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How's this done?

Also, I am wondering, are all subsets of $\mathbb{R}$ cardinally equivalent to each other? If not, why not?

Asaf Karagila
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JohanLiebert
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    What do you know about cardinality? How can you obtain $(0,1)$ from $[0,1)$? As for your other question, have you considered the singleton subsets, ${a}$? Its cardinality is 1. What about the rationals? They are countable. What about any interval? They have cardinality of $(0,1)$ or equivalently the cardinality of $\mathbb{R}$. – snar Oct 01 '13 at 13:25
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    Hint: $(0,1) = \left((0,1)\setminus{\frac1n\colon n=2,3,\ldots}\right)\cup {\frac1n\colon n=2,3,\ldots}$. – njguliyev Oct 01 '13 at 13:26
  • Related and related. In the latter, you can see how one can actually construct such a map. – Cameron Buie Oct 01 '13 at 13:34
  • @snarski so then, does that mean all uncountable subsets of $\mathbb{R}$ are cardinally equivalent? – JohanLiebert Oct 01 '13 at 13:37

4 Answers4

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Consider $f:[0,1)\to(0,1)$ such that $$ f(x)=\begin{cases} 1-\frac{1}{n+1} & \text{if $x=1-1/n$ for some natural number $n$}\\ x &\text{otherwise} \end{cases} $$ You can prove that $f$ is bijection. So two sets have same cardinality.

Hanul Jeon
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  • Whoa, whoa, whoa boss, how did you contrive that monstrosity of a function?! Please explain. – JohanLiebert Oct 01 '13 at 13:29
  • @JohanLiebert I saw this function at the past. – Hanul Jeon Oct 01 '13 at 13:36
  • @JohanLiebert: The fundamental idea is an important one-sweeping the uncomfortable item(s) under the rug by moving everything down the line. It is the same idea as the bijection $[0,\infty)$ to $[1,\infty)$ by taking $n$ to $n+1$. We just need to find a countable series of points and shift them all by $1$ – Ross Millikan Oct 01 '13 at 13:59
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More generally, if $X$ is infinite and $x\in X$ then there is a bijection between $X$ and $X\setminus \{x\}$.

Let $x_0=x,x_1,...\in X$ be an infinite sequence of distinct elements of $X$. Then define $$f(x)=\begin{cases} x_{n+1} & \text{if } x=x_n\\ x &\text{otherwise} \end{cases}$$

This is the same as totori's answer, he just picks a specific sequence $x_n=1-\frac{1}{n}$. Shuchang's answer is a variation, since his map goes the other direction, but he is basically using $x_0=0, x_n=2^{-n}\text{ otherwise}$.

Thomas Andrews
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Define a mapping $f$ from $(0,1)$ to $[0,1)$ by $$f(x)=\left\{\begin{matrix}0,&x=\frac12\\ \frac1{2^n},&x=\frac1{2^{n+1}}\\ x,&\text{otherwise}\end{matrix}\right.$$ The mapping $f$ is bijective, which implies $\text{card}(0,1)=\text{card}[0,1)$.

Shuchang
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  • I need the reasoning on how you contrived this function to be explained, please. Just seeing this answer, there's no way I could come up with my own answer to a similar question on an exam. I have no idea how you came up with. What matters to me is not the answer to the question, but understanding how to solve it myself. – JohanLiebert Oct 01 '13 at 13:31
  • Yes, The difference between them is only a point right? Do you know Hilbert hotel problem? The solution is translation. You need to move this single "guest" into some room and then move the guest in that room into another room again and again. – Shuchang Oct 01 '13 at 13:37
  • @Johan: Henning's answer here discusses how to come up with such functions in general, by using an injection each way to construct a bijection. – Cameron Buie Oct 01 '13 at 13:44
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The question of bijections have been answered.

The second question, however, has gone mainly unnoticed. The answer is no. Almost trivially, $\varnothing$ is a subset of $\Bbb R$ and it is certainly not of the same cardinality as $\Bbb R$ itself.

Moreover, there are finite sets, and countably infinite sets which are also of different cardinalities from one another.

One can ask, if so, if $A\subseteq\Bbb R$ is uncountable, is it necessarily of the same cardinality as $\Bbb R$? The positive answer to this question is known as "The Continuum Hypothesis", but the answer is that there is no provable answer. It is consistent with the usual axioms of mathematics that the answer is "yes", and it is consistent that the answer is "no".

Namely, there are universes of set theory in which the answer is positive, and no uncountable set of real numbers has a smaller cardinality than $\Bbb R$; but there are universes of set theory where the answer is no and there is an example for such set. But do note that these example are not sets which can be explicitly written by a formula, we can simply prove they exist in one way or another from set theoretical assumptions.

Asaf Karagila
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