Interior points of the closure of A

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Homework Help Overview

The discussion revolves around properties of interior points and closures in the context of metric spaces. Participants are examining the relationship between the set of interior points of a subset and the closure of that subset.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Some participants explore the implications of taking a set A as open and test the statement regarding interior points with examples. Others are attempting to prove a theorem related to the intersection of an open set and the closure of a subset, discussing both directions of the proof.

Discussion Status

The discussion includes attempts to prove specific theorems and clarify definitions. Some participants have made progress in one direction of a theorem, while others express uncertainty about proving the opposite direction. There is an ongoing exploration of definitions and properties related to closures and interior points.

Contextual Notes

Participants are working within the constraints of a metric space and discussing the implications of open sets and their intersections with arbitrary subsets. There is mention of using contradiction in proofs, and some participants are questioning the clarity of certain statements made in the discussion.

golriz
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Is it true?
" Set of interior points of the closure of A equals the set of interior points of A. "
 
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golriz said:
Is it true?
" Set of interior points of the closure of A equals the set of interior points of A. "

What did you try already?

Maybe you can start by taking A open. Then your statement says that A equal the interior of the closure of A. Try it with some easy/less easy examples of open sets first.
 
I want to prove this theorem:
" Let U be an open subset of a metric space X,and A be an arbitrary subset of X. Prove that the intersection of U and closure of A is empty if and only if the intersection of U and A be empty. "
I've proved one direction of this theorem:

If the intersection of U and closure of A is empty then the intersection of U and A is empty too.
The closure of A is equal to the union of A and the set of all limit points(accumulation points) of A. Then we can use this definition of the closure of A. then after substitution, we have:
[ intersection of A and U ] U [intersection of U and the set of limit points of A] = empty set
so it says that both the sets [ intersection of A and U ] and [intersection of U and the set of limit points of A] should be empty.
 
But I don't know how to prove the opposite direction of this question.
I have tried:

using contradiction, it means the intersection of U and cl(A) is not empty.
 
golriz said:
I want to prove this theorem:
" Let U be an open subset of a metric space X,and A be an arbitrary subset of X. Prove that the intersection of U and closure of A is empty if and only if the intersection of U and A be empty. "
I've proved one direction of this theorem:

If the intersection of U and closure of A is empty then the intersection of U and A is empty too.
The closure of A is equal to the union of A and the set of all limit points(accumulation points) of A. Then we can use this definition of the closure of A. then after substitution, we have:
[ intersection of A and U ] U [intersection of U and the set of limit points of A] = empty set
so it says that both the sets [ intersection of A and U ] and [intersection of U and the set of limit points of A] should be empty.

This can be made easier. Just use that A\subseteq \overline{A}.

For the other direction, use the definition of closure. And use that A is an open set that doesn't intersect U.
 
Then the intersection of set of interior points of U (that equals to U, since U is open) and set of interior points of A equals to the set of interior points of the intersection of (U and A). And the set of interior points of (U and cl(A) ) equals to intersection of U and the set of interior points of cl(A) that is not empty. can I use:
" Set of interior points of the closure of A equals the set of interior points of A. "
to show the contradiction?
 
sorry!
what is?
" Just use that A\subseteq \overline{A}. "
 
oh! at first it was typed with vague characters! but now it is in correct form
 
Thank you very much!for your help
 

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