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I have a few questions to ask. They are simple, but the purpose is to make sure I'm staying on track.
First, the definitions that I will use.
T2-Space or Hausdorff Space, find http://http://mathworld.wolfram.com/T2-Space.html" here.
Separable Space, find http://mathworld.wolfram.com/SeparableSpace.html" here.
Let X be the underlying set.
The questions or statements are...
1 - The Space of Reals is a T2-Space. This seems rather obvious since you can just construct the intervals for x and y (x<>y) such that they do not overlap. Hence, two neighbourhoods that are disjoint.
2 - The Discrete Topology is separable if and only if X itself is countable.
Note: A- is the intersection of the collection of closed subsets in X that contain A. Therefore, if A is closed, A- = A.
Note: A is dense in X if and only if A- = X.
Note: <> means not equal to.
Now, using the notes above.
X can be the only possible solution in this case because every subset is closed, therefore A- <> X, unless A=X. Therefore, if X is countable, it is separable.
Is that right?
3 - The Indiscrete Topology is separable if and only if there is countable subset in X. If X is the set of real numbers, then it is separable because the naturals numbers is a countable subset in R, and it is dense because N- = X, because X is the only set that contains N.
4 - The Space of Reals is separable. Since the rational numbers are countable and it is a subset of R, and R is the only closed set that contains Q then Q- = R. Hence, it is separable.
5 - The Finite Complement Topology is separable. If X is countable, then we are done. If X is not countable, then we can create a countable subset (call it A) within X such that X is the only closed set that contains A.
Therefore, A~N (cardinality) because if this were not true, then A is finite, which is closed and A- <> X, so it is not dense. (A can not have higher cardinality of N because then it wouldn't be countable like we constructed.)
Is my thinking right?
EDIT: I am going to re-read my post later. I might have some mistakes that I don't know yet.
First, the definitions that I will use.
T2-Space or Hausdorff Space, find http://http://mathworld.wolfram.com/T2-Space.html" here.
Separable Space, find http://mathworld.wolfram.com/SeparableSpace.html" here.
Let X be the underlying set.
The questions or statements are...
1 - The Space of Reals is a T2-Space. This seems rather obvious since you can just construct the intervals for x and y (x<>y) such that they do not overlap. Hence, two neighbourhoods that are disjoint.
2 - The Discrete Topology is separable if and only if X itself is countable.
Note: A- is the intersection of the collection of closed subsets in X that contain A. Therefore, if A is closed, A- = A.
Note: A is dense in X if and only if A- = X.
Note: <> means not equal to.
Now, using the notes above.
X can be the only possible solution in this case because every subset is closed, therefore A- <> X, unless A=X. Therefore, if X is countable, it is separable.
Is that right?
3 - The Indiscrete Topology is separable if and only if there is countable subset in X. If X is the set of real numbers, then it is separable because the naturals numbers is a countable subset in R, and it is dense because N- = X, because X is the only set that contains N.
4 - The Space of Reals is separable. Since the rational numbers are countable and it is a subset of R, and R is the only closed set that contains Q then Q- = R. Hence, it is separable.
5 - The Finite Complement Topology is separable. If X is countable, then we are done. If X is not countable, then we can create a countable subset (call it A) within X such that X is the only closed set that contains A.
Therefore, A~N (cardinality) because if this were not true, then A is finite, which is closed and A- <> X, so it is not dense. (A can not have higher cardinality of N because then it wouldn't be countable like we constructed.)
Is my thinking right?
EDIT: I am going to re-read my post later. I might have some mistakes that I don't know yet.
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