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Ok so, I was given given a proof of this by my lecturer, but I'm not understanding it fully,, and I was wondering if this suffices as a proof.

Proof:
It is enough to show that $(0,1)$ is uncountable.
Let $x \in (0,1)$ and write $x$ in it's decimal expansion. i.e. $x= 0.a_1a_2a_3...$, for $a_i \in \{0,1,2....9\}$ for $i \in \Bbb N$
Next assume that $(0,1)$ is countable, that is $(0,1) =\{x_1, x_2, x_3,... \}$. We write
$x_1 = 0.a_{11}a_{12}a_{13}...$
$x_2 = 0.a_{21}a_{22}a_{23}...$

This is where I differ from my lecturer. He says:
let $y= b_1b_2b_3...$, where
$b_i = 2$ if $a_{ii} \neq 2$ and
$b_i = 3$ if $a_{ii} = 2$ for $i \in \Bbb N$
Then $y \in (0,1)$, but $y \neq x_i$, for all $i \in \Bbb N$
This contradiction proves that $(0,1)$ is uncountable.

What I'm saying is:
Let $b_i = a_{ii} + 1$.
Then $y$ will differ from $x_1$ in the first decimal position, from $x_2$ in the second decimal position,... and from $x_n$ in the the $n^{th}$ decimal position. Hence, the list $\{ x_1, x_2,... \}$ can never be an exhaustive list and so $(0,1)$ is uncountable.

Just want to know what you think. Thanks.

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    $\begingroup$ What happens when $a_{ii} = 9$? Though, in general, the proof could just say $b_i \neq a_{ii}$. $\endgroup$ – Karolis Juodelė May 28 '14 at 15:40
  • $\begingroup$ @KarolisJuodelė Ok, so if I say that then, the proof is acceptable? $\endgroup$ – Crockett May 28 '14 at 15:42
  • $\begingroup$ $\Bbb R$ is a complete metric space, hence uncountable by Baire's theorem. $\endgroup$ – Pedro Tamaroff May 28 '14 at 16:40
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Your proof doesn't work in case $a_{ii} = 9$. Perhaps you could include $b_i = 0$ if $a_{ii} = 9$.

The reason the instructor used $2$ and $3$ is to avoid the problem of having an infinite string of $9$'s.

To guarantee that each $x$ has a unique representation, you can require that each sequence $\{a_{ij}\}$ does not end in an infinite string of $9$'s. If it did, you could represent the number $x$ with a finite decimal representation. The problem is that the number you construct using the $b_i$ could end in an infinite string of $9$'s which will have a different representation from every other number in the list, but could in fact equal one of those numbers. For instance, you might construct $0.4999999\ldots$ which already appeared in the list as $0.5000000\ldots$. Avoiding $9$'s altogether circumvents this problem.

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