# Homework Help: Infimum of a set.

1. Aug 30, 2010

### ╔(σ_σ)╝

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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 ?

Last edited: Aug 31, 2010
2. Aug 31, 2010

### ╔(σ_σ)╝

Did I do something incorrectly ?

3. Aug 31, 2010

### Office_Shredder

Staff Emeritus
Looks good to me

4. Aug 31, 2010

### ╔(σ_σ)╝

Great! Thanks.