Prove $\inf (A + B) = \inf A + \inf B$

Question

Possible Duplicate:
How can I prove $\sup(A+B)=\sup A+\sup B$ if $A+B=\{a+b\mid a\in A, b\in B\}$

Prove:

If $A$ and $B$ are nonempty subsets of $\mathbb{R}$ and are both bounded above, then $$\operatorname{lub}(A+B)=\operatorname{lub}(A)+\operatorname{lub}(B).$$

I am thinking I need to utilize the fact that $A+B=\{a+b: a \in A, b \in B\}$, I'm just not sure how this gives a proof.

Thanks.

You can find some good starting points on how to format mathematics on the site here and here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. — Zev Chonoles, Feb 04 '13 at 05:54
Thanks, my LaTeX knowledge is passable, but it definitely needs some work — Peej Gerard, Feb 04 '13 at 06:01
Possible duplicates: http://math.stackexchange.com/questions/82680/ http://math.stackexchange.com/questions/4551/ http://math.stackexchange.com/questions/30918 — mrf, Feb 04 '13 at 08:30

providence · Accepted Answer · 2013-02-04T08:19:08.550

First we show that $\sup(A + B) \leq \sup(A) + \sup(B)$:

Let $\gamma$ in $A + B$ be arbitrary. We see that $$\gamma = a + b \quad \text{where} \quad a \in A \text{ and } b \in B$$ Now, we know that $$\sup(A) \geq a \quad \text{and} \quad \sup(B) \geq b$$ so it follows that $$\gamma = a + b \leq \sup(A) + \sup(B)$$ Since $\gamma$ was arbitrary, this holds for all elements of $A+B$. We must have $$\sup(A + B) \leq \sup(A) + \sup(B)$$

Next, we show that $\sup(A + B) \geq \sup(A) + \sup(B)$:

Well, assume the contrary. Then we have $$ \sup(A + B) < \sup(A) + \sup(B)$$ This implies that for all $\gamma \in A + B$ there exists some $a \in A$ and some $b \in B$ such that $\gamma < a + b$. But $(A + B) \ni \sigma = a + b$, so this is a contradiction.

Putting these two together, we see that $\sup(A + B) = \sup(A) + \sup(B)$ as we wanted.

Prove $\inf (A + B) = \inf A + \inf B$

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