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Is there a bijection between $\mathbb{Z}\times\mathbb{Z}\times\dots$ for countably infinitely many $\mathbb{Z}$'s and $\mathbb{R}$? That is, is $\mathbb{Z}\times\mathbb{Z}\times\dots$, repeated countably infinitely many times, uncountable?

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The question after "that is" is not equivalent to the question in the first sentence; not all uncountable sets have the cardinality of $\mathbb R$. – joriki Dec 6 '12 at 9:33
What other sets do you know to be uncountable? What other sets do you know to be "bijectable" with $\mathbb{R}$? – arjafi Dec 6 '12 at 9:36

First note that as joriki commented there are many uncountable sets which are not of the same size as the real numbers. There might be uncountable sets which are strictly smaller than the real numbers.

To both your questions, however, the answer is yes. Note that:

$$\mathbb{Z\times Z\times Z\times\dots = Z^N}\\\mathbb{R\times R\times R\times\dots = R^N}$$

Now we have this: $$|\mathbb R|=2^{|\mathbb N|}\leq|\mathbb Z|^{|\mathbb N|}\leq|\mathbb R|^{|\mathbb N|}=2^{|\mathbb{N\times N}|}=2^{|\mathbb N|}=|\mathbb R|$$

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Can we consider any other countable set instead of $\mathbb Z$? Is the problem remained valid? Thanks Asaf. I am a disciple of you here. – Babak S. Dec 6 '12 at 9:56
Obviously we can. Any set below the cardinality of $\mathbb R$, as a matter of fact. – Asaf Karagila Dec 6 '12 at 10:01

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