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

Question

If $A,B$ non empty, upper bounded sets and $A+B=\{a+b\mid a\in A, b\in B\}$, how can I prove that $\sup(A+B)=\sup A+\sup B$?

Answer 1

Set $x=\sup A$, $y=\sup B$. Given $a\in A, b\in B$ we have $a\le x, b\le y$ so $a+b\le x+y$ so it's an upper bound. Now take $\varepsilon > 0$ and find $a,b$ such that $a>x-\varepsilon/2, b>y-\varepsilon/2$ and you have $a+b > x+y - \varepsilon$. That suffices (since it means that every potential "smaller upper bound" $x+y-\varepsilon$ is not really an upper bound).

Answer 2

Show that $\sup(A+B)$ is less than or equal to $\sup(A)+\sup(B)$ by showing that the latter is an upper bound for $A+B$. Then show that $\sup(A)+\sup(B)$ is less than or equal to $\sup(A+B)$ by showing that $\sup(A+B)$ is an upper bound for $A+\sup(B)$ and that $\sup(A+\sup(B)) = \sup(A)+\sup(B)$.

Answer 3

The following proofs are based on Question 2 here. The nub is to prove two inequalities: $$\sup(A+B) \geq \sup A+ \sup B \tag{1}$$

$$\sup(A + B) \leq \sup A + \sup B \tag{2}$$

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Proof of $(2)$:

By definition of $A + B$ and $\sup(A + B)$, for all $a \in A$ and $b \in B$,

$$\color{seagreen}{a + b} \color{seagreen}{\leq \sup (A + B)}.$$ Subtract $b$ from both sides:

$$a = \color{seagreen}{a + b} - b \color{seagreen}{\leq \sup (A + B)} - b.$$

Hence, if we fix $b \in B$, then $\color{seagreen}{\sup (A + B)} - b$ is an upper bound for $$\color{seagreen}{A + B} - B = A.$$

And so by definition of $\sup A$, for every $b \in B$, $$\sup A \leq \sup (A+ B) − b.$$ Rearrange the previous inequality: $\color{magenta}{b} \leq \sup(A +B) − \sup A$ for all $b \in B$.

Hence, $\sup(A +B) − \sup A$ is an upper bound for any $\color{magenta}{b}$.

By the definition of supremum, the previous inequality means: $$\color{magenta}{\sup B} \leq \sup(A + B) − \sup A \iff \sup A + \sup B \leq \sup(A + B).$$

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Proof of $(1)$:

Since $\sup A$ is an upper bound for $A$, $a \leq \sup A$ for all $a \in A$. Then $b \leq \sup B$ for all $b \in B$. Hence $a + b \leq \sup A + \sup B$ for all $x \in A$ and $y \in B$. Hence, $\sup A + \sup B$ is an upper bound for $A + B$. Hence, by definition of supremum, $\sup A + \sup B \geq \sup(A + B)$.

Answer 4

Another proof for $\sup(A + B) \geq \sup A + \sup B$.

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Proof when $\sup A + \sup B$ is finite:

Posit $e > 0.$ Then there exists $a \in A$ and $b \in B$ such that $a > \sup A − \frac{e}{2}$ and $b > \sup B − \frac{e}{2}$. Then $a + b \in A + B$. Ergo, $$\color{seagreen}{\sup(A + B)} \geq a + b \color{seagreen}{> \sup A + \sup B - e} \implies \color{seagreen}{ \sup(A + B) > \sup A + \sup B - e }.$$ Since $e > 0$ is arbitrary, $\sup(A + B) \geq \sup A + \sup B$

Proof when $\sup A + \sup B$ is infinite:

Since the sets here are nonempty, the suprema here are not equal to $-\infty$, so we're not in danger of encountering the undefined sum $-\infty +\infty$. If $\sup A + \sup B = + \infty$, then at least one of the suprema, say $\sup B$, equals $+\infty$. Select some $a_0 \in A$. Then $$\sup(A+B) \geq \sup(a_0 + B) = a_0 + \sup B = + \infty,$$ so $\sup(A+B) \geq \sup(A) + \sup(B)$ holds in this case. (Source)

Answer 5

$$\forall ~~a\in A,~~b\in B, a\le \sup A~~ \text{and }~~b\le \sup B \implies a+b \le a\le \sup A +\sup B$$

hence $$\sup A +B\le \sup A +\sup B$$ Now we know there exist $a_n\in A$ and $b_n\in B$ such that $$\sup A=\lim_{n\to\infty} a_n\qquad and \qquad\sup B =\lim_{n\to\infty}b_n$$

then $$\sup A+\sup B =\lim_{n\to\infty} a_n+\lim_{n\to\infty} b_n =\lim_{n\to\infty} a_n+b_n \le \sup A+B$$

whence $$\sup A+\sup B = \sup A+B$$