Supremum of Sumset (Proof Writing)

Question

Given $A,b\subseteq\mathbb{R}$, define the set $A+B=\lbrace a + b | a\in A, b\in B\rbrace$. I would like to prove that $\sup(A+B)=\sup(A)+\sup(B)$, but in a specific way.

Here is what I have done so far:

If $s=\sup(A)$ and $t=\sup(B)$, then $s+t\geq a+b$ for all $a\in A$ and $b\in B$, so $s+t$ is an upper bound for $A+B$. Next, let $u$ be some other upper bound for $A+B$. Then, $u\geq a+b$, and $u-a\geq b$ for all $b\in B$. It follows that $u-a$ is an upper bound to $B$, so $u-a\geq t$.

I am then asked to conclude that $s+t=\sup(A+B)$, but am unsure of how to do this. I understand that the goal is to state $u\geq s+t$, as this would show that $s+t$ is the least upper bound.

Answer 1

Let us use the sequential characterisation of the supremum

$$s=\sup (A) \implies s=\lim_\infty a_n $$ with $$(a_n)\in A^{\Bbb N}$$

by the same, $$t=\lim_\infty b_n$$

thus $$s+t $$ is an upper bound of $A+B $ and

$$s+t=\lim_\infty (a_n+b_n) $$

with $(a_n+b_n)\in (A+B)^{\Bbb N} $.

We are done.

Answer 2

Rephrasing:

$\sup(A+B) = \sup(A) + \sup(B).$

1) $\sup(A+B) \le \sup(A) +\sup(B).$

Let $t:= \sup(A); s:=\sup(B).$

For $a\in A$, and $b \in B:$

$a \le t;$ $b\le s $

$\rightarrow :$

$ a + b \le t+s$, or

$\sup(A+B) \le \sup(A)+ \sup(B).$ (The easier part).

2) $\sup(A) + \sup(B) \le \sup(A+B)$

As you suggest :

Let $u$ be an upper bound of $a+b$, i.e.

$a+ b \le u $. Fix $a$ for the moment.

Then: $b \le u - a$, for all $b.$

$u-a$ is an upper bound for $b.$

Hence $u-a \ge s$, since $s$ is the least upper bound for $b$.

Now: $u - s \ge a$, for $a \in A.$

Hence $u-s$ is an upper bound for $a$,

$u-s \ge t$ , since $t$ is the least upper bound for $a$.

$u \ge s+t.$

Since $u$ is an upper bound for $a+b$, and

$u \ge s+t,$

we have for the least upper bound of $a+b$:

$\sup(A+B) \ge s+t.$

Answer 3

This is perhaps not quite an answer to your question, but it may still be helpful.$% \require{begingroup} \begingroup \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\subcalch}[1]{\\ \quad & \quad #1 \\ \quad &} \newcommand{\subcalc}{\quad \begin{aligned} \quad & \\ \bullet \quad & } \newcommand{\endsubcalc}{\end{aligned} \\ \\ \cdot \quad &} \newcommand{\Ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} \newcommand{\when}{\Leftarrow} %$

Here is the simplest definition of $\;\sup\;$ I know: for any upper-bounded $\;C\;$, $$ \tag{0} \sup(C) \le z \;\equiv\; \langle \forall c : c \in C : c \le z \rangle $$ for all $\;z\;$. (You are implicitly already using parts of this definition, it looks like.)

Now, let's show that the left and right hand sides of $$ \tag{1} \sup(A+B) = \sup(A) + \sup(B) $$ have the same upper bounds, starting with the most complex side, viz. the right hand side.

In other words, for every $\;z\;$, we calculate $$\calc \sup(A) + \sup(B) \le z \op\equiv\hint{arithmetic: rearrange -- to prepare for definition $\Ref{0}$} \sup(A) \le z - \sup(B) \op\equiv\hint{apply definition $\Ref{0}$} \langle \forall a : a \in A : a \le z - \sup(B) \rangle \op\equiv\hint{arithmetic: rearrange -- to prepare for definition $\Ref{0}$} \langle \forall a : a \in A : \sup(B) \le z - a \rangle \op\equiv\hint{apply definition $\Ref{0}$} \langle \forall a : a \in A : \langle \forall b : b \in B : b \le z - a \rangle \rangle \op\equiv\hint{arithmetic: rearrange -- to prepare for definition of $\;+\;$} \langle \forall a,b : a \in A \land b \in B : a + b \le z \rangle \op\equiv\hint{definition of $\;+\;$} \langle \forall c : c \in A+B : c \le z \rangle \op\equiv\hint{apply definition $\Ref{0}$} \sup(A+B) \le z \endcalc$$

Therefore (using $\;\langle \forall z :: c \le z \equiv d \le z \rangle \;\equiv\; c = d\;$) we have proved $\Ref{1}$.

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