Proving Open Sets and Open Balls in Normed Spaces

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In summary: The converse is not necessarily true, as there could be points in S that are not limits of any sequence in X \setminus S.
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sigh1342
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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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  • #2
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 [itex]x \in S[/itex] such that for all [itex]r > 0[/itex], [itex]B(x,r) \not\subset S[/itex], and show that it follows that [itex]X \setminus S[/itex] 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 [itex]x \in S[/itex] there exists [itex]r > 0[/itex] such that [itex]B(x,r) \subset S[/itex]. Why does it follow that [itex]x \in S[/itex] cannot be the limit of any sequence in [itex]X \setminus S[/itex]?
 

FAQ: Proving Open Sets and Open Balls in Normed Spaces

What is an open set?

An open set is a subset of a metric space where any point within the set has a corresponding neighborhood that is also contained within the set. In other words, there are no boundary points in an open set.

How is an open set different from a closed set?

An open set contains all its interior points, while a closed set contains its boundary points. This means that an open set does not include its boundary, while a closed set does.

What is the definition of an open ball?

An open ball is a set of points within a given distance from a specific point, called the center. The distance is defined by a positive real number, known as the radius. An open ball is denoted as B(x, r), where x is the center point and r is the radius.

How is an open ball related to an open set?

An open ball is a specific type of open set. It contains all points within a given distance from a center point, without including the boundary points. This means that an open ball is a subset of an open set.

What is the significance of open sets and open balls in topology?

Open sets and open balls are fundamental concepts in topology, as they are used to define the topological structure of a space. They allow for the definition of continuity, convergence, and other important properties in mathematics and science.

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