# Counterexample for the following statement

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.

• What is $X$, the Tychonoff space? What is the first space? Not all spaces are nice enough to be a topological ring. Or do you mean that given any $Y$, there is a Tychonoff space $X$ so that $B(Y,\mathbb R)\cong B(X,\mathbb R)?$ – Thomas Andrews Apr 11 at 15:06
• @ThomasAndrew Yes, I meant exactly what you said at the end of your comment. – Dog_69 Apr 11 at 18:35

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 $$\mathbb{R}$$.
• I can't believe you. Are you really saying me that my idea was fine? Because after seeing the proof of this theorem and talk about it with Henno Brandsma I was sure I was wrong. You will make me very happy if you say me are right. By the way, what do you think about my proof? I understand yours, and it's much simpler and much more elegant. In fact, I'm quite annoyed with mylsef because I knew the result you are using. But, do you think my proof actually proves what I wanted, that any continuous function with respect continuous with respect to $X$ must remain continuous with respect to $X^*$? – Dog_69 Apr 11 at 19:10
• I don't understand this sentence: "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 $f^{-1}(V)\subset U$." . – Angela Richardson Apr 12 at 4:43
• Well, I think that is because it is wrong. The idea is that I have removed some open sets, so maybe some functions which were continuous are continuous no more. But, given a function $f$ which was continuous, I study its continuity in the ''pathological points'', those where $x$ and $X\setminus U$ couldn't be separable (in fact this is my main doubt: is t enough to sudy continuouty only in these points?). So, suppose the answer is not, i.e. the $f$ we have picked is not continous in $x$ (for $x$ a ''pathological point'')... – Dog_69 Apr 12 at 10:19
• ...The sentence you are asking me for is the negation of continuity in $x$. If $f$ isn't continuous, then there exists a neighbourhood $W$ of $f(x)$, such that, for each neighbourhood $V$ of $x$, $V\subseteq f^{-1}(W)$. However, that doesn't mean $f^{-1}(W)$ must be contained in $V$. It is true I an consider $f^{-1}(W)\cap V$, but I guees $$\{a\in\mathbb R: f^{-1}(a)\in f^{-1}(W)\cap V\}$$ needn't to be a neighbourhood of $f(x)$. Hence my claim was wrong. – Dog_69 Apr 12 at 10:20
• There's a typo in my last comment, I meant $V\not\subseteq f^{-1}(W)$. Moreover, I din't mention it but, if I don't fix my proof showing that actually there exists $W$ such that $f^{-1}(W)\subseteq V$ (which I guess may not be true), it is wrong. – Dog_69 Apr 17 at 10:32