This was part of a three part question where I was supposed to prove two sets have equal cardinality by finding bijections. I've created a bijection $f: \Bbb Z \Rightarrow 2\Bbb Z$ by $f(x)=2x$. I've created a bijection $g: (0,1) \Rightarrow (4,50)$ by $g(x)=46x+4$. I think those are both correct. My last question is finding a a bijection between (0,1) and [0,1]. I've seen this question several times on this board, but I've yet to understand them and I can't really go back and ask any questions to the original posters. I know a bijective function exists between (0,1) and $\Bbb R$, but I don't think that helps me here.
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Do you know that the rationals are countable? Let
$\{q_i:i\in\omega\}$ enumerate $\mathbb Q\cap (0,1)$, and
$\{p_i:i\in\omega\}$ enumerate $\mathbb Q \cap [0,1]$.
Map $q_i\to p_i$ and let every irrational number map to itself.
Forever Mozart
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We have learned that the rationals are countable and that |$\Bbb N$| = |$\Bbb Q$|. What does your w represent? And also, this may be a dumb question, but what do you mean by enumerate? – user21 Feb 15 '17 at 04:39
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you can replace my $\omega$ with $\mathbb N$. Enumerate = assign a number to each element (using some bijection, for instance $q_i=q(i)$ where $q:\mathbb N \to \mathbb Q\cap (0,1)$ is a bijection). – Forever Mozart Feb 15 '17 at 04:48
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Okay. I guess i'm not really understanding what you mean by $q_i$ or $p_i$. You and @fleablood are saying the same thing, I think, but I'm not understanding them. – user21 Feb 15 '17 at 05:02
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It's counter intuitive but push 0 to $a_1$. And push $1$ to $a_2$. Then create an infinite sequence $a_i $ and push $a_i $ to $a_{i+3}$. Meanwhile for any $x $ not in the sequence, map $x $ to $x $.
Example: Let $a_n = 1/n$. Let $f (0)=1/2$. Let $f (1)= 1/3$. Let $f (1/n)=\frac 1 {n+2} $. If $x \ne \frac 1n $ for any natural $n $, let $f (x)=x $.
That's easy to be shown to be bijective.
Another, perhaps more common, example is to let $\{a_n\}=\mathbb Q $ be an sequence of all the rationals. Map $0\rightarrow a_1$ and map $1\rightarrow a_2$ and $a_i \rightarrow a_{i+2} $ while every irrational gets mapped mapped to itself.
fleablood
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In your second example, you're mapping $0 \Rightarrow a_1$. What is $a_1$? Just a random rational number? – user21 Feb 15 '17 at 05:01
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Well, not random but ... arbitrary. You know Q is countable. That means you can make a sequence {$a_n $} that lists all the rational numbers one after another. There's an infinite number of ways to do this. Pick one. Which ever one you pick... stick to it. The "diagonal" is perhaps the most well known: 1,1/2,2,1/3,3,1/4,2/3,etc. – fleablood Feb 15 '17 at 05:08
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Could we not just map (0,1) $\Rightarrow$ (0,1) and then do something else with 0 and 1? Sorry if that sounds dumb, I'm just confused about that. – user21 Feb 15 '17 at 05:33
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No, because everything was taken and there'd be nowhere left to push them to. – fleablood Feb 15 '17 at 05:52
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Are you fimiliae with the infinite hotel? How do you map $N \cup {a} $ to $N $ when one of them seems to have one more point? Well, you map $a $ to 1, and then every $k $ to $k+1$. This is much the same. You simply make an infinite countable list of terms, push shove the extra two terms in the beginning and push every other one two terms down. What if the set is uncountable. Well, just push a string of countable ones down and leave the rest alone. – fleablood Feb 15 '17 at 06:00
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Here's a list of every rational in between 0 and 1: 1/2,2/3,1/4,3/4,1/5,2/5,3/5,4/5,1/6,5/6,1/7 etc. So map 0 to 1/2, 1 to 2/3, 1/2 to 1/4, 2/3 to 3/4, etc. Leave the irrationals alone. – fleablood Feb 15 '17 at 06:05
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HINT: Find first the bijection between sets $\{1/n:n\in\mathbb{N}\}\cup\{0\}$ and $\{1/n:n\in\mathbb{N},\ n\geq2\}$
Przemysław Scherwentke
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I think I understand what you're saying. The first set is just [0,1] and the second is (0,1). Are you saying that I should find the bijection $f: [0,1] \Rightarrow (0,1)$ instead of the other way around? – user21 Feb 15 '17 at 04:45
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@johnie4usc Not exactly, but both mine sets are countable, so bijection exists, and the completions to $[0,1]$ and $(0,1)$, respectively, are identical, so identity is the remaining part of a bijection. – Przemysław Scherwentke Feb 15 '17 at 04:49
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I guess I'm not really following what you're saying. You're just saying to map every element of (0,1) to itself to form a bijective function to [0,1]? I think that $f:(0,1) \Rightarrow (0,1)$ can just be defined by $f(x)=x$, correct? I don't know how to map the 0 and 1 which aren't included in (0,1). Maybe I'm just thinking about this completely wrong. – user21 Feb 15 '17 at 04:58
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@johnie4usc Not, not $(0,1)$, but $(0,1)\setminus{1/n:n\in\mathbb{N},\ n\geq2}$. THEN $f(x)=x$ is a bijection (identity). The remaing part of the bijection is in the hint. – Przemysław Scherwentke Feb 15 '17 at 05:41
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Consider how to map {-1,0,1,2,3,4...} to {1,2,3,4,5,6...} you'd simply do, k to k+2. Supposed you had to do {-1,0}U (0,infty) ... well you have to fit in the extra two points in somehow. So k to k+2 for the integers and all else stay the same. But what about {0,1}U (0,1). Well you know the rationals are countable BUT IT DOESN'T MATTER HOW THEY ARE COUNTABLE. It' enough that you know they are there must be some list, where they are listed one after another. Simply use that list even if you don't know what it is. – fleablood Feb 15 '17 at 06:15
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An "enumeration" is a listing of a countable or finite set in ordered and indexed or "enumerated" by a number. Let the rational numbers strictly between 0 and 1 be the list {$a_1,a_2,a_3,..... $}. We may not know how to describe the list but we know it exists because that is what countable means. Likewise let the rationals between 0 and 1 inclusively be the list {$b_1,b_2,b_3....$}. This 8s a different list. We don't need to know what it is, we just need to know there is such a list. So we map $a_k <=> b_k$. That's all there is two it. – fleablood Feb 15 '17 at 06:22