Suppose we have the following:
$X = A + B $ where $X, A,$ and $B$ are any sets. $A + B = \{ a + b : a \in A , \; \; \; b \in B \} $ Can we conclude that $\sup X = \sup A + \sup B $ ?
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Yes. One way of defining the supremum of a set $X$ is that $\sup X$ is the unique number satisfying:
- $x\leq \sup X$ for all $x\in X$,
- there exists a sequence $(x_n)\subseteq X$ such that $\sup X=\lim_{n\to\infty} x_n$.
Now, suppose $X=A+B$, and let us check that $\sup A+\sup B$ satisfies the two conditions above. First, we obviously have that $a+b\leq \sup A+\sup B$ for all $a\in A$ and $b\in B$. Second, since $\sup A$ and $\sup B$ are supremums of $A$ and $B$ we know that there exists sequences $(a_n)\subseteq A$ and $(b_n)\subseteq B$ such that $$ \sup A=\lim_{n\to\infty}a_n,\quad\text{and}\quad \sup B=\lim_{n\to\infty} b_n. $$ But then $x_n=a_n+b_n$ is a sequence from $X$ satisfying $\lim_{n\to\infty} x_n=\sup A+\sup B$. Therefore, we conclude that $\sup X=\sup A+\sup B$.
Stefan Hansen
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Thanks for your reply. I Would like to ask you if the result holds if $A$ and $B$ are disjoint ?? – Sep 30 '13 at 06:00
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I haven't used that $A$ and $B$ should be disjoint, and hence it holds regardless. – Stefan Hansen Sep 30 '13 at 06:16