Topological space which is homeomorphic to its square

Question

If $X$ is an infinite set, we can consider it as a topological space with the discrete topology, and it has the property that $X$ and $X\times X$ are homeomorphic. Does the property that $X$ is homeomorphic to $X\times X$ imply that $X$ has the discrete topology? Or are there other examples of spaces that satisfy this?

Answer 1

Spaces like that are often $0$ dimensional (base of clopen sets): many standard spaces are like that:

$X=\mathbb{Q}$ in the usual topology obeys $X^2\simeq X$, as both are countable metrisable spaces without isolated points.

$X=\mathbb{P}$ (the irrationals in the usual topology) likewise, but now because they both are separable, completely metrisable, zero-dimensional, nowhere locally compact spaces. (Or because maybe you know that $P \simeq \mathbb{N}^\mathbb{N}$)

$X=C$, the Cantor set, because $C^2$ and $C$ are both compact zero-dimensional metric spaces without isolated points.

$X=C\setminus \{p\}$ (for $p \in C$) as well, as this $X$ is the unique (up to homeomorphism) zero-dimensional crowded (no isolated points) metric space that is not compact, but is locally compact.

Many infinite-dimensional spaces from analysis: $\ell^p$ for all $p$, and $\ell^\infty$ as well and $c_0$ and $c$ (all basically topologically the same example $\Bbb R^\mathbb{N}$, except $\ell^\infty$ which is not separable).

Ugly spaces like any infinite space in the indiscrete (trivial) topology you forgot to mention.

The space can be metric and one-dimensional too: Erdős space is such a space, the irrational variant is even completely metrisable.

Answer 2

The Cantor space $C = \{0,1\}^\mathbb{N}$ clearly also has this property, but its topology isn't discrete.

Answer 3

For arbitrary separable infinite-dimensional Hilbert space $H$ you have a linear (even unitary) homeomorphism $H \sim H \times H$, which can be built using any countable orthonormal base in $H$.

Answer 4

Let $Z$ be any space and $A$ any infinite set. Then $X = \prod_{\alpha \in A} Z_\alpha$, where $Z_\alpha = Z$, has the property $X \times X \approx X$.