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Prove that $\not\exists$ a topological space $X$ such that $\mathbb{R}$ is homeomorphic to $X\times X$.

My approach was to prove by contradiction: say $f:\mathbb{R}\rightarrow X\times X$ be such a homeomorphism. Then we look at $g_i:=\pi_i\circ f$ for $i=1,2$. So, $f=g_1\times g_2$. We know both $g_1$ and $g_2$ are open surjective maps. Clearly $g_i$ can't be one-one but I can't advance from here. If we can show that $\exists r_1,r_2\in \mathbb{R}$ such that $g_i(r_1)=g_i(r_2)$, then we are done. Any help? Any algebraic topology answer is most welcome as well.

Partha
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1 Answers1

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General fact:

if $X$ and $Y$ are connected spaces and $A\subsetneq X$ and $B \subsetneq Y$ are proper subspaces of them, then $(X\times Y) \setminus (A\times B)$ is also connected.

Suppose now that $X$ were a space such that $$\Bbb R \simeq X \times X$$

Note that $X$ is connected (as a continuous image of $\Bbb R$) and is uncountable. It follows from the first general fact ( taking $A=\{p\}, B=\{q\}$) that for any $(p,q) \in X \times X$, $X \times X \setminus \{(p,q)\}$ is connected, while for any $x \in \Bbb R$, the space $\Bbb R\setminus \{x\}$ is not connected. This contradicts the supposed homeomorphism ( which would be preserved when removing a point and its image). So no such $X$ can exist.

Henno Brandsma
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