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 :)

  • $\begingroup$ $\mathbb N$ is a simple counterexample for the third one. $\endgroup$ – Dejan Govc Dec 10 '13 at 10:24
  • $\begingroup$ $[0,1]^\infty{}{}$ $\endgroup$ – user98602 Dec 12 '15 at 7:06
  • $\begingroup$ @MikeMiller Drat, beat me to it! (As a good further exercise, it's easy to show that you can't get a simultaneous counterexample to $A$ and $C$.) $\endgroup$ – Noah Schweber Dec 12 '15 at 7:08
  • 1
    $\begingroup$ @NoahSchweber: I had the advantage of not using words :) $\endgroup$ – user98602 Dec 12 '15 at 7:08
  • $\begingroup$ Why $\mathbb{Q}$ in the discrete topology? The usual subspace topology, works, but it's not quite trivial to see that $\mathbb{Q} \times \mathbb{Q} \approx \mathbb{Q}$, but $\mathbb{Z}$ in the discrete = subspace topology is a better example (admittedly homeomorphic to the rationals in the discrete topology, but the latter does not directly embed into the reals). The irrationals or the Cantor set also work, etc. $\endgroup$ – Henno Brandsma Dec 12 '15 at 8:37

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.

  • 2
    $\begingroup$ ... of infinite cardinality. But $[0,1]^{\mathbb N}$ works as counterexample to the first and second statement. $\endgroup$ – martini Dec 10 '13 at 9:04
  • $\begingroup$ @martini : about the third statement? :( $\endgroup$ – user87543 Dec 10 '13 at 9:06
  • $\begingroup$ As said by martini, my space must be of infinite cardinality... $\endgroup$ – user87543 Dec 10 '13 at 9:06
  • $\begingroup$ @martini, thanks I had overlooked that, I will correct this. $\endgroup$ – Carsten S Dec 10 '13 at 9:07
  • 1
    $\begingroup$ You are welcome. Cantor space is the first space I think of when I see $X\times X\approx X$. $\endgroup$ – Carsten S Dec 10 '13 at 9:17

Fix your favorite compact connected space $X$ - then $X^\omega$ is compact, connected, and homeomorphic to its own square.


Take the Cantor set. This gives a counterexample for B and C. For A, as Mike said in the comments, take the set $[0,1]^\infty$.


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 $J$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy