Is the following true about sups and infs?

[AxiomOfChoice]

I am trying to determine whether the following is true.

Let $A\subset \mathbb R$, and let $|A| = \{|a|:a\in A\}$. Then $\sup |A| = \max \{|\inf A|, |\sup A|\}$.

I think I could figure this out on my own, but there is something else demanding my attention right now...so, hopefully this isn't too tough. Thanks!

 

[micromass]

Seems true... But it still needs to be proven. How would you begin such a proof?

 

[Dr. Seafood]

I'm on the phone now, but I can't help but read this :p

I think, show different cases: if |inf A| < |sup A|, show that that sup|A| = |sup A|; etc.

 

[HallsofIvy]

First note that, for any a in A, $inf(A)\le a\le sup(A)$ so that $|a|\le max(|sup(A)|, |inf(A)|)$ so that "max(|sup(A)|, |min(A)|)" is an upper bound. Now, to show it is the least upper bound, show that there must be a in A such that |a| is arbitrarily close to either |sup(A)| or |inf(A)|, whichever is larger.