# Proving $(0,1)$ is not countable

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 $f$?

• Commented Mar 22, 2012 at 17:40
• math.stackexchange.com/questions/18969/… Commented Mar 22, 2012 at 17:43
• 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
Commented Mar 22, 2012 at 17:44
• @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"? Commented Mar 22, 2012 at 17:50
• I down-voted, since Googling "Proving (0,1) is not countable" returns pages and pages of proofs Commented Mar 22, 2012 at 19:16

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.

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.

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.