2

I need a really quick way of showing there's a bijection from $\mathbb{Q}\cap(a,\ b)$ to $\mathbb{Q}$ for any real numbers $a < b$. I attempted a few ways but I'm drawing a blank right now .Nothing I've worked on is fruitful (it either goes nowhere or is much too complicated) so I'm omitting it from the question. Any simple ideas? Mainly I'm looking for something that can be rigorously justified and explained in a matter of no more than three to four lines.

Clarification: I don't need to construct a bijection, I just need to show that there is one.

Clarification 2: The context I'm working in doesn't have a definition of "countable", so I can't just say both sets are countable unfortunately.

Clarification 3: We know that there exists a bijection from $\mathbb{Q}$ to $\mathbb{N}$. Our construction was essentially that you can list all elements of $\mathbb{Q}$ in a grid and spiral outwards from the origin, ignoring duplicates, and assigning the next natural to the next unique rational in the spiral path.

Andrés E. Caicedo
  • 79,201
user3002473
  • 8,943
  • An explicit bijection or a proof that a bijection exists? – Alexis Olson Oct 15 '16 at 21:20
  • @AlexisOlson A proof that a bijection exists. – user3002473 Oct 15 '16 at 21:20
  • 4
    They both are countable sets. – Jack D'Aurizio Oct 15 '16 at 21:22
  • @JackD'Aurizio Hmm, you're right, but I don't think that's acceptable for the context I'm working in (we don't know what "countable" means yet). I would've said that if I could've. I'll clarify the question further, sorry for the confusion. – user3002473 Oct 15 '16 at 21:23
  • Then you may just show that there is an injective map from $\mathbb{Q}\cap(a,b)$ to $\mathbb{Q}$ and an injective map from $\mathbb{Q}$ to $\mathbb{Q}\cap (a,b)$. See https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem – Jack D'Aurizio Oct 15 '16 at 21:25
  • This is an even more non-trivial fact. –  Oct 15 '16 at 21:35
  • 1
    Do you know how to construct a bijection between $\mathbb{Q}$ and $\mathbb{N}$ and, if yes, how? That would frame your question better in terms of what you can use vs. what not. – dxiv Oct 15 '16 at 21:38
  • @dxiv Updated the question to include that. – user3002473 Oct 15 '16 at 21:40
  • I feel that this is essentially a duplicate of the proposed question. If there is a reasonable argument, I'd be happy to assist in reopening the question (assuming it hasn't been reopened by the time I'm back). In either case, I feel that the linked thread should be helpful. – Asaf Karagila Oct 15 '16 at 22:58
  • @AsafKaragila One difference is that the $(a,b)$ here is not assumed to have rational endpoints necessarily. Because of that, bijections established for $(0,1)$ do not automatically translate to an answer to this question. Whether the OP finds the missing step trivial to fill in, or not, depends on background and context. – dxiv Oct 15 '16 at 23:55

5 Answers5

4

Let's map $[0,+\infty)\cap \mathbb Q$ bijectively onto $[0,1)\cap \mathbb Q$ with $$ f(x) = 1-\frac{1}{x+1} $$ and map $(-\infty,0) \cap \mathbb Q$ bijectively onto $(-1,0)\cap \mathbb Q$ with $-f(-x)$.

This has the advantage that it is the restriction of a homeomorphism $(-\infty,+\infty)\to (-1,1) $ to the rationals.

Fron this we can do any $(a,b)$ with rational endpoints.

GEdgar
  • 111,679
  • Just a (minor) nitpick about the last line, but the question doesn't assume that $a,b \in \mathbb{Q}$. – dxiv Oct 15 '16 at 22:18
1

Hint based on Clarification 3:

We know that there exists a bijection from $\mathbb{Q}$ to $\mathbb{N}$. Our construction was essentially that you can list all elements of $\mathbb{Q}$ in a grid and spiral outwards from the origin, ignoring duplicates, and assigning the next natural to the next unique rational in the spiral path.

$\mathbb{Q} \cap (a,b)$ is clearly a subset of $\mathbb{Q}$, and it is clearly infinite. Take then the previously defined "grid" and remove from it any rationals that fall outside $(a,b)$. By the same "spiral" argument, you can define a bijection between the numbers left unremoved i.e. $\mathbb{Q} \cap (a,b)$ and $\mathbb{N}$. Once you have the bijections $f:\mathbb{Q} \to \mathbb{N}$ and $g:\mathbb{Q} \cap (a,b) \to \mathbb{N}$, you have the bijection $g \circ f^{-1} :\mathbb{Q} \cap (a,b) \to \mathbb{Q}$.

dxiv
  • 76,497
0

The intersection $\mathbb Q \cap (a,b)$ is countable.

0

1) since $\Bbb Q $ is dense in $\Bbb R $, $\mathbb{Q}\cap (a,\ b)$ is an infinite sized set

2) $\mathbb{Q}\cap (a,\ b) \subset \Bbb Q $

3) then by (1) the same bijection you constructed from $\Bbb Q \to \Bbb N $ also maps all elements of $\mathbb{Q}\cap (a,\ b)$ onto $\Bbb N $. And since (1) shows there are an infinite number of elements then such a bijection exists.

Phillip Hamilton
  • 1,181
0

We can describe $Q:=\mathbb{Q}\cap(a,b)$ as $$ Q=\{\frac{r}{s}:r\in\mathbb{Z}, s\in\mathbb{Z}^+,gcd(|r|,s)=1,as<r<sb\}, $$ where the condition $gcd(|r|,s)=1$ ensures that we are counting each rational in $Q$ only once. This implies that $Q$ can be written as the disjoint union $$ Q=\bigcup_{s\in\mathbb{Z}^+}Q_s, $$ where $Q_s:=\{\frac{r}{s}\in Q:r\in\mathbb{Z}\}$. Clearly every $Q_s$ is finite (and nonempty if $s(b-a)>1$) so you might write the elements (if any) of $Q_1$ in a row, the elements (if any) of $Q_2$ in a second row, and so on. This arrangement allows you to Cantor-diagonalize in the same fashion that the diagonal argument for $\mathbb{Q}$, concluding that the cardinal of $Q$ is the same as that of $\mathbb{N}$, and so that that of $\mathbb{Q}$ (which by definition means that there exists a bijection between $Q$ and $\mathbb{Q}$).