Accumulating point and closures

[oumaima1]

Please help me in this question
I failed to solve it!
Considering (E,X) a topological space,consider F a closed of E and x an accumulating point of F,show that x is an element of F ^^

 

[HallsofIvy]

What did you try? You are told that F is a closed subset of E. What is the definition of "closed subset"? You are told that x is an accumulating point of F. What is the definition of "accumulating point"? (Does your text actually use "accumulateing point" or the more common "accumulation point"?)

 

[oumaima1]

What did you try? You are told that F is a closed subset of E. What is the definition of "closed subset"? You are told that x is an accumulating point of F. What is the definition of "accumulating point"? (Does your text actually use "accumulateing point" or the more common "accumulation point"?)

Yes,please it's accumulation point sorry
the definition i am told is:
Definition :consider (E,X) a topological space
A part F of E is told closed if ∁F is an open subset of E.We say also that F is a closed subset of E.
Definition:
Consider (E,X) a topological space ,A⊂E and x∈E,we say x is an accumulation point of A if:

∀O∈X,x∈O⇒(O∖{ x })∩A≠∅

 

[HallsofIvy]

Yes,please it's accumulation point sorry
the definition i am told is:
Definition :consider (E,X) a topological space
A part F of E is told closed if ∁F is an open subset of E.We say also that F is a closed subset of E.
Definition:

So F is closed if and only if its complement is open.

Consider (E,X) a topological space ,A⊂E and x∈E,we say x is an accumulation point of A if:

∀O∈X,x∈O⇒(O∖{ x })∩A≠∅

This isn't quite correct. if "O∈X" then O is a member of X, not a subset so "x∈O" makes no sense. I think you mean "for \(\displaystyle O\subset X\)" and you mean O to be an open set, right?

The best way to prove this is by "contradiction". Suppose x is NOT in F. Then x is in the complement of F which is open- x is an interior point of the complement of F which, in turn, means that there exist an open set, a subset of complement of F, containing x.

 

[oumaima1]

So F is closed if and only if its complement is open.


This isn't quite correct. if "O∈X" then O is a member of X, not a subset so "x∈O" makes no sense. I think you mean "for \(\displaystyle O\subset X\)" and you mean O to be an open set, right?

The best way to prove this is by "contradiction". Suppose x is NOT in F. Then x is in the complement of F which is open- x is an interior point of the complement of F which, in turn, means that there exist an open set, a subset of complement of F, containing x.

Many thanks it worked