$ \newcommand{\R}{\mathop{\mathbf R}} \newcommand{\FN}{\mathop{\mathfrak N}} $

The exercise is in p. 99 of the book. It says the following:

Let $X$ be a topological space, and let $B(X,\R)$ denote the space of all bounded continuous real-valued functions on $X$. Then, there exists a Tychonoff space $Y$ such that $B(X,\R)$ and $B(Y,\R)$ are isomorphic.

The author also gives a hint:

Hint. Weaken the topology on $X$ to obtain a completely regular space with the same ring of functions. Then, identify points to get a Tychonoff space.

I think the hint is enough to prove the result.

We start weakening the topology on $X$ as follows: an open set $U$ will be open in this new space $X^*$ if and only if it is open in $X$ and, for each point $x\in U$, there exists a continuous function which separates $x$ and $X\setminus U$.

We have to prove that this is actually a topology. Consider two ''new'' open sets $U_1$ and $U_2$; for each $x\in U_1\cap U_2$, we need to define a continuous function $f:X\rightarrow \R$ such that $f(x)=0$ and $f(X\setminus U_1\cup X\setminus U_2)= 1$. For that, consider the functions $f_1(x)$ and $f_2(x)$ which separates $x$ from $X\setminus U_1$ and $X\setminus U_2$ respectively, and define $$f(x)=\min \{1,|f_1(x)|+|f_2(x)|\}\text{.}$$ $f$ is continuous because the arguments of the minimum are; furthermore $f(0)=0$ and, if $x\notin U_1\cap U_2$, then at least one of the $f_i$ is $1$. Hence $U_1\cap U_2$ is open in $X^*$.

Now consider a family of open sets $\{U_i\}_{i\in I}$; for each $i\in I$: $$ X \setminus \bigcup_{j\in I} U_j = \bigcap_{j\in I} X\setminus U_j \subseteq X\setminus U_i\text{.}$$

Thus, for each $x$ in the union, it is enough to consider $U_i$ containing $x$ and $f_i$ the function that separates $x$ and $X\setminus U_i$.

These two steps, together with the fact that $X$ is a topological space, show that $X^*$ is another topological space, and that the topology on $X^*$ is coarser than the topology of $X^*$.

The last thing we have to show is that the set of all bounded continuous real-valued functions remains unchanged. So, suppose that we have removed some open set $U$ of $X$; it is because, at least for one $x\in X$, there was no continuous function separating $x$ and $X\setminus U$. Then, is a continuous function with respect to $X$ still continuous at $x$? Suppose the answer is not, i.e. for some such a function $f$ we can open neighbourhood $V$ of $f(x)$ such that $f^{-1}(V)\subset U$. Without lost of generality, we assume that $f(x)=0$; it means $0\in V$. Then, we can work with an $\epsilon$-ball centred at $0$, $W=(-\epsilon,\epsilon)$. But the function $$ \tilde f(y)= \begin{cases} \displaystyle\frac{1}{\epsilon} f(y), & \mbox{if } y\in f^{-1}((-\epsilon,\epsilon)) \\ \displaystyle1, & \mbox{otherwise}. \end{cases}$$

is a continuous function that separates $x$ and $X\setminus U$. It is clear that $f$ does, and to see that $f$ is continuous we can consider a net $\{x_i\}_{i\in I}$ converging to some $x'\in f^{-1}(\epsilon)$ (with respect to the topology of $X$); then

$$\tilde{f}(x_i)= \begin{cases} \displaystyle\frac{1}{\epsilon}f(x_i), & \mbox{if } x_i\in f^{-1}((-\epsilon,\epsilon)) \\ \displaystyle1, & \mbox{otherwise} \end{cases}$$

and the net $\{\tilde{f}(x_i)\}_{i\in I}$ converges to $1$, since $f$ was supposed to be continuous with respect to $X$. So, we have constructed a continuous function which separates $x$ and $X\setminus U$, which is impossible by hypothesis. The conclusion is, if $f$ is continuous(in $X$) but for some open set $U$ and some $x\in U$, there is no continuous function separating $x$ and $X\setminus U$, then there is no neighbourhood of $f(x)$ such that $f^{-1}(V)\subset U$. In particular, $B(X,\R)\subseteq B(X^*,\R)$ (the other inclusion is trivial because the topology of $X^*$ was finer).

To make the space $X^\ast$ Tychonoff, we just take the quotient by the equivalence relation $$ x\sim y \Longleftrightarrow \FN (x) = \FN (y)$$

(the so-called Kolmogorov quotient), and denote by $Y$ the quotient space. Now, a bounded real-valued function $f$ on $Y$ will be continuous if and only if $\mu\circ f\in B(X^*,\R)=B(X,\R)$, i.e. $B(Y,\R)$ and $B(X,\R)$ are isomorphic (and the isomorphism is precisely the quotient map $\mu$, whose inverse is such that to each $f\in B(X,\R)$ it assigns the map $[x]\mapsto f(x)$).

