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Recall that a countable set $S$ implies that there exists a bijection $\mathbb{N} \to S.$

Now, I consider $(0,1).$ I want to prove by contradiction that $(0,1)$ is not countable.

First, I assume the contrary that there exists a bijection $f,$ and I can find an element in $S,$ but not in the range of $f.$ But I can't find such element. How can you construct such $f$?

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    See http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument2012-03-22
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    http://math.stackexchange.com/questions/18969/why-are-the-reals-uncountable2012-03-22
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    Does it really deserve two (well, while I was typing the 3rd appeared) downvotes without even stating the reason? Giving 1 upvote just to cancel the effect.2012-03-22
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    @jason: And another upvote. But for the sake of hopelessly not with it people like me, would it be possible to use "I" and not "i"?2012-03-22
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    If this problem arises in the context of a course, there may already be a theorem in your course that comes close to what is needed here. I do not imagine that you would be expected to discover the diagonal argument.2012-03-22
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    One more method: Try drawing a circle centred at (1/2,1/2) with radius 1/2. Every point in (0,1) can be mapped to the circle which in turn can be projected to the real line from $(-\infty,\infty)$ i.e. $\mathbb{R}$, which is uncountable.2012-03-22
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    I down-voted, since Googling "Proving (0,1) is not countable" returns pages and pages of proofs2012-03-22
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    @Inquest: Yes, but how do you prove that $\Bbb{R}$ is uncountable then?2013-12-10

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