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Using lemma and proof by contradiction to show that $\mathbb{N}$ is countable

Given the set $A= \{a_n\}_{n=0}^{\infty}$ where $a_n = n+1$. I have to show that $\mathbb{N}$ is countable.

My thoughts: I know from working with proof by contradiction that I can do something like this:

By induction, $a_n$ is an odd number
So, by transitivity of relation $\{2k+1\}_{k=0}^n\subset A$
Since $A$ is a relation, $A$ is countable. Thus, $\mathbb{N}$ is countable.

I have a few questions.

Is this the right way to think about it? This is my first time using lemma and proof by contradiction (I’m really trying to get into proof by contradiction), and I’m not sure if this is actually the right way to think about it.
Can I just say that $A$ is countable, and then proof by contradiction is redundant?
Is there a more elegant way to think about this?

A:

The usual setup of a proof by contradiction is that you are assuming the lemma to be false, and trying to establish a contradiction by presenting a counterexample to the lemma. In your case, you have assumed that the set in question is countable, and that an arbitrary element $n$ in that set is an odd number. Now you show that it is not true that all the elements are odd. Since you have disproven your lemma, you have succeeded in establishing a contradiction.
As for the question about whether the proof by contradiction is redundant, you are essentially right to say that it is. Without the proof by contradiction, we might be able to disprove this lemma, but we could not even formally state that we could disprove it, since we have not proven that the set in question

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