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I would just like to check if my proof is correct in the direction that if that function exists then $A$ is countable.

We know that $A$ is infinite, then let $A=\{a_n\}_{n=1}^{\infty}$. I think stating it like this becomes trivial instantly, so I am not sure if I am allowed to write out set $A$ like that, but I cannot see why not. $1-1$ (injection) implies that each element in the codomain of $f$ gets mapped to at most one element in the domain. Then let us define that function $f:A \rightarrow \mathbb{N}$ as follows: $f(a_n)=n$, where $n \in \mathbb{N}$. Then $A$ is countable.

Am I allowed to set the form of $f$ like I did? I think I am, since the statement says "there exists" and not "for all" (i.e. I am allowed to pick a form of $f$).

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    Do you know Cantor-Schröder-Bernstein? – 2017-02-19
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    I didn't. But now I do – 2017-02-19
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    You cannot write $A=\{a_n\}_{n=1}^{\infty}$ unless you already know that $A$ is countable. – 2017-02-19
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    Can you find a sequence so that $\mathbb{R} = \left\{ a_n \right\}_{n=1}^{\infty}$? – 2017-02-19
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    @max Ok, I see why I cannot write it like that now.. – 2017-02-19

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If you write $A=\{a_n\}_{n=1}^{\infty}$ you're already assuming $A$ is countable.

More properly: a set $A$ is infinite if and only if there exists a 1-1 mapping $f\colon\mathbb{N}\to A$ (requires some form of the axiom of choice).

If $A$ is infinite and there is also a 1-1 mapping $g\colon A\to\mathbb{N}$, then the Cantor-Schröder-Bernstein theorem tells you that $A$ has the same cardinality as $\mathbb{N}$, that is, it is countable.

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    Cantor-Schröder-Bernstein theorem requires a function $h: \mathbb {N} \rightarrow A$ that is 1-1 for there to exist a bijection. I cannot see how to prove that there is such – 2017-02-20
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    @isquared-KeepitReal That map exists because $A$ is infinite. – 2017-02-20
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    but $\mathbb {R }$ is infinite too, and there is no such map – 2017-02-20
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    @isquared-KeepitReal $\mathbb{N}$ is a subset of $\mathbb{R}$. There is no 1-1 mapping $\mathbb{R}\to\mathbb{N}$, but this is the reverse direction and, indeed, $\mathbb{R}$ is not countable. – 2017-02-20
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    I was talking about this direction: $f: \mathbb {N} \rightarrow \mathbb {R}$. Also, $\mathbb{N}$ is a subset of $\mathbb {Q}$ and yet both are countable. – 2017-02-20
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    @isquared-KeepitReal And where's the problem? For *every* infinite set $A$ there exists a 1-1 mapping $f\colon\mathbb{N}\to A$. – 2017-02-20
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    yes, precisely. I am trying to figure out why this is the case. Because as I said before, take $\mathbb {R}$ it is infinite. But is there $f: \mathbb {N} \rightarrow \mathb{R}$? No, and so there must be something that I am missing – 2017-02-20
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    @isquared-KeepitReal 1-1, with other terminology, is *injective*; there is no assumption on it to also be surjective. – 2017-02-20