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Q. i.prove that if A is open and B is arbitrary subset of $R^n$ then A+B ={x+y : $x\in A$, y $\in B$ } is open.

ii.show that if A and B are closed subsets of R then A+B need not be closed.

my doubt:

in this question do i have to show that there exist a e>0 s.t ball of radius 'e' is contained in the set? or any other approach simpler than this? what about second part?I think it demands a counter example.

Foggy
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2 Answers2

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For 1) just write the the sum as a union of translations of open sets. Then all that's left to prove is translations of open sets are open. This is easy for intervals, so it's easy in general.

For 2) consider $A=\{-n| n\in \mathbb{N}\}$ and $B=\{n+\frac{1}{n+1} | n\in \mathbb{N}\}$. Then $0\notin A+B$ as the sum of an integer and a non-integer cannot be an integer. But $0$ is a limit point. Both sets are clearly closed. So we're done.

Also today the naturals shouldn't include $0$ for this to work.

Zach Stone
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  • sum as a union of translations of open sets? how do u do that – Foggy Sep 05 '15 at 07:02
  • It's clearly true if $B$ is a singleton. That's just the definition of translation and a trivial union. If B were two elements, the sum is the union of two different translations of A. Now write it in general. – Zach Stone Sep 05 '15 at 07:05
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Hint for b.

If $A$ and $B$ are closed and bounded then $A+B$ is closed. So you need to look at unbounded closed sets.

mathcounterexamples.net
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