# 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!