# Metric Spaces and Topology

• Oct 13th 2009, 03:07 PM
CollegeMathKid
Metric Spaces and Topology
A and B are any sets in a metric space (X, d)
Prove that
(A U B )' = A' U B'

' meaning interior.

And then prove that

(A U B ) closure = A closure U B closure

I know that you have to prove it both ways, and that the proofs will be relativley similar, and I know that they are true. I'm just really confused on where to get started
• Oct 13th 2009, 04:23 PM
Jose27
Quote:

Originally Posted by CollegeMathKid
A and B are any sets in a metric space (X, d)
Prove that
(A U B )' = A' U B'

' meaning interior.

This is not true: Take $\displaystyle X= \mathbb{R}$, $\displaystyle A=(0,1) \cap \mathbb{Q}$ $\displaystyle B= (0,1) \cap (\mathbb{R} - \mathbb{Q} )$ then $\displaystyle int(A)=int(B)=\emptyset$ but $\displaystyle int(A \cup B)=(0,1)$
• Oct 13th 2009, 04:42 PM
Plato
Quote:

Originally Posted by CollegeMathKid
A and B are any sets in a metric space (X, d)
Prove that
(A U B )' = A' U B', ' meaning interior.

That is false.
Let $\displaystyle A=[1,2]~\&~B=[2,3]$ then $\displaystyle 2\in(A\cup B)^o=(1,3)$.

BUT $\displaystyle 2\not\in A^o\cup B^o=(1,2)\cup(2,3)$.
• Oct 13th 2009, 05:59 PM
redsoxfan325
However, you should be able to prove that $\displaystyle int(A)\cup int(B)\subseteq int(A\cup B)$.