On exercise 14H from Willard's 'General Topology' book

Question

$ \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 $[x]\mapsto f(x)$).

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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.

EDIT:

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.

Answer 1

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".