# Closure of intersection is a subset of intersction of closures

• May 12th 2011, 12:01 AM
magus
Closure of intersection is a subset of intersction of closures
I just want to check that this proof is correct. This is from a section on neighbourhoods.

$\overline{(A\cap B)}\subseteq \bar{A}\cap \bar{B}$

It suffices to prove that the set of limit points of the intersection is a subset of the intersection of the set of limit points.

$(A\cap B)' \subseteq A' \cap B'$

Let p be a limit point of $(A\cap B)$. Then there exist neighbourhoods in $A$ or $B$ that contain that point and other points. In fact every neighbourhood in $(A\cap B)$ does. This means that the above statement is correct. From there we may compute

$(A\cap B) \cup (A\cap B)' \subseteq (A\cap B) \cup A' \cap B'$
$\overline{(A\cap B)}\subseteq \bar{A}\cap \bar{B}$

As required.

Is this correct?

There's also a second question where you have to prove that there is not equality and to come up with an example. My example is kind of inelegant and I was wondering if someone could show me another.

Mine is this.

Take two sets that are the union of two circles each $B((1,1),1)\cup B((\frac{1}{2 },-\frac{1}{2 } ),3/4)$ and $B((-1,1),1)\cup B((-\frac{1}{2 },-\frac{1}{2 } ),3/4)$

The two larger circles share a limit point at $(0,1)$ which is outside of the intersection.

Is there a better example for this?
• May 12th 2011, 02:29 AM
Plato
Let $A = (0,1)\;\& \,B = (1,2)$.
Then $\overline A \cap \overline B = \{ 1\} \;\& \;\overline {A \cap B} = \emptyset$
• May 12th 2011, 02:47 AM
magus
Ahh that's much more simple. Thank you.

In retrospect that means that I could have just used the two large circles. Oh well.