If a nonempty subset of integers is bounded from below, it has a minimum

Question

Let $A$ be a non-empty subset of $\mathbb Z$. Suppose there exists $s \in \mathbb Z$ such that $s \le a$, for all $a \in A$. Show that $A$ has a minimum.

I was assuming induction would be used for this proof since that is what we just covered, but it doesn't seem to apply. The main thing throwing me off is that $s$ is in $\mathbb{Z}$ not $A$.

Let $\Bbb{Z}^-$ be the set of negative integers. If $A \cap \Bbb{Z}^-$ is empty, then $A \subset \Bbb{N}$ and your claim is just the well ordering principle. If $A \cap \Bbb{Z}^-$ is non-empty, then by your hyphotesis it is finite (you can prove this by contradiction). Hence $A \cap \Bbb{Z}^-$ has a minimum (see reference below). Now show that this minimum works as a minium for $A$. Reference: http://math.stackexchange.com/questions/548806/a-finite-set-always-has-a-maximum-and-a-minimum — Reveillark, Feb 24 '15 at 23:08
possible duplicate of Every non-empty subset of the integers which is bounded above has a largest element. — , Feb 25 '15 at 00:37
^ this is a mirror image of the problem stated here; just reverse the inequalities. — , Feb 25 '15 at 00:38

score 1 · Answer 1 · answered Feb 24 '15 at 23:12

1

Consider the bijection from $\mathbf Z \rightarrow\mathbf Z$ defined by $x\mapsto x-s$. Then $0$ is less than every element of $A'=A-s$. $A'$, as a non-empty subset of $\mathbf N$, has a smallest element $n_0$. Then the reciprocal bijection, $x\mapsto x+s$ maps $n_0$ to the smallest element of $A$, $n_0 +s$.

answered Feb 24 '15 at 23:12

Bernard

179,256

We haven't covered bijections yet. Thanks for the input though. – Zooey Feb 24 '15 at 23:26

score 0 · Answer 2 · answered Feb 24 '15 at 23:10

0

Suppose $A$ has no minimum. Then for every $n\in\mathbb{Z}$ there must exist some $a\in A$ such that $a<n$. In particular, there must exist some $a\in A$ such that $a<s$, which is a contradiction.

answered Feb 24 '15 at 23:10

Auslander

1,309

2

That really does make sense I just can't believe it would be that simple. We just learned induction so kind of figured that's what was going to be used. It's only excercises for now but I need to get this stuff down for the test coming up. I swear my professor isn't teaching this class right. One day we are doing induction the next it's logic and then the next day we are back at induction. I looked on ratemyproffesor to late, if I saw his reviews I probably wouldn't have taken the class. – Zooey Feb 24 '15 at 23:24

If a nonempty subset of integers is bounded from below, it has a minimum

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