$X$ is homeomorphic to $X\times X$ (TIFR GS $2014$) Question is :
Suppose $X$ is a topological space of infinite cardinality which is homeomorphic to $X\times X$. Then which of the following is true:


*

*$X$ is not connected.

*$X$ is not compact

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

*None of the above.


I guess first two options are false.
We do have possibility that product of two connected spaces is connected.
So, $X\times X$ is connected if $X$ is connected. So I guess there is no problem.
We do have possibility that product of two compact spaces is compact.
So, $X\times X$ is compact if $X$ is compact. So I guess there is no problem.
I understand that this is not the proof to exclude first two options but I guess the chance is more for them to be false.
So, only thing I have problem with is third option.
I could do nothing for that third option..
I would be thankful if some one can help me out to clear this.
Thank you :)
 A: The Cantor set is a counter-example to the second and third statement. Note that the Cantor set is homeomorphic to $\{0,1\}^{\mathbb N}$, hence it is homeomorphic to the product with itself.
An infinite set with the smallest topology (exactly two open sets) is a counter-example to the first statement. Martini gives a better counter-example in a comment.
A: Take the Cantor set. This gives a counterexample for B and C. For A, as Mike said in the comments, take the set $[0,1]^\infty$.
A: Fix your favorite compact connected space $X$ - then $X^\omega$ is compact, connected, and homeomorphic to its own square.
A: 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 $J$.
