Homeomorphism of product of topological spaces

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?

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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

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).