(adsbygoogle = window.adsbygoogle || []).push({}); Topology Proof: AcBcX, B closed --> A'cB'

1. The problem statement, all variables and given/known data

Prove:

AcBcX, B closed --> A'cB'

and where the prime denotes the set of limit points in that set

X\B is the set difference

2. Relevant equations

Theorem:

B is closed <--> For all b in X\B, there exists a neighborhood U of b with UcX\B

3. The attempt at a solution

Okay, so I am having no problem showing that the set of limit points of B is contained in B, if B is closed. I feel that there must be a way to extend this fact to show that A'cB', but I am not sure exactly how to do this.

From the theorem:

Since B is closed, there exists a neighborhood U of b with UcX\B, for all b in X\B

then, the intersection of B and U must be empty, AND the intersection of B and (U\{b}) must also be empty.

From the definition of limit point, if b is contained in X\B, then b cannot be contained in B'

or, for all x in B', x also is contained in B.

This is where I have gotten stuck.

Could I possibly show that B closed implies A closed, then use the same logic to show that A'cAcB ?

Thanks

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Topology Proof: AcBcX, B closed -> A'cB'

**Physics Forums | Science Articles, Homework Help, Discussion**