Please, can you verify my proof? Also, any suggestion/comment/whatever (including alternative proofs) to make the proof simpler and better will be grateful. Thanks.


As Henno Brandsma suggested me, I read the proof of this theorem in the book Rings of continuous functions, of Leonard Gillman and Meyer Jersion. However, the proof is not what I expected. Instead of weakening the topology first and then taking the quotient, first the authors take the quotient with a suitable equivalence relation (two points are the equivalent if and only if their values under the set of all continuous real-valued functions are the same). But they don't endow this space with the quotient topology, but with respect to the ''projections'' of the set of continuous real-valued functions onto the quotient. As a consequence, they point out that the set-theory quotient map need not to be a topological quotient map.

I understand the proof, but I can't see why it is equivalent to mine or the one given by Henno Brandsma because, in our case, the the quotient map is actually a quotient map in the topological sense. I would like some clarification to this.

  • $\begingroup$ What is \FN meant to mean? $\endgroup$ – Henno Brandsma Mar 14 at 21:04
  • $\begingroup$ Oh, I'm sorry, I meant the set of all neighbourhoods. I'll correct if right now. $\endgroup$ – Dog_69 Mar 14 at 21:10

First of all, in your proof attempt it's not clear that taking the Kolmogorov quotient preserves the set of bounded continuous functions, and if it does, it requires a proof.

To the idea as Gilman and Jerison do it in Rings of Continuous Functions, the go-to text in this area:

Let $\mathcal{F}=B(X,\mathbf{R})$ be the original set of bounded continuous functions on $X$. Then define an equivalence relation on $X$ by:

$$x \sim y \text{ iff } \forall f\in \mathcal{F}: f(x)=f(y)$$

and define the standard map $q: X \to X/{\sim}$ that sends $x$ to its class $[x]$, and note that to each $f \in \mathcal{F}$ we can associate a unique $\hat{f}:X/{\sim} \to \mathbf{R}$ such that $\hat{f} \circ q = f$: just define $\hat{f}([x])=f(x)$ and note that by definition of $\sim$ this does not depend on the representative $x$ of the class $[x]$. This is just set theory, no topology at all. We'll call $X/{\sim}$: $Y$ from now on.

Then we define $\mathcal{F}'=\{\hat{f}: f \in \mathcal{F}\} \subseteq \mathbf{R}^Y$ and we give $Y$ the weak topology w.r.t. $\mathcal{F}'$.

A standard characterisation of continuity of spaces in a weak topology gives us first that $q:X \to Y$ is continuous and second that $B(Y,\mathbf{R})= \mathcal{F'}$ and the map $f \to \hat{f}$ is a ring isomorphism between that set and $B(X,\mathbf{R})=\mathcal{F}$. Also $Y$ is Tychonoff by Willard 14.12, without any more work.

So no quotient topologies, just a weak topology on a "quotient set".

  • $\begingroup$ I meant Kolmogorv quotient. Thanks again. And yes, in fact I proved a similar result you referred to: If I'm not wrong, a space is completely normal if and only if the topology on $X$ is the initial with respect to the set of all bounded real-valued functions (it is proven in Willard). From this, it is easy to achieve what you say. Your proof is much shorter and clearer (even more if I had already proven that result). I thought that, but I try following the hint. $\endgroup$ – Dog_69 Mar 14 at 21:18
  • $\begingroup$ And what do you think about my proof? $\endgroup$ – Dog_69 Mar 14 at 21:22
  • $\begingroup$ @Dog_69 It's not yet a complete proof, it needs some filling in of details. But this is the idea. This is the way Gilman and Jerrison (Rings of Continuous functions) show it in one of the early chapters, where they show they can reduce the study of $C(X)$ to completely normal spaces. Check it out, it's a very good book on this branch of topology. I think your proof does not yet work as it stands now. $\endgroup$ – Henno Brandsma Mar 14 at 21:29
  • $\begingroup$ I'll do it. But, meanwhile, could you sketch briefly what it remains to show, please? Because I tought it was completed. $\endgroup$ – Dog_69 Mar 14 at 21:30
  • $\begingroup$ I have been reading the proof of this theorem and it's different for I expected. The authors consider first the quotient set and they endow it with the initialtopology with respect to $\mu(C(X))$. Furthermore, they say the map $\mu$ (which is the quotient map from a set-theoretical point of view) needs not to be a quotient map. But in your proof and in mine (even if it is incomplete), $\mu$ is. Could you explain this a little, please? $\endgroup$ – Dog_69 Mar 15 at 19:27

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