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I am trying to find an answer to this problem "Show that the sets $\mathbb Q$ of rational numbers and $A = \{x \in\mathbb Q | 0 \le x \le 1\}$ have the same cardinality"

So, I know we would have to find a bijection from $\mathbb Q$ to $A$. I was thinking of this one :

$f(x) = \begin{cases} \frac{x}{x+1}, & \text{if $x$ > - 1} \\[2ex] 1, & \text{if $x$ = -1} \\[2ex] ?, & \text{if $x$ < -1} \\ \end{cases} $

which I guess it would be well defined (since $x < x+1, \frac{x}{x+1} < 1$) and injective ( $$\frac{x}{x+1} = \frac{x'}{x'+1} => xx' + x = x'x + x' => x =x'$$). But now I can't find how to define the rationals under -1 to make f surjective !

I tried also to find a bijection from $\mathbb N\times\mathbb N$ to $A$ (we already saw in class that $|\mathbb N\times \mathbb N| = |\mathbb N| = |\mathbb Q|$)

So far, the closest answer on Internet I could find to my question was on this topic : https://www.physicsforums.com/threads/cardinality-of-an-interval-of-rationals.432232/ , but I can't make sense of it, so I am doing a new topic !

Thank you very much for your help

J. W. Tanner
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Progranma
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    Note any infinite set has cardinality at least as great as the cardinality of $\Bbb N$. Note that $\Bbb Q$ has the same cardinality as $\Bbb N$. Note then that the cardinality of a subset of $\Bbb Q$ has cardinality at most as great as the cardinality of $\Bbb Q$ and thus of $\Bbb N$. So, your set's cardinality is simultaneously greater than or equal and less than or equal to the cardinality of $\Bbb N$. – JMoravitz Dec 12 '19 at 13:28
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    "So, I know we would have to find a bijection from $\Bbb Q$ to $A$" No, if you aren't asked to explicitly find a bijection, there is no reason to. You are better off just finding injections in each direction and applying the Cantor-Schroder-Bernstein theorem. Explicitly writing a bijection can be needlessly confusing. Simply proving one exists is often far easier. – JMoravitz Dec 12 '19 at 13:29
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    Oh wow, thank you so much !! that is so clear ! aah, I am so dumb, I spent 2 hours trying to find a bijection – Progranma Dec 12 '19 at 13:31
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    @JMoravitz There is no need to use the result in your first comment: any infinite subset of a countable set is countable. Unlike the general result you indicate (which uses the axiom of choice), this one doesn't, and it also avoids having to invoke the Cantor-Schroder-Bernstein theorem. – Andrés E. Caicedo Dec 12 '19 at 14:36

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