Proving a set has a lower bound

Question

Is there a specific way to prove that a set has a lower bound (or an upper bound for that matter)?

A lot of the examples I see just end up stating that there exists a lower bound and I'm not really sure where that comes from... Any help would be appreciated!

Answer 1

Think about the definition of lower bound.
Suppose $A\subset\mathbb{R}$ and let $\delta\in\mathbb{R}$

If $\delta$ is a lower bound of $A$, then $\forall x\in A, \delta \le x$

Prove that there is such a delta.

Answer 2

Suppose you have a set $S \subseteq \mathbb{R}$ and real number $a \in \mathbb{R}$. To say "$a$ is a lower bound of $S$" is to say that $a \le s$ for all $s \in S$. To prove this, you show exactly that: start with an arbitrary $s \in S$, use the definition of membership of $S$, then via some logical path, prove that $a \le s$.

To take your example, take $S = \lbrace x > 0 : x \in \mathbb{R}\rbrace$, and $a = 0$. Note that this is indeed quite obvious, and the proof is going to rightfully feel like we are just stating the obvious!

Start by supposing $s \in S$. What does this tell us; what is the definition of $S$? This means that $s > 0$, which, by definition of $>$, means that $s \ge 0$ and $s \neq 0$. Therefore, $s \ge 0$. So, in total, we've shown that for all $s \in S$, we have the inequality $s \ge 0$, which is what we need to show a lower bound. Therefore $0$ is a lower bound for $S$.

How do you prove a lower bound exists? Often this just involves finding an explicit number that bounds your set below. Sometimes it's a non-constructive argument by contradiction, where a lower bound is assumed to exist, and a contradiction is reached (e.g. proving the boundedness of continuous functions). It's not always easy to extract a concrete lower bound from doing this.