Can open sets and closures intersect in a topological space?

[Shaji D R]

Suppose A and B are open sets in a topological Hausdorff space X.Suppose A intersection B is an empty set. Can we prove that A intersection with closure of B is also empty? Is "Hausdorff" condition necessary for that?

Please help.

 

[pasmith]

Suppose A and B are open sets in a topological Hausdorff space X.Suppose A intersection B is an empty set. Can we prove that A intersection with closure of B is also empty? Is "Hausdorff" condition necessary for that?

Please help.

Given that A and B are disjoint, the only way A can intersect with the closure of B is if there exists a \in A such that a is a limit point of B.

But that's impossible: A is an open neighbourhood of a which contains no points in B. Hence a is not a limit point of B.

This holds whether or not X is Hausdorff.

 

[economicsnerd]

Rephrasing pasmith's argument without reference to individual points...

Let X be any topological space, and suppose A,B\subseteq X are disjoint and A is open. Then:
-X\setminus A is closed because A is open.
-X\setminus A \supseteq B because A,B are disjoint.
-As the closure of B, the set \bar B is the smallest closed set that contains B.
-In particular, \bar B \subseteq X \setminus A.
Rephrasing the last point, A,\bar B are disjoint sets.

 

[Shaji D R]

Thank you very much

 

[ibdsm]

A \cap B empty

If A and B are disjoint, the B is a subset of the complement of A.

If A is open, its complement is closed.

Hence, in this case, the closure of B is contained in the complement of A.

Hence, A and the closure of B are disjoint.

There is no need for the ambient space to be Hausdorff. There is no need for B to be ooen.

 

[JG89]

If A and B are disjoint, the B is a subset of the complement of A.

If A is open, its complement is closed.

Hence, in this case, the closure of B is contained in the complement of A.

Hence, A and the closure of B are disjoint.

There is no need for the ambient space to be Hausdorff. There is no need for B to be ooen.

The space has to be Hausdorff and B has to be open. A = (0,1) and B = [1,0] is a counter example to what you say

 

[ibdsm]

1. [1,0] is the empty set, as there are no real numbers which are both at least 1 and at most 0. Hence your "counter-example" fails.

2. By definition,
(i) a subset of a topological space is closed if and only if it is the complement of an open set
(ii) the closure of a subset, B, of a topological space is the smallest closed subset of the space which contains B.

There is no question of being a Hausdorff space or even a T1 or T0 space for this.

 

[JG89]

Wow, I can't believe I wrote that. I was dead tired and my counter-example I was trying to type out was A = (0,1), B = [1,2]. But even that fails.

My apologies, I'll try not to post when I'm half awake anymore lol.

 

[ibdsm]

I'm glad I'm not the only one to blunder at times!