# Metric Spaces and Topology

• Oct 13th 2009, 04:07 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, 05:23 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 $X= \mathbb{R}$, $A=(0,1) \cap \mathbb{Q}$ $B= (0,1) \cap (\mathbb{R} - \mathbb{Q} )$ then $int(A)=int(B)=\emptyset$ but $int(A \cup B)=(0,1)$
• Oct 13th 2009, 05:42 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 $A=[1,2]~\&~B=[2,3]$ then $2\in(A\cup B)^o=(1,3)$.

BUT $2\not\in A^o\cup B^o=(1,2)\cup(2,3)$.
• Oct 13th 2009, 06:59 PM
redsoxfan325
However, you should be able to prove that $int(A)\cup int(B)\subseteq int(A\cup B)$.