Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am stuck on a problem about homeomorphic topological spaces and can't go on... So the problem is:

If we have that $X_{1} \times X_{2}\simeq Y_{1} \times Y_{2}$ (the product of topological spaces X1 and X2 is homeomorphic to the product of Y1 and Y2), to prove is that the components might not be homeomorphic. There is a hint: Let's consider $X_{1}=X_{2}=Y_{1}=\mathbb{N}$ and $Y_{2}= \left \{ p \right \}$ with the discrete topology.

Okay, we know from the definition that in discrete topology all sets are open, this means that {p} is open too...I don't understand how to prove that the components might be not homeomorphic...can somebody explain me?

Thanks in advance.

share|cite|improve this question
The point of the hint is that you can prove $\mathbb{N} \times \mathbb{N}$ is homeomorphic to $\mathbb{N} \times \{p\}$. Now, is $Y_1 = \mathbb{N}$ homeomorphic to $Y_2 = \{p\}$ or not? – Shaun Ault Nov 12 '12 at 12:38
As an aside, another example is $\mathbb{R}^n \times \mathbb{R}^k \simeq \mathbb{R}^p \times \mathbb{R}^q$ with $n + k = p = q$, for different choices of $n,k,p,q$. – Christopher A. Wong Nov 12 '12 at 12:47
It should be not homeomorphic to {p}, but i don't know how to prove it :( how to find a function with its inverse not continious? – Lullaby Nov 12 '12 at 12:48
A homeomorphism is not only continuous and open but it is a bijection. Obviously, there is no bijection between $\{p\}$ and $\Bbb N$. – Berci Nov 12 '12 at 13:18
The hint is making things harder than they need to be. Here is an easier hint $3 \times 4 = 2 \times 6$. – Rob Arthan Nov 12 '12 at 13:31
up vote 2 down vote accepted
  1. Since no map $f:\{p\}\to\Bbb N$ can be surjective (other elements than $f(p)$ are not the images of anybody along $f$), there cannot be a homeomorphism between them as topological spaces (as a homeomorphism must be bijective).

  2. On the other hand, $\Bbb N\times \Bbb N \simeq \Bbb N$ as topological spaces (because both are discrete and has the same cardinality: countably infinite).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.