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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 $f$?

jason
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    See http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument – Ross Millikan Mar 22 '12 at 17:40
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    http://math.stackexchange.com/questions/18969/why-are-the-reals-uncountable – Asaf Karagila Mar 22 '12 at 17:43
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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. – SBF Mar 22 '12 at 17:44
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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"? – André Nicolas Mar 22 '12 at 17:50
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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. – André Nicolas Mar 22 '12 at 17:57
  • 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. – Inquest Mar 22 '12 at 18:22
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    I down-voted, since Googling "Proving (0,1) is not countable" returns pages and pages of proofs – William Mar 22 '12 at 19:16
  • @Inquest: Yes, but how do you prove that $\Bbb{R}$ is uncountable then? – Beni Bogosel Dec 10 '13 at 23:03

3 Answers3

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Assume that $(0,1)$ is countable. Then you can write $[0,1]=(x_n)_{n \geq 0}$. Do the following steps:

  • split $[0,1]$ into three equal parts $[0,1/3],[1/3,2/3],[2/3,1]$. Then $x_0$ is not in one of the given intervals. Denote it by $[a_0,b_0]$.
  • split $[a_0,b_0]$ into three equal parts $I_1,I_2,I_3$. Then $x_1$ is not in one of the given intervals. Denote it by $[a_1,b_1]$.
  • inductively construct an interval $[a_{n+1},b_{n+1}]\subset [a_n,b_n]$ such that $x_{n+1} \notin [a_{n+1},b_{n+1}]$ and $b_{n+1}-a_{n+1}=\frac{1}{3}(b_n-a_n)$.

Since $[a_n,b_n]$ is a decreasing sequence of compact intervals with $b_n-a_n \to 0$ their intersection is a point $C \in [0,1]$. If $[0,1]=(x_n)$ then there exists $m$ such that $x_m=C$. But then $x_m \notin [a_m,b_m]$ and therefore it cannot be in the intersection of all intervals. Contradiction.

Beni Bogosel
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This is the famous Cantor's Diagonal Argument.

The bijection $f$, which we have assumed to exist, can map any positive integer to a value in $(0,1)$ (and since it's a bijection, none of the points in $(0,1)$ are left over). Also, the points in $(0,1)$ can be treated as a number line so that each point on the line is a value between $0$ and $1$ which we can write out as a decimal.

Imagine writing these decimals out in order according to the integer they map to. So perhaps $1 \mapsto 0.5$ and $2 \mapsto 2/3$ and $3 \mapsto 1/\pi$:

 1: 0.50000000...
 2: 0.66666666...
 3: 0.31830988...
 ..etc..

Now read down the diagonal (marked in bold above) and pick a different digit than what you see. For instance, if you see a $5$, use "$6$", if you see anything else use "$5$". That gives us a series of digits... in this case it starts out "$0.655...$".

Clearly, this series of digits is not in the list, since it differs from each item in the list by at least one decimal place. Clearly it is a number in the range $(0,1)$. Since we used $5$ and $6$ it also doesn't have a repeating "$99999...$" or "$00000...$" (which is a way $2$ different sequences of digits could represent the same number). So our assumption that $f$ was a bijection must have been false.

learner
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mcherm
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This question has already been answered, however I will write another approach to the problem using measure theory, for future reading.

Let $\mu^{*}$ the outer measure and suppose that $(0,1)$ is countable, so we have that $$\mu^{*}((0,1))=0$$ But, also since $$\mu^{*}((0,1))=\mathcal{l}(0,1)=1-0=1$$ but $1\not=0$. Therefore, $(0,1)$ is not countable.