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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.

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    The hint is making things harder than they need to be. Here is an easier hint $3 \times 4 = 2 \times 6$.2012-11-12

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  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).