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I am trying to prove that an infinite subset $T \subset \mathbb{N}$ is countable using the Well-Ordering Principle. The strategy is to define $f: \mathbb{N} \to T$ as follows. We let $f(1)$ be the least element of $T$, $f(2)$ be the least element of $T \setminus f(1)$, and upon defining $f(1), \ldots, f(k)$ in this manner, we define $f(k+1)$ to be the least element of $T \setminus \{f(1), \ldots, f(k)\}$.

It's clear to me, intuitively, that this is a bijection. I just can't figure out how to prove it. So, by definition, upon defining $f(m+1)$, I require $f(m+1) \neq f(m)$, so if $a \neq b$, and WLOG $a < b$, then $f(a) \neq f(b)$ and, in fact, $f(b) > f(a)$ by construction. So $f$ has to be injective. As for surjectivity, I can say, "well, I've exhausted all of $\mathbb{N}$ by induction and $T \subset \mathbb{N}$, so I've definitely defined it for all of $T$. It sounds like I'm assuming my conclusion because I'm likely to break injectivity by exhausting $\mathbb{N}$. But, I guess if I prove injectivity/surjectivity separately, that isn't a huge problem. But then, how do I know I've exhausted all of $T$ as the image of some $n \in \mathbb{N}$?

Can anyone help me sort this out?

Brad G.
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1 Answers1

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Suppose that $x \in T \setminus f(\mathbb{N}) \not = \emptyset$, then for all $k \in \mathbb{N}$, we have $f(k) < x$. Hence, we have an infinite subset of the natural numbers that has an upper bound. Use this contradiction to argue that $T \setminus f(\mathbb{N})$ must be empty.

Mike Desgrottes
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  • I'm not sure if I fully understand why this is a contradiction, though, as $T$ could be bounded above, right? I must be missing something. (Or is that impossible since $T$ is infinite? If it's bounded above, it's finite.) – Brad G. Aug 21 '21 at 03:43
  • The fact that T is an infinite subset of the natural numbers means that it cannot be bounded above. – Mike Desgrottes Aug 21 '21 at 03:51
  • That makes sense. If my argument for injectivity correct? – Brad G. Aug 21 '21 at 03:52
  • You have to add more details but the idea is good. You have to show that $f(m) = f(n)$ implies that $m = n$. You can show that by induction if you've shown that $f(m + 1) \not = f(m)$ for all $m \in \mathbb{N}$. Note that if $n < m$, then $f(n) = \inf T \setminus {f(1),...,f(n-1)}$ and $f(m) \in T \setminus {f(1),...,f(n -1)}$. Hence, $f(n) < f(m)$. – Mike Desgrottes Aug 21 '21 at 03:58
  • I was trying to prove the contrapositive $m \neq n$ implies $f(m) \neq f(n)$. My updated attempt: Let $a \neq b \in \mathbb{N}$ and, without loss of generality, suppose $a < b$. Then, by definition, $f(b) = \min\left(T \setminus f(1), \ldots, f(a), \ldots, f(b-1) \right)$, so $f(b) > f(a)$, i.e., $f(b) \neq f(a)$. – Brad G. Aug 21 '21 at 03:59
  • It might be easier to just prove that $f$ is strictly decreasing. – Brad G. Aug 21 '21 at 04:00
  • Yea, It's correct. – Mike Desgrottes Aug 21 '21 at 04:03