An exercise of Rudin's Real and Complex Analysis says:
Is it true that every compact subset of $\mathbf{R}^1$ is the support of a continuous function? If not, can you describe the class of all compact sets in $\mathbf{R}^1$ which are supports of continuous functions? Is your description valid in other topological spaces?
I proved that, in every metric space, a set (not necessarily compact) $A$ is the support of a continuous function if and only if it is the closure of an open set.
However, I don't think that this characterization holds for every topological space.
So, in order to find a counter-example, I tried to use the discrete and the trivial topologies. However, the discrete topology is metrizable and this characterization holds for topological spaces with the trivial topology.
Is this result true for any topological space? If not, what would be a counter-example?
Best Answer
Here is a counter-example (compact and Hausdorff).
I will use ordinal topology (see e.g. first uncountable ordinal and Willard's General topology). Here's a crash-course:
The first uncountable ordinal is an uncountable totally ordered set $\omega_1$ with the property that all of its downsets $\left\{x\in\omega_1:x<\alpha\right\}$ (where $\alpha\in\omega_1$) are countable. Such a set exists and is unique up to order isomorphism.
The second uncountable ordinal is the set $\Omega=\omega_1\sqcup\{\omega_1\}$, the union of $\omega_1$ with a new point. It is also an ordered set with $\omega_1=\max\Omega$. With the order topology, $\Omega$ is a compact Hausdorff, non-second countable space. Every continuous function from $\Omega$ to any first-countable space is constant on a neighbourhood of $\omega_1$.
Let $X_1$ and $X_2$ be two copies of the second countable ordinal with maxima $x_1$ and $x_2$, respectively, and let $X$ be the space obtained by gluing $x_1$ and $x_2$ (i.e., the quotient of the disjoint union $\Omega\times\left\{1,2\right\}$ by identifying $(\omega_1,1)\sim(\omega_1,2)$).
Call the image of $x_i$ in $X$ just $x_0$. Then $X$ contains $X_1$ as a regular closed subset. the interior of $X_1$ is just $\left\{x\in X_1:x<x_0\right\}$, and $X_1$ is the closure of this set. Similarly for $X_2$.
However, $X_1$ is not the support of any continuous function $f\colon X\to \mathbb{R}$. Indeed, if $X_1$ were the support of $f$, then $f=0$ on $X\setminus X_1$, which is the interior of $X_2$, so $f=0$ on $X_2$ as well. In particular $f(x_0)=0$. However $f$ is constant on a neighbourhood of $x_0$, by general properties of the first uncountable ordinal (and because $\mathbb{R}$ is first countable). So the support of $f$ is strictly smaller than $X_1$, a contradiction.