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I am not getting a fact whether if $X\,\times\,Y$ is homeomorphic to $Y\,\times\,Y$ then is it true that $X$ is homeomorphic to $Y$? Actually I want to know the result for say $X=[0,1]$ and $Y=[0,1)$.

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You have asked 4 questions on a trot. Could you let us know what these questions are for? If it is homework, kindly tag them so. –  user17762 Feb 12 '11 at 19:58
    
@Sivaram Ambikasaran: no it is not a homework question,I am solving a question paper of an entrance examination.if you know the answer please help. –  elinor Feb 12 '11 at 20:18
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You can deduce that the if ... then statement is false from your example. Clearly in that example $X$ isn't homeomorphic to $Y$, so you just have to show that $X \times Y$ is homeomorphic to $Y \times Y$ (which it is). –  joriki Feb 12 '11 at 20:40

2 Answers 2

I could be wrong, but apparently no, that is not true. For example, $\mathbb{R}^{\omega} \times \mathbb{R}^{\omega}$ seems to be homeomorphic to $\mathbb{R} \times \mathbb{R}^{\omega}$, both homeomorphic to $\mathbb{R}^{\omega}$ but clearly, $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^{\omega}$. (Here $\mathbb{R}^{\omega}$ is the countable product of $\mathbb{R}$ in the product topology.)

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A slightly simpler example would be that $\{1\}\times\mathbb{N}$ is homeomorphic to $\mathbb{N}\times\mathbb{N}$ but $\{1\}$ is not homeomorphic to $\mathbb{N}$ :-) –  joriki Feb 12 '11 at 20:36

A more advanced example was given by Bing (reference) and shows that the problem can be far more intrincate than a cardinality argument (whether of the spaces or the indexing set).

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