The form of a set of positive integers that contains $k+1$ whenever it contains $k$

Question

I've seen there are lots of proofs by mathematical induction on the site, yet I haven't seen a problem like mine (or at least couldn't pattern my problem after the other).

"Let $m$ be the smallest element of set $A\subseteq\mathbb{N}$. If $k+1$ is the element of $A$ whenever a natural number $k$ is, then $A=\{n\in\mathbb{N}:n\geq m\}$."

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I think I should create another set for which only the first part of the problem is true, but not $$A=\{n\in\mathbb{N}:n\geq m\}.$$ Then conclude from the former information that this is really true.

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(1) Let's assume $a=m$; since $m\in A\subseteq\mathbb{N}$, $a$ is a natural number in $A$.

(2) Let $K\subseteq\mathbb{N}$, let $m$ be its smallest element, and for all $k\in K$ there's $k+1$ in $K$. For this set (1) also holds, since some $k=m$ must be the element of $K$.

(3) If $m$ is the smallest element of $K$, then $m+1$ is the next element in the set ($m+1\in K$), and so on. This is true because of the assumption (2) "for any natural $k\in K$ there's $k+1\in K$". Therefore now we know that $K$ contains all natural numbers $k$, starting from $m$: $$K=\{k\in\mathbb{N}:k\geq m\}.$$

Is this a good proof? If someone could drop a hint I'd be glad.

Answer 1

You're making it more complicated than it needs to be. I'm not at all sure what part (1) is about, for example.

Note that since $m$ is the smallest element of $A$ and $A\subseteq\Bbb N,$ then we can immediately conclude that $$A\subseteq\{n\in\Bbb N:n\ge m\}.\tag{$\star$}$$ This is because if $a\in A,$ then $a\in\Bbb N$ since $A\subseteq\Bbb N,$ and $a\ge m$ since $m$ is the least element of $A;$ thus, $a\in\{n\in\Bbb N:n\ge m\},$ and since this holds for all $a\in A,$ then $(\star)$ holds.

All that remains is to show that if $n\in\Bbb N$ such that $n\ge m,$ then $n\in A,$ from which the reverse inclusion of $(\star)$ follows, and so we have equality. That's where induction comes in handy. The idea is to note that if $n\in\Bbb N$ and $n\ge m,$ then either $n=m$ or $n=m+k$ for some positive integer $k.$ Proceed by induction on $k$ to show that all such $n$ are elements of $A.$

Let me know if you have any trouble, or just want to check your reasoning.

Answer 2

I don't know how to merge accounts, and that one was made without login. (roro) I tried, but it says my profile link is invalid... whatever. Now I cannot comment on that answer. I'm sorry I'm doing exactly what is against the rules (replying to the answer in "an answer", but I figured, because I've already chosen the best one, this won't hurt).

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I didn't see that you used "$\subseteq$" instead of "$=$". ...

Natural number is a positive integer. Set of natural numbers is a set with $1$ as a starting number, and the next number is bigger by $1$ from the previous.