Does every topological space $X$ have a “functional Hausdorffization” (or “Urysohnization”), namely, does the inclusion functor from the full subcategory of functionally Hausdorff spaces to that of topological spaces have a left adjoint? • Question 1b: If so, is it given by quotienting by the equivalence relation “$f(x) = f(y)$ for all continuous real-valued functions $f$ on $X$” or is there some subtlety?

Edit (2019-06-08): The paper “A universal factorization theorem in topology” (Canad. Math. Bull. 9 (1966) 201–207), by R. Sharpe, M. Beattie & J. Marsen defines an equivalence relation $T_{3/2}$ (it is not clear to me whether this is a typo for $T_{3\frac{1}{2}}$, and whether they may be under the same confusion as mentioned above) on any topological space by $x\,\mathbin{T_{3/2}}\,y$ when for every continuous $f\colon X\to [0,1]$ we have $f(x)=f(y)$, i.e., the relation defined above. Calling just $R$ this equivalence relation, according to this paper's main theorem, the crucial point in answering questions 1ab positively is to check that $R = \lim R$ in the notation of the paper, where $\lim R$ is the equivalence relation defined by $x\mathbin{(\lim R)}y$ when for every continuous $f\colon X\to Z$ with $Z$ such that $R$ is just equality (i.e., in our case, every functionally Hausdorff $Z$) we have $f(x)=f(y)$; but this is trivial: so it appears that questions 1ab have a positive answer.

Question 2a: Even if complete regularization is not given by a quotient, is there still a unique coarsest equivalence relation $R$ on any topological space such that $X/R$ is completely regular? • Question 2b: If so, can we describe $R$ concretely, and can we describe the natural continuous map $X' \to X/R$ (where $X'$ is complete regularization as defined above)?

  • 2
    $\begingroup$ For questions 0a - 0c : Taking the quotient by the equivalence relation defined by continuous $\mathbb{R}$-valued functions is not enough to "Tychonoffify" a space, you must also coarsen the topology to the one generated by the cozero sets (the preimages of open sets under continuous maps to $\mathbb{R}$), or equivalently the topology as a subspace of the Stone-Čech compactification, as you say. $\endgroup$ Jun 8, 2019 at 18:36
  • $\begingroup$ Gro-Tsen, is there a reason you haven't accepted K.P's answer. It seems to completely answer your question. $\endgroup$
    – Tyrone
    Dec 13, 2022 at 14:03
  • $\begingroup$ @Tyrone I had forgotten to. This is now done. $\endgroup$
    – Gro-Tsen
    Dec 13, 2022 at 14:30

1 Answer 1


0a Correct, except for one point: is $X$ is not completely regular then there is no $\beta X$, so the third equivalence in `functionally Hausdorff' does not exist.

0b Not quite, the paper mentioned in your edit maps $X$ into a Tychonoff cube and lets $X'$ be the image, see 0a: there is no $\beta X$ available.

0c That is indeed erroneous; it is the quotient of the set $X$ endowed with the (weaker) topology generated by the continuous functions.

1a This is correct, as you point out below the question.

1b There is no subtlety, every continuous $f:X\to[0,1]$ will factor through that equivalence relation.

2a Assuming this definition of coarse: the coarsest equivalence relation on the underlying set (all points equivalent) gives a one-point quotient, which is certainly completely regular. In that case the answer to 2b is an obvious yes.

2b Let $X$ be the deleted corkscrew; its complete regularization $X'$ is just the set underlying set $X$ with a weaker topology: that generated by the continuous functions. It has many completely regular quotients: for every $\alpha<\omega_1$ and $n<\omega$ let $C(\alpha,n)$ be the set of points $(i,\beta,m)$ with $i\in\mathbb{Z}$, $\beta<\alpha$ and $m>n$. The set $C(\alpha,n)$ is clopen in the corkscrew-without-$\{a^+,a^-\}$, hence so is $\{a^+\}\cup C(\alpha,n)$ in $X$. Identifying that set to a point yields a completely regular quotient. There is no finest equivalence relation among these and their common refinement of all these yields the identity relation, which does not yield a completely regular quotient.

  • 2
    $\begingroup$ Much as I esteem your expertise in general topology, I don't know why you insist "there is no $\beta X$". Some of us use $\beta X$ for the Gelfand spectrum of $C_b(X)$. This is the left adjoint to the forgetful functor from compact Hausdorff spaces to topological spaces. When restricted to completely regular spaces it agrees with any other construction of $\beta X$, such as Stone's or Čech's, as it has the same universal property. The map $X \rightarrow \beta X$ is then injective iff $X$ is functionally Hausdorff and an embedding iff $X$ is Tychonoff. $\endgroup$ Oct 2, 2019 at 14:02
  • $\begingroup$ @Robert Furber: I'm old school $\endgroup$
    – KP Hart
    Oct 16, 2019 at 12:24

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