This question arises trying to solve exercise 14H of Willard's General Topology book. That exercise asks us to proof that given any topological space, there exists another space which is Tychonoff ($T_{3\frac{1}{2}}$) such the ring of bounded real-valued continuous functions, $B(X,\mathbb R)$,is isomorphic to the first. The hint given in the exercise suggests us to weaken the topology and then identify points. And that is precisely what I did (you can see my try here): I defined a new space in which open sets where those such that:
- They were open in the original space and,
- For each point belonging to this set, there exists a continuous function separating the point and the complementary.
Now, I claim that this operation remains the ring $B(X,\mathbb R)$ unchanged. Below I post my reasoning. I ask you if it is correct (I guess it is not). However, I have tried to find some counterexample, but I have not found any. The one I tried was in the following space:
$\mathbb R$ with the following topology:
If $x\neq 0$, the neighbourhoods of $x$ are as usual.
If $x=0$, then the neighbourhoods are of the form $U\setminus (U\cap\{1/n\}_n)$, where $U$ is any standard neighbourhood.
This space is presented in example 14.2 of Willards book. The author proves it is Hausdorff but not $T_3$; hence neither Tychonoff. However, I did not find any function belonging to the same ring but not to the second.
Here I post my argument:
I 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 find an 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
$$\tilde f(y)=\begin{cases} \frac{1}{\epsilon} f(y) & \mbox{if } y\in f^{−1}((−ϵ,ϵ))\\ 1, & \mbox{otherwise} \end{cases} $$
is a continuous function that separates $x$ and $X\setminus U$. It is clear that $\tilde f$ does, and to see that it 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
$$ \tilde f(x_i)=\begin{cases} \frac{1}{\epsilon} f(x_i), & \mbox{if } x_i\in f^{−1}((−ϵ,ϵ))\\ 1, & \mbox{otherwise} \end{cases} $$
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. Hence I conclude that, 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, $B(X,\mathbb R)\subseteq B(X^*,\mathbb R)$ (the other inclusion is trivial because the topology of $X^*$ was finer).
Thanks
I would like to mention that Henno Brandsma answered my previous question as it is done in the book Rings of Continuous Functions, written by Gilman and Jerison. However, given the hint, it seems to me Willard is thinking on an alternative proof, and that is what I am focused in.
Your construction does indeed leave $B(X,\mathbb{R})$ unchanged.
It is obvious that removing open sets will not add any functions to $B(X,\mathbb{R})$. The only circumstance in which removing an open set will remove functions from $B(X,\mathbb{R})$ is if the open set which is being removed is the preimage of an open subset of the reals under a function $f \in B(X,\mathbb{R})$.
If some open set $U$ is removed from $X$, it is because there exists a point $x \in U$ which is not separable from $X \setminus U$, i.e. if for any $f\in B(X,\mathbb{R})$ there exists a $y\in X \setminus U$ such that $f(x)=f(y)$. Then the preimage of any subset of the reals either contains $x$ and hence contains a point $y$ outside $U$ or does not contain $x$ and hence does not contain the whole of $U$. Therefore $U$ cannot be the preimage of an open subset of $\mathbb{R}$.