# Proving the uncountability of $[a,b]$ and $(a,b)$

I am trying to prove that $[a,b]$ and $(a,b)$ are uncountable for $a,b\in \mathbb{R}$. I looked up Rudin and I am not too inclined to read the chapter on topology, for his proof involves perfect sets. Can anyone please point me to a proof of the above facts without point-set topology?

I am thinking along these lines:

$\mathbb{R}$ is uncountable.If we can show that there exists a bijection between $(a,b)$ ad $\mathbb{R}$ we can prove $(a,b)$ is uncountable.But I am not sure how to construct such a bijection.

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Have you read about Cantor's diagonalization argument to show uncountability of real numbers? –  Rankeya Dec 6 '12 at 3:50
Also, does Rudin really use topology to show uncountability of $(a,b)$ and $[a,b]$? –  Rankeya Dec 6 '12 at 3:52
No, Baby Rudin does not use topology. It's in the topology chapter, and yes perfect sets are in the chapter, but his proof merely involves diagonalization. –  Jebruho Dec 6 '12 at 4:11
–  Martin Sleziak Dec 6 '12 at 7:58

Yes, you need only find a bijection from each interval to the Reals to prove the intervals are uncountable.

There are many bijections from an open interval $(a, b)\to \mathbb{R}$, e.g.

$g(x) = \cot\left(\large\frac{\pi}{2}x\right)$ is a bijection $g: (0, 1)\to \mathbb{R}$.

Now, we need to find a bijection from the closed interval $[a, b]\to \mathbb{R}$, and we can do this by first showing that there exists a bijection from the closed interval $[a, b]$ to the open interval $(a, b)$.

Taking the interval: $[0,1]$. Define $f(x)$ as following: $$f(x) = \left\{ \begin{array}{1 1} \frac{1}{2} & \mbox{if } x = 0\\ \frac{1}{2^{n+2}} & \mbox{if } x = \frac{1}{2^n}\\ x & \mbox{otherwise} \end{array} \right.$$

Then $f: [0, 1] \to (0, 1)$ is a bijection.

Now, compose: $g\circ f = g(f(x)): [1, 0] \to \mathbb{R}$, and you have your bijection.

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If you're willing to accept that $\Bbb{R}$ is uncountable, finding a bijection between $(a, b)$ and $\Bbb{R}$ is actually overkill. It suffices to note that $\Bbb{R}$ may be written as the union of countably many intervals of length $b-a$, and countable unions of countable sets are countable.

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+1 Short and sweet. –  JavaMan Jan 9 '13 at 2:22

My strategem would be to prove firstly this more general result, whose name I'd like to ask if anybody knows? Also, is there a question on MSE about this theorem?

Let $S$ be an infinite set such that $|S| > |\mathbb{N}|$. Also, let $C \subset S$ be countable. Then:
$1.$ $|S - C|$ is infinite.
$2.$ $|S| = |S - C|$

Subsequently, after you prove that $|(a,b)| = |\mathbb{R}|$, apply this theorem with $C = \{a,b\}$ and $S = [a,b]$ to prove that $|\; [a,b] \; | = |(a,b)| = |\mathbb{R}|$.

This stronger catholic result can be used frequently, such as to prove further equalities of cardinalities in addition to those above: $|\; [a,b] \; | = |(a,b)| = |\; [a,b) \; | = |\; (a,b] \; |= |\mathbb{R}|.$

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We could do this conventionally by using Cantor's diagonal slash argument. This is almost certainly the best way for those new to (un)countability to understand why it's true. But once you have a little more familiarity with the subject matter, I think that the following proof is both unusual and entertaining.

Let's prove this by playing a game! Actually, let's let Bob and Alice play a game, using the interval $[a,b]$. (although it works equally well if they play with $(a,b)$

Suppose we have the interval $[a,b]$ and a set $S$. In the game, Alice begins, and she chooses some $a_1 \in(a,b)$ to act as a new lower bound and tells Bob what it is. Bob then chooses some $b_1 \in (a_1,b)$ to act as a new upper bound and tells Alice. Alice then picks $a_2 \in (a_1,b_1)$, and they carry on taking turns in this manner, producing smaller and smaller intervals.

Now, $a_n$ is an increasing, bounded sequence in $\mathbb{R}$ so we know it converges, say to $a' \in (a,b)$.We say that Alice wins the game if $a'\in S$ and Bob wins otherwise. A question arises, given a set $S$, is it possibly for Alice or Bob to find a winning strategy so that they always win?

Suppose $S$ is countable. Then we can enumerate $S = \{s_1,s_2,s_3,...\}$. In this case, Bob can always win (I'll let you think about why he can always find a winning strategy! It's a little tricky but should be do-able with a little familliartiy with countability).

But what if $S = [a,b]$? Well then clearly Alice wins, because $a' \in(a,b) \subset [a,b]$.

So $[a,b]$ cannot be countable!

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This is awesome. –  Erwin Dec 6 '12 at 8:59
I just realized one thing.If we use the Schroder-Bernstein theorem,which I studied just now, and we use the fact that $(a,b)$ is an injection into $[a,b]$ and $[a,b]$ is an injection into $(a,b)$.That way, there is a bijection between $[a,b]$ and $(a,b)$. –  Erwin Dec 6 '12 at 9:08
@Erwin That's very true, and it shows that the cardinality of $[a,b]$ and $(a,b)$ are the same, but it's not needed for the proof I gave, and is in fact stronger than just showing $[a,b]$ and $(a,b)$ are uncountable. This is because we can show if one of them is countable, the other one is too.(More generally, the countable union of countable sets is countable). If we have a bijection $f:(a,b)\rightarrow \mathbb{N}$ say, define $g:[a,b]\rightarrow \mathbb{N}$ by $g(a) = 1, g(b) = 2$ and $g(x) = f(x)+2$ for all $x\in(a,b)$. This is a bijection.The argument is similar for $[a,b]$ countable. –  Tom Oldfield Dec 6 '12 at 17:54

I would consider using Cantor's diagonalization argument. It is a pretty efficient proof and not too difficult to understand.

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$$\tan \left( \frac{\pi}{(b-a)} (x-\frac{a+b}{2})\right)$$

Basically $f(x)=\frac{\pi}{(b-a)} (x-\frac{a+b}{2})$ is the linear function such that $f(a)=-\frac{\pi}{2}$ and $f(b)=\frac{\pi}{2}$.

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If (a, b) is countable, then we have a bijection between Q and (a, b). Q bijects with any infinite set of rational numbers (c, d). Thus, we can encode all points of (a, b) by a countably infinite list of all rational numbers as follows:

1-x.kfj...

2-y.fli...

3-i.opq... . . .

Find the first non-zero digit of "a" in its infinite decimal expansion. Put 0s down on paper until you reach this digit position. Then choose the numeral from the next digit position from rational number 1 and use it basically as follows... if we have a=.00700... and the first rational number on our list is 1.234000..., then we choose 0, since 7 comes in the third position after the decimal point, and 0 lies in the fourth position. So, our number so far is .0000. Then choose the numeral in the next digit position from rational number 2, suppose it's .00005454... Then we have .00005, and so on. We end up with a number greater than "a", yet less than "b", nowhere on our list of rational numbers. Thus, there exists no bijection from Q to (a, b), and thus (a, b) is uncountable. Since (a, b) is uncountable, [a, b] is uncountable also.

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