algebra-geometry duality

Question

For topological spaces $S$ and $T$, denote by $C(S)$ and $C(T)$ the corresponding algebras of continuous real-valued functions. What are the necessary conditions that we need to impose on $S$ and $T$ so that the following is true: $C(S)\cong C(T)$ implies $S \cong T$ ?

Answer 1

For this answer I shall assume all spaces are Tychonoff since when dealing with rings of continuous functions, it suffices to deal with Tychonoff spaces. The property that you need is realcompactness. In this answer, I will describe realcompactness, but for more information on realcompactness, one should consult the book Rings of Continuous Functions by Gillman and Jerison.

A space is said to be realcompact if it can be embedded as a closed subspace of some product $\mathbb{R}^{I}$ for some set $I$. Like compactness, there is a notion of a realcompactification. If $X$ is a space, then a set of the form $f^{-1}[\{0\}]$ is called a zero set. The complement of a zero set is known as a cozero set. If $X$ is a completely regular space, then the Hewitt-realcompactification $\upsilon X$ of $X$ is the intersection of all cozero sets $U\subseteq\beta X$ with $X\subseteq U$. A completely regular space is realcompact if it is equal to its realcompactification. It is well known that if $X,Y$ are completely regular spaces, then $C(X)\simeq C(Y)$ if and only if $\upsilon X\simeq\upsilon Y$. In particular, if $X,Y$ are realcompact spaces and $C(X)\simeq C(Y)$, then $X\simeq Y$.

The spaces that people deal with in mathematics are usually realcompact. A space of cardinality below the first measurable cardinal is realcompact if and only if it can be given a compatible complete uniformity. Measurable cardinals are extremely large, and their existence cannot be proven using the standard axioms of set theory, so restricting our attention to sets of non-measurable cardinality is not much of a restriction.

Every paracompact space whose cardinality is below the first measurable cardinal is realcompact since paracompact spaces can always be given a complete uniformity. Furthermore, every Lindelof space is realcompact.

On the other hand, the space $\omega_{1}$ with the ordinal topology is not realcompact. Furthermore, in this answer I give an example of a space $X$ with $|X|<|\upsilon X|$.

Realcompactness can also be characterized in terms of the Baire $\sigma$-algebras. If $X$ is a completely regular space, then the Baire $\sigma$-algebra on $X$ is the $\sigma$-algebra on $X$ generated by the zero sets. A completely regular space is realcompact if and only if every $\sigma$-complete ultrafilter on the Baire $\sigma$-algebra is principal. In fact, the Hewitt realcompactification of completely regular space can be put into a one-to-one correspondence with the $\sigma$-complete ultrafilters on the Baire $\sigma$-algebra as follows. Suppose $X$ is a completely regular space, and $\mu$ is a probability measure on the Baire $\sigma$-algebra of $X$ where $\mu(R)=0$ or $\mu(R)=1$ for each $R$ in the Baire $\sigma$-algebra. Then the mapping $C(\beta X)\rightarrow\mathbb{R},f\mapsto\int f|_{X}d\mu$ is a continuous linear function, so there is a point $x_{\mu}\in\beta X$ where $\int f|_{X}d\mu=f(x)$ for all $f\in C(\beta X)$. Furthermore, the mapping $\mu\mapsto x_{\mu}$ is a one-to-one correspondence between the measures corresponding to $\sigma$-complete ultrafilters on the Baire $\sigma$-algebra of $X$ and the Hewitt realcompactification $\upsilon X$ of $X$.