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

[FrancisD]

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

Homework Statement

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

Homework Equations

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

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

 

[FrancisD]

Okay, so I was able to prove that if AcBcX, then A'cB'.

Then, A'cB'cB