Every metric space is regular

1. Aug 23, 2010

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

To prove that every metric space is regular. :)

3. The attempt at a solution

So, a regular space satisfies the T1 and T3-axioms.

For T1: Let a, b be two distinct points of a metric space (X, d). Then d(a, b) > 0, and let r = d(a, b)/2. Then the open ball K(a, r) is a neighborhood of a which doesn't contain b, and K(b, r) is a neighborhood of b which doesn't contain a.

For T3: Let A be a closed subset of X, and let b be in X\A. Since we proved that T1 holds, for every x in A and for b there exists a neighborhood of Ux which doesn't contain b. A is a subset of the union U of these neighborhoods (U is a neighborhood of A). Now, define r = min{d(x, b)/2: x is in A}. Then K(b, r) and U are disjoint.

I hope this works.

2. Aug 23, 2010

Office_Shredder

Staff Emeritus
You probably should justify why r in your second argument is non-zero.

Also, your U is picked too generally. Knowing that singletons are closed, we could have picked U to be everything except for b, which obviously wouldn't work. You need to pick the balls surrounding the points in A more carefully

Last edited: Aug 23, 2010
3. Aug 24, 2010

OK, here are three other attempts.

i) T3: Let r = d(A, b) = inf{d(a, b) : a is in A}, where b is in X\A. Since A is closed, Cl(A) = A, and d(A, b) = 0 if and only if b is in the closure of A (i.e. in A), which contradicts our assumption, so d(A, b) > 0. Take r = d(A, b)/2, and take the union U of the family of open balls {K(x, r) : x is in A}. Then U and K(b, r) are disjoint.

ii) Since every metric space is Haussdorf (one can easily show this in a similar manner like showing that it is T1), for every x in A, and for b in X\A, we can find disjoint neighborhoods of x and b respectively, Ux and Vx. The intersection of all Vx and the union of all Ux are disjoint neighborhoods of A and b, respectively.

iii) Since every metric space is normal (i.e. it's T1 and T4), it's regular, too (since, for every two disjoint closed sets one can find disjoint neighborhoods of these sets, and a singleton is closed in a metric space). Of course, this proof assumes that normality is proven :)

4. Aug 24, 2010

ystael

Your argument (i) suffices, assuming you can take it as known that $$d(b, A) = 0$$ if and only if $$b \in \overline{A}$$.

5. Aug 24, 2010

Yes, I went through the proof of that argument, it was part of another assignment. Thanks