Need someone to verify my logic

[quasar_4]

Homework Statement

Find the supremum and infimum of the following sets and test whether these sets have a maximum or minimum:

(a) { |x|/(1+|x|) s.t. x is in R}

(b) { x/ (1+x) s.t. x > -1}


Homework Equations

Order Axioms and Field Axioms for the real numbers; infimum and supremum definitions; min and max definitions.

The Attempt at a Solution

I just don't see why these aren't effectively the same in terms of their supremums and infimums, min and max. Since |x|> 0 for nonzero x, |x| will always be positive, so this ratio will always be positive. Furthermore, it follows immediately that |x| < 1 + |x| for the same reason. Since the denominator is always bigger, the ratio is always less than 1. So I would argue that the supremum of (a) is 1. Now since we can include any x in R, the set also contains 0, and none of these can ever be less than 0. So 0 should be the infimum and serve as the minimum element of the set (since it is contained in the set).

Now I can see how (b) would differ for all x in R, but we are now restricting x > -1, i.e.,
x = 0, ... So it would seem that we have the exact same set - no negative x, the ratio is always less than 1, etc. So wouldn't this be exactly the same? What am I missing, if not?

 

[HallsofIvy]

I would think the are very obviously NOT the same. If x= -1.00001, what is x/(x+ 1)?

 

[quasar_4]

oh yeah, I guess there are a bunch of negative reals between -1 and 0... wow that should have been obvious. I knew I was missing something simple. It always seems to be the case for me in analysis...

since we could go "closer and closer" each time to -1, does that mean that this set doesn't have a lower bound? I mean, if you put in something like -0.99, then we get -0.99/(1-0.99) = -99, but then we could try x= -0.9999 and get something even smaller, and then we could keep going forever.
I'm sorry if it's trivial.. I'm pretty dumb when it comes to this stuff.