MHB Is the Supremum of the Sum of Two Sets Equal to the Sum of their Suprema?

  • Thread starter Thread starter alexmahone
  • Start date Start date
alexmahone
Messages
303
Reaction score
0
Let $A$ and $B$ be sets of real numbers and write $$C=\{x+y:x\in A,y\in B\}.$$ Find a relation among $\sup A$, $\sup B$, and $\sup C$.

My attempt:

I'm assuming the answer is $\sup C=\sup A+\sup B$.

$x\le \sup A\ \forall x\in A$
$y\le \sup B\ \forall y\in B$
$\implies x+y\le \sup A+\sup B$ $\forall x\in A,\ y\in B$

So, $\sup A+\sup B$ is an upper bound for $C$.

Suppose $\sup C\neq\sup A+\sup B$.

$\implies \exists l<\sup A+\sup B$ such that $x+y\le l$ $\forall x\in A,\ y\in B$

$\implies x\le l-y\ \forall x\in A,\ y\in B$

So, $l-y$ is an upper bound of $A\ \forall y\in B$

I feel that I'm on the right track but I don't know how to get a contradiction. Any suggestions?
 
Last edited:
Physics news on Phys.org
Prove that $\sup A+\sup B-\varepsilon$ is not an upper bound of $C$ for any $\varepsilon>0$. For this find a $c\in C$ such that $c>\sup A+\sup B-\varepsilon$.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K