Closed Subset Addition in Metric Spaces: Real Analysis Homework Help

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SUMMARY

The discussion centers on the question of whether the sum of two closed subsets E and F of the real numbers R, defined as E + F = { e + f | e ∈ E, f ∈ F }, is necessarily closed. Participants highlight that E + F is closed if at least one of the sets E or F is compact. Examples provided include A = {..., -4, -3, -2, -1} and B = {1 + 1/2, 1 + 1/2 + 1/3, ...}, illustrating the complexity of the problem and the necessity for deeper analysis.

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  • Understanding of metric spaces and the definition of closed sets.
  • Familiarity with compactness in real analysis.
  • Knowledge of the properties of real numbers and their subsets.
  • Ability to manipulate and analyze sequences and series in mathematical contexts.
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  • Study the properties of compact sets in metric spaces.
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  • Explore examples of closed sets and their sums in various contexts.
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Students and educators in real analysis, mathematicians focusing on metric spaces, and anyone seeking to deepen their understanding of the properties of closed sets and their sums in mathematical contexts.

Mr_Physics
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Homework Statement



Let E, F be two closed and non-empty subsets of R, where R is seen as a metric space with the distance d(a,b)=|a-b| for a,b ϵ R.

Suppose E + F := { e+f |e ϵ E, f ϵ F}. Is is true that E+F has to be closed?

Homework Equations





The Attempt at a Solution



I'm not sure how to start this one.
 
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This is not an easy one, by far!

But consider:

[tex]A=\{...,-4,-3,-2,-1\}[/tex]

and

[tex]B=\{1+1/2,1+1/2+1/3,1+1/2+1/3+1/4,...\}[/tex]

Note that E+F is closed if one of E or F is compact...
 
micromass said:
This is not an easy one, by far!

But consider:

[tex]A=\{...,-4,-3,-2,-1\}[/tex]

and

[tex]B=\{1+1/2,1+1/2+1/3,1+1/2+1/3+1/4,...\}[/tex]

Note that E+F is closed if one of E or F is compact...

I'll agree it's not easy. It does take some head scratching. Here's another one to think about on a different line. Take A=Z (the integers) and B=Z*sqrt(2). micromass's suggestion is really pretty clever though once you figure it out.
 
Last edited:

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