Let $\Bbb R$ be endowed with the standard topology. Clearly, we can trivially represent $$ \Bbb R \cong \Bbb R^0\times \Bbb R \tag{1} $$ and also, there is not such topological space $X$ that $\Bbb R \cong X^2$. Thus, I wonder whether necessary $$ \Bbb R \cong X\times Y $$ implies that $X = \Bbb R^0$ and $Y = \Bbb R$ (or vice-versa). With $\Bbb R^0$ I mean the singleton topological space $1$.
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What does the notation $\mathbb R^0$ mean? – Rudy the Reindeer Apr 02 '13 at 08:39
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1The projections onto $X$ and $Y$ must mean that both are connected. So the argument seems to reduce to the one showing $\Bbb R$ is not $X^2$. – Asaf Karagila Apr 02 '13 at 08:41
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1@MattN.: sorry, that's just a singleton – SBF Apr 02 '13 at 08:42
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2t.b.'s answer to your previous question would seem to work mutatis mutandis in this case as well. – user642796 Apr 02 '13 at 08:45
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@ArthurFischer: you're right indeed, if you post it as an answer, I'll accept it – SBF Apr 02 '13 at 08:49
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t.b.'s answer to your previous question would seem to work mutatis mutandis in this case as well:
If $\mathbb{R} = X \times Y$, then both $X$ and $Y$ are path connected. If both $X$ and $Y$ have cardinality $> 1$ it follows that removing any single point from $X \times Y$ yields a path connected space, but the real line clearly does not have this property.
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