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Countable and Normal

A) Reals with the "usual topology." Is there a way to prove this space is normal other than just saying it is normal because every metric space is normal?
b) Reals with the "K-topology:" basis consists of open intervals (a,b)and sets of form (a,b) - K where K = {1, 1/2, 1/3, ... } Why connected? Why 2nd countable?

Solution Preview

Let X be the space of reals.

a) Proof:
Suppose A and B are two nonintersected closed set in X. For any x in X, we define d(x,A) as the distance from x to A since X is a metric space and the distance is well-defined. Now if x is in A, then d(x,A)=0. If x is not in A, then d(x,A)>0. We can similarly define d(x,B), the distance from x to B. So d(x,A)+d(x,B)>0 since A and B have no intersections. We also know d(x,A) and d(x,B) are continous functions.
Now we define f(x)=[d(x,A)-d(x,B)]/[d(x,A)+d(x,B)], then we know f(x) is continous. Moreover, f(A)={-1} and f(B)={1}. So f(X) is in [-1,1]. Next we define g(x)=[f(x)+1]/2. Then g(A)={0}, g(B)={1}. g is continous. Therefore, by the Urysohn Lemma, X satisfies T4 ...

Solution Summary

This shows how to create statements regarding metric spaces and being connected, second countable, and normal.