Open sets and closures

1. Jan 27, 2014

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?

2. Jan 27, 2014

pasmith

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.

3. Jan 27, 2014

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.

4. Jan 28, 2014

Shaji D R

Thank you very much

5. Feb 4, 2014

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.

6. Feb 5, 2014

JG89

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

7. Feb 5, 2014

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.

8. Feb 5, 2014

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.

9. Feb 5, 2014

ibdsm

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