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Can someone take a look at the proof I've constructed below? Does this work and does anyone have any suggestions?

Suppose $A=\{(m,n)\in\mathbb{N}\times\mathbb{R}:n=\pi m\}$. Is it true that $\vert\mathbb{N}\vert=\vert A\vert$?

My answer:

Yes, this is true.

$Proof.$ Let us consider $f:\mathbb{N}\rightarrow A$ defined as $f(m)=(m,\pi m)$.

We see that $f$ is injective since $f(m)=f(n)$ gives $(m,\pi m)= (n,\pi n)$.

Now to see that $f$ is surjective take any $x=(b,b\pi)\in A$ and observe that $f(b) = x$.

But then $f$ is a bijection between $\mathbb{N}$ and $A$. It follows that $\vert\mathbb{N}\vert=\vert A\vert$. $\Box$

ClownInTheMoon
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    It seems to be a perfectly valid proof. But in the injectivity part, you want to prove that $m=n$, which of course follows from $(m,\pi m)=(n,\pi n)$ – u1571372 Oct 28 '16 at 21:45
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    It has one blemish: in the proof that $f$ is injective you stopped a little sooner than you really should. Specifically, you really ought to say explicitly that $(m,\pi m)=(n,\pi n)$ implies that $m=n$, since that’s the conclusion that you’re actually trying to reach. – Brian M. Scott Oct 28 '16 at 21:47

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This is a valid proof, with one suggestion of improvement on proving injectivity, defined as if $f(m) = f(n)$ is true, then $m = n$:

$f(m) = f(n) \Leftrightarrow (m, \pi m) = (n, \pi n) \Rightarrow m = n.$

Phoebe Liu
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