Homeomorphism between $X$ and $X \times X$ with different topologies I have to solve this problem subdivided in three parts:


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*Let $X$ be an infinite set with the discrete topology. Prove that $X$ is homeomorphic to $X\times X$.

*If $X$ has the trivial topology, is it still true that $X$ is homeomorphic to $X\times X$? 

*Find another example of a space $X$ homeomorphic to $X\times X$ in which $X$ has not the trivial topology neither the discrete one.
I tried to solve it in this way:


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*Let's define the homeomorphism $$\phi: X \to X\times X \\ x \mapsto (x,x)$$ In this way every open set in $X\times X$ is mapped to an open set through $\phi^{-1}$ because $X$ has the discrete topology in which every subsset is open

*I would use the previous homeomorphism because the unique open set which contains $(x,x)$ is $X\times X$ and $\phi^{-1}$ maps $X\times X$ in $X$ which is the unique open set which contains $X$.
Is my reasoning right?


*For this question I have no idea how to proceed. I know that I cannot use the euclidean topology because, for example, $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^2$. Have you any ideas?

 A: 
  
*
  
*Let $X$ be an infinite set with the discrete topology. Prove that $X$ is homeomorphic to $X\times X$.
  

Unfortunately your $\phi: x\mapsto (x,x)$ is not a homeomorphism. The $\phi^{-1}$ simply doesn't exist because $\phi$ is not onto. For example if $a\neq b$ then there is no $x$ such that $\phi(x)=(a,b)$.
To solve that problem you need some set theoretic background, namely:

Lemma 1. Let $X$ be an infinite set. Then there exists a bijection $f:X\to X\times X$.

You can find a proof for that in some set theoretic books. That is not trivial at all. Note that it doesn't have to be a homeomorphism (and indeed, in some cases it cannot be, for example when $X=\mathbb{R}$ as you've noted yourself). There's no notion of topology in the lemma above.
Now we take another important fact:

Lemma 2. Let $X,Y$ be two discrete spaces. Then any function $f:X\to Y$ is continuous.

I leave as an exercise. Note that Lemma 2 applies to trivial topologies as well. Combining these two we have:

Corollary. Any infinite discrete space $X$ is homeomorphic to $X\times X$.




  
*If $X$ has the trivial topology, is it still true that $X$ is homeomorphic to $X\times X$? 
  

For the trivial topology note that $X$ has exactly two open subsets: $\emptyset, X$. How many open subsets $X\times X$ has? Since every open subset in product topology is of the form $\bigcup U\times V$ then we can easily verify that a nonempty combination of $\emptyset$ and $X$ can only give $X\times X$. Meaning topology on $X\times X$ is trivial. Combining it with Lemma 1 easily shows that they have to be homeomorphic.



  
*Find another example of a space $X$ homeomorphic to $X\times X$ in which $X$ has not the trivial topology neither the discrete one.
  

One such example would be rationals $\mathbb{Q}$ with Euclidean topology. This is due to

Theorem (Sierpiński). Every countable, metrizable space without isolated points is homeomorphic to $\mathbb{Q}$.

