Can we recover a space from its continuous functions?

Question

Let $X$ be a topological space and let $\mathcal{F}(X,\Bbb R)$ be the set of continuous function from $X$ to $\Bbb R$.

Can we recover the topology of $X$ by only the knowledge of $\mathcal{F}(X,\Bbb R)$?

That is, can we determinate whether or not a subset $U$ of $X$ is open only by using $\mathcal{F}(X,\Bbb R)$?

If not, is there an example of two distinct topologies $\mathcal{T_1}$ and $\mathcal{T}_2$ on a set $X$ such that $$\mathcal{F}_{\mathcal{T}_1}(X,\Bbb R)=\mathcal{F}_{\mathcal{T}_2}(X,\Bbb R)\,?$$

I know from this question that we can indeed recover the topology of $X$ if we consider instead continuous functions from $X$ to $\{0,1\}$ with topology $\big\{\emptyset,\{1\},\{0,1\}\big\}$. But what about the case $X=\Bbb R$?

Answer 1

There is a counterexample with $X=\{1,2\}$; since there aren't many topologies on that space you should have no trouble finding two that give the same continuous real-valued functions.

If you know that $X$ is a compact Hausdorff space then you can recover the topology from $\mathcal F(X,\Bbb R)$. Because the space of continuous functions is a Banach algebra, and it turns out that the maximal ideal space is $X$.

Answer 2

Consider $X = \{ 0 , 1\}$ with $\tau_1 = \{ \emptyset , \{1 \}, \{0,1\}\}$ and $\tau_2 = \{ \emptyset , \{0,1\} \}$.
With the first topology we have that a function $f: X \to \mathbb{R}$ is open if the pre-image of an open set is open.
Now suppose we have $f(1) \neq f(0)$, then we can consider $A = B ( f(2), \frac{1}{2} |f(1)-f(2)|) \subset \mathbb{R}$, which is clearly open, however $f^{-1}(A) = \{2 \}$ is not open, so $f$ is not continuous.
If $f(1)=f(0)$ we have $f{-1}(U) =X$ for all open $U \subset X$.

For the second topology, we can consider the same set $A$ if $f(1) \neq f(0)$ to conclude that $f$ has to be constant to be continuous.

So for both topologies we have $\{ f : X \to \mathbb{R} \mid f \textrm{ is constant.}\}$ is the set of continuous functions.

Answer 3

In this answer I gave the construction in Eric van Douwen’s paper A regular space on which every continuous real-valued function is constant, Nieuw Arch. Wisk. $30$ ($1972$), $143$-$145$, which actually gives a ‘machine’ for starting with a $T_3$ space having two points that cannot be separated by a continuous real-valued function and producing from it a $T_3$ space on which all continuous real-valued functions are constant. In this answer I gave a construction of a $T_3$ space with two points that cannot be separated by a continuous real-valued function; the example is due to John Thomas, A Regular Space, Not Completely Regular, The American Mathematical Monthly, Vol. $76$, No. $2$ (Feb., $1969$), pp. $181$-$182$. Quite a few others are known.

Take any space $\langle X,\tau\rangle$ produced by van Douwen’s machine, and let $\tau'$ be the indiscrete topology on $X$; then $\langle X,\tau\rangle$ and $\langle X,\tau'\rangle$ have the same continuous real-valued functions, namely, the constant ones. For that matter, we can take $\tau'$ to be the cofinite topology on $X$, thereby making the space $T_1$.