Does every bounded total ordered set have a supremum/infimum?

Question

My question is really simple. I know intuitively every bounded total ordered set has supremum and infimum but I don`t know how to prove it formally. Must the set be complete?

Answer 1

Given that you tagged the question as "real-analysis" I will assume we are talking about $\Bbb R$.

We will prove that if $A \in \Bbb R$ is bounded, then $A$ has a supremum. The proof for the infimum is the same but for the set $B = -A$.

Remember that $\sup A = s \iff s \geq A \wedge \forall_{\epsilon > 0} ]s-\epsilon, s] \cap A \neq \emptyset$ where, by $s \geq A$ I mean that $s$ is an upper bound for $A$.

Let $m$ be an upper bound on $A$. Of course $m$ need not be the supremum, for example $m = 1, A = ]-4, 0[$, but we can use $m$ to find the supremum. Also, let us assume $A$ is non-empty.

Consider the set of all upper bounds of $A$; call it $M$. Obviously $m$ is in that set so that set is non-empty as well.

Let $f(x)$ be a function such that $f(x) = \epsilon_x$, the bigger $\epsilon$ such that $]x-\epsilon, x] \cap A = \emptyset$. Let us define $f(x) = 0$ if $x \in A$. Because $A \neq \emptyset$, we have $a \in A: f(a) = 0$. If we have $f(M) = 0$, you can show that $M$ is the supremum. Suppose $f(m) = k > 0$. Because $f$ is continuous, then $g(x) = f(x) - \frac{k}{2}$ also is, and $g(a) < 0, g(m) > 0$ so by the MVT, the function $g$ attains the value $0$ for some $s, a < s < m$ and we can show that $s$ is the supremum.