About homeomorphism of two sets

Question

$X$ is a topological space of infinite cardinality which is homeomorphic to to $X \times X$.Then

a) $X$ is not connected

b) $X$ is not compact

c) $X$ is not homeomorphic to a subset of $\mathbb{R}$

d) None of the above

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the answer is d and i don't know how $\mathbb{N}$ is a counterexample of a,b and c. Now under discrete topology it is homeomorphic to $\mathbb{N}$ since any bijective function from $\mathbb{N}$ to $\mathbb{N}\times \mathbb{N}$ is continuous and inverse also continuous under discrete topology (inverse map of any open subset is open). But how consider a and b???

Answer 1

As you said, $\mathbb{N}$ is a counterexample for c).

Let $Q = \prod_{n=1}^\infty I_n$ where $I_n = [0,1]$. This space is known as the Hilbert cube. It is compact and connected and satisfies $Q \times Q \approx Q$.

Note that you cannot find a single counterexample $X$ for a) and c). Assume $X$ would be such a space. Then $X$ must be homeomorphic to a connected subset of $\mathbb{R}$, that is, homeomorphic to an interval $J$. But $J \times J$ is not homeomorphic to $J$.

$Q$ is a counterexample for both a) and b), and the Cantor set (see Henno Brandsma's answer) is a counterexample for both b) and c).

Answer 2

It's d);

a) is refuted by $X = \mathbb{R}^\omega$. b) and c) by the Cantor set.