Proving Open Sets and Open Balls in Normed Spaces

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SUMMARY

The discussion focuses on proving that a set S in a normed space X is open if and only if for every point x in S, there exists an open ball B(x; r) with radius r > 0 such that B(x; r) is a subset of S. Participants explore the contrapositive approach, suggesting that if such an open ball does not exist for some x in S, then the complement X\S is not closed. The conversation emphasizes the relationship between open sets and closed sets in the context of normed spaces.

PREREQUISITES
  • Understanding of normed spaces
  • Familiarity with the concepts of open and closed sets
  • Knowledge of sequences and limits in mathematical analysis
  • Basic proficiency in mathematical logic and proofs
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  • Study the properties of open and closed sets in metric spaces
  • Learn about the definitions and implications of convergence in normed spaces
  • Explore the concept of the contrapositive in mathematical proofs
  • Investigate the relationship between open balls and neighborhoods in topology
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Mathematics students, particularly those studying real analysis or topology, as well as educators seeking to deepen their understanding of open sets in normed spaces.

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Homework Statement

Show that S is open if and only if ∀x ∈ S, ∃ a open ball B(x; r)(r > 0) such that B(x; r) ⊂ S
And what we have is , let X be normed space, S ⊂X , Then S is close if and only if $$∀{x_{n}}⊂X, s.t. x_{n}->x \in X$$, x∈ S. A set S is open if and only if X\S is close.


The attempt at a solution
I try to use following approach.
"=>" since S is open, there exist some sequence s.t. its limit is not in S(otherwise X/S is not close), then for any x, we can construct the convergent sequence such that its limit is not in S.(I don't know it is true or not). So by the limit definition. ∃ a open ball B(x; r)(r > 0) such that B(x; r) ⊂ S.

"<=" I try to prove X\S is close, so that X is open. $$∀{x_{n}}⊂X, s.t. x_{n}->x \in X$$ , x should be in X\S, but I have no idea to prove it.um.. Can anyone give me some ideas? :)
 
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sigh1342 said:
Homework Statement

Show that S is open if and only if ∀x ∈ S, ∃ a open ball B(x; r)(r > 0) such that B(x; r) ⊂ S
And what we have is , let X be normed space, S ⊂X , Then S is close if and only if $$∀{x_{n}}⊂X, s.t. x_{n}->x \in X$$, x∈ S. A set S is open if and only if X\S is close.


The attempt at a solution
I try to use following approach.
"=>" since S is open, there exist some sequence s.t. its limit is not in S(otherwise X/S is not close), then for any x, we can construct the convergent sequence such that its limit is not in S.(I don't know it is true or not).

I'm not sure this even makes sense.

I would prove this by contrapositive: Assume that there exists an x \in S such that for all r &gt; 0, B(x,r) \not\subset S, and show that it follows that X \setminus S is not closed.

"<=" I try to prove X\S is close, so that X is open. $$∀{x_{n}}⊂X, s.t. x_{n}->x \in X$$ , x should be in X\S, but I have no idea to prove it.um.. Can anyone give me some ideas? :)

Your assumption is that for all x \in S there exists r &gt; 0 such that B(x,r) \subset S. Why does it follow that x \in S cannot be the limit of any sequence in X \setminus S?
 

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