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I have homework for tommorow to prove that: $$ \inf(A+B) = \inf A + \inf B $$

I tried to do this:

assume a ∈ A

$a >= \inf A$ (for every a ∈ A)

assume b ∈ B

$b >= \inf B $ ( for every b ∈ B)

therefore: a+b >= infA + infB (for every a,b ∈ A+B)

infA + infB is a lower bound of A+B

I asked my teacher about this solution, and he said that it's only proof that it's a lower bound and not a Inf. how can I proof that?

please use simple mathmatics because I am not so smart :) I know that there are 2 question like this in the site, but I understood only stage 1 (which I wrote).

Bernard
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Silas2033
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1 Answers1

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There are two properties that a number must have in order to be considered an infimum (greatest lower bound). You have indeed only used the single property of being a lower bound, but you have not included the property that makes a number a greatest lower bound. You only need to modify your proof slightly.

A number, $\alpha$, is called the greatest lower bound, or infimum, of a set of numbers A if and only if $\alpha$ is a lower bound of A, and $\alpha$ is the greatest member of the set of lower bounds of A. In symbolic notation, we require:

1) For each $a\in A$, $\alpha \leq a$.

and

2) If $b \leq a$ for all $a\in A$, then $b\leq \alpha$.

Only a number $\alpha$ that satisfies both of these properties can be called an infimum, or greatest lower bound, of A.

So your modified proof would look like this:

Assume $\inf A$ exists and $\inf B$ exists. Then, if $a\in A$, we must have

1) $a\geq \inf A$

and

2) for each number $c$, if $c\leq a$ for all $a\in A$, then $c\leq \inf A$.

Similarly, if $b\in B$, then

3) $b\geq \inf B$

and

4) for each number $c$, if $c\leq b$ for all $b\in B$, then $c\leq\inf B$.

Assuming that $A + B$ is the set of all numbers of the form $a + b$ with $a\in A$ and $b\in B$, we can add (1) and (3) above to get the inequality:

5) If $a\in A$ and $b\in B$, then $a + b \geq \inf A + \inf B$.

Therefore, we know that if $c\in A + B$, then $c \geq \inf A + \inf B$, so $\inf A + \inf B$ is a lower bound of $A + B$. Thus, $\inf A + \inf B \leq \inf(A + B)$. Now we must show it is the greatest lower bound.

To do this, we will show that $\inf A + \inf B \geq \inf(A + B)$.

For each $\epsilon > 0$, choose $a\in A$ such that $a - \inf A < \frac{\epsilon}{2}$, and choose $b\in B$ such that $b - \inf B < \frac{\epsilon}{2}$.

Then, adding the two inequalities tells us that, for all $\epsilon > 0$, there exists a number $(a+b)\in (A + B)$ such that $$(a + b) - (\inf A + \inf B) < \epsilon$$

Therefore, for each $\epsilon > 0$, $$\inf(A + B) \leq a + b < \inf A + \inf B + \epsilon$$

Thus, $\inf(A + B) \leq \inf A + \inf B$ and $\inf(A + B) \geq \inf A + \inf B$. By the trichotomy law, this implies $\inf(A + B) = \inf A + \inf B$.

Ronald Lett
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