In my analysis textbook, the definition of a "complete" set is as follows: Let $S$ be an ordered field. Then $S$ is complete if $sup(A) \in S$ for every nonempty $A \subseteq S$ where $A$ is bounded above. It is left as an exercise to show the existence of greatest lower bounds. Here is my unfinished attempt (I want to show that $inf(A) = -sup(-A)$ as suggested by the book):
Suppose $S$ is complete, and $A \subseteq S$ is nonempty and bounded below.
Then $-A \subseteq S$ is also nonempty and bounded above.
So, $\exists c_0 \in S$ st $c_0 \geq x$$\forall x \in -A$ and $c_0 \leq c$ for all other upper bounds $c$
$c_0 \geq x$ for all $x \in -A$ implies $-c_0 \leq -x$ for all $-x \in A$.
Hence, $c_0$ is a lower bound for $A$.
Now, if $c$ is an arbitrary upper bound for $-A$, $c \geq x$$\forall x \in -A$
So, $-c_0 \leq -x$$\forall -x \in A$.
Also, we knew from before that $c_0 \leq c$, so $-c_0 \geq -c$
This means $-c_0 = -sup(-A) = inf(A)$.
My issue is that I never was able to show that if every lower bound is of the form $-c$ where $c$ is an upper bound of A, so the logic from the 2nd to last line to the last line of the proof seems faulty. Also, my textbook has not yet stated that being bounded below implies the existence of $inf(A)$ so I can't start the proof with "let $α=inf(A)$ as was done in another post of a similar problem.
Thanks in advance!