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 Post subject: AnalysisPosted: Wed Feb 07, 2018 4:48 pm

Joined: Mon Feb 05, 2018 4:22 am
Posts: 9
prove that (0,1) is uncountable

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 Post subject: Re: AnalysisPosted: Wed Feb 07, 2018 7:49 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 838
Location: Larisa
Asis ghosh wrote:
prove that $(0,1)$ is uncountable

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.

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 Post subject: Re: AnalysisPosted: Wed Feb 07, 2018 7:50 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 838
Location: Larisa
Also worth noting Cantor's diagonal argument

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