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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 $X=\Bbb R$?

Batominovski
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user275913
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3 Answers3

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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 $X$.

David C. Ullrich
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    It is more generally possible for Tychonoff spaces, because the complements of zero sets form a base for the topology. – Niels J. Diepeveen Oct 01 '15 at 16:53
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    Depending on what we mean by "recovering" the topology from the space of continuous functions. Yes, the sense in which you're using the term seems likely what the OP meant. But (as you doubtless know) the cool thing about the case of a compact Hausdorff space is we can recover the topology of $X$ from $C(X)$ in a much stronger sense: If $C(X_1)$ and $C(X_2)$ are just isomorphic as rings then $X_1$ and $X_2$ are homeomorphic, – David C. Ullrich Oct 01 '15 at 17:15
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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 $T_1$.

Community
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Brian M. Scott
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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.

Hetebrij
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