Infimum of a Set: Proving inf(A+B) = inf(A) + inf(B)

[╔(σ_σ)╝]

Homework Statement

Let A \subset R and B \subset R be bounded below and define A+B = {x+y| x\in A and y \inB}. Is it true that inf (A+B) = inf(A) + inf(B) ?

Homework Equations

The Attempt at a Solution

First I proved that inf(A+B) exist by doing the following.

inf(A) \leq x and inf(B) \leq y for all x and y so,

x- inf(A) + y-inf(B) \geq 0

x + y \geq inf(A) + inf(B)

So the set A +B is bounded below and therefore has a greatest lower bound; that is, inf(A+B) exist.

My next step was to show that inf(A+B) = inf(A) + inf(B).

To do this, I showed did the following: Suppose there is a number \alpha which is a lower bound of A+B and \alpha > inf(A) + inf(B) . I tried to derive a contradiction.\alpha > inf(A) + inf(B) and since \alpha is a lower bound...
\alpha\leq x+ y for all x and y ( which are elements of A and B, respectively )

\alpha = inf(A) + inf(B) + \epsilon
\alpha = inf(A) + \frac{\epsilon}{2} + inf(B)+\frac{\epsilon}{2}

We know that since inf(A) and inf(B) exist there is an x_{0}\in A and y_{0}\in B such that

x_{0} < inf(A) + \frac{\epsilon}{2} and

y_{0} < inf(B) + \frac{\epsilon}{2}.

So this means

\alpha > x_{0} + y_{0} which is a contradiction since \alpha is a lower bound.

From this it is clear that
inf(A+B) = inf(A) + inf(B)How is this ? Is it clear enough ?

 

[╔(σ_σ)╝]

Did I do something incorrectly ?

 

[Office_Shredder]

Looks good to me

 

[╔(σ_σ)╝]

Great! Thanks.