If a topological space is completely regular, can we assume the seperating function to be bounded?

Question

I am working through Conways functional analysis p.137.

A completely regular space $X$ is a topological space in which for each point $x$ and closed set $A$ disjoint to $\{x\}$ there exists a continuous function $f \in C(X)$ with $f(x) = 1$ and $f(y) = 0$ for all $y \in A$.

The first lines of the proof of proposition 6.1 suggest that $X$ is at least T1 if not T2. Let's suppose $X$ is T2, then for each pair of points $x \neq y$ we find such a seperating function.

Can we w.l.o.g. suppose this function to be bounded?

My idea is to work with something like cutoff functions which live on two disjoint open neighbourhoods (using T2) $O_x$ and $O_y$ of $x$ and $y$ but I am not sure why something like this should exist.

Answer 1

Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be the function defined by $g(x) = 0$ if $x \le 0$ , $g(x) = x$ for $x \in [0,1]$, $g(x) = 1$ for $x \ge 1$. This is continuous, and if $f$ is any separating function, $g \circ f$ is a bounded one.

Answer 2

Yes. Just define $g(x)=\max(0,\min(1,f(x)).$

Answer 3

Yes, you can suppose the function to be bounded, since you have homeomorphisms from $\mathbb R$ to intervals in $\mathbb R$, for example. So, you can construct a homeomorphism $F: \mathbb R\to(-1,2)$ such that $F(1)=1$ and $F(0)=0$ (there are infinitely many such functions).

Then, take your function $f$ and look at $g(x)$ defined as $g(x)=F(f(x))$.