Is the Sum of Two Closed Sets in R^n Always Closed?

Click For Summary
SUMMARY

The sum of two closed sets in R^n, denoted as A+B = {x+y | x in A and y in B}, is not always closed. A counterexample is provided where A = B = {0}, leading to A+B = {0}, which is closed, but also shows that x+y can be in (A+B)^c when x and y are chosen from the complements of A and B. The discussion emphasizes the importance of finding counterexamples to test the validity of mathematical statements, particularly when both sets are infinite.

PREREQUISITES
  • Understanding of closed sets in topology
  • Familiarity with R^n and its properties
  • Knowledge of converging sequences and their limits
  • Basic concepts of set operations and complements
NEXT STEPS
  • Explore the properties of closed sets in topology
  • Study counterexample techniques in mathematical proofs
  • Investigate the implications of finite versus infinite sets in R^n
  • Learn about the closure of sets and their algebraic operations
USEFUL FOR

Mathematicians, students studying topology, and anyone interested in the properties of closed sets and their operations in R^n.

teacher2love
Messages
2
Reaction score
0
Let A, B in R^n be closed sets. Does A+B = {x+y| x in A and y in B} have to be closed?

Here is what I've tried. Let x be in A^c and y in B^c which are both open since A & B are closed. So for each x in A^c there exists epsilon(a)>0 s.t. x in D(x, epsilon(a) is subset of A^c. For each y in B^c there exists epsilon(b)>0 s.t. y in D(y, epsilon(b)) is a subset of B^c.

Can I add the two together to get x+y in (A+B)^c to show that there exist epsilon > 0 s.t. D(x+y, epsilon) is in (A+B)^c. Thus (A+B)^c is open => (A+B) is closed.

Thanks for your help.
 
Physics news on Phys.org
It is often much easier to work with sequences when you can!

A set is closed if the limit of every converging sequence of elements of that set lies in the set.
 
teacher2love said:
Can I add the two together to get x+y in (A+B)^c to show that there exist epsilon > 0 s.t. D(x+y, epsilon) is in (A+B)^c. Thus (A+B)^c is open => (A+B) is closed.

It's not true that x+y will be in (A+B)^c. (almost any example you can think of will show this, but to take a simple one, let A=B={0}, x=-y=1).

A good strategy on these types of problems (where you're not sure if the given statement is true or false) is to start by trying to find counterexamples. If you find one, you're done, and if not, try to see what's preventing you from finding one.

For example, you might notice that you can't find any counterexamples when one of the sets is finite. Well, this is just because, as is easy to prove, A+B is closed when one of A or B is finite. So you can continue, now looking only at examples where both A and B are infinite.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 58 ·
2
Replies
58
Views
5K
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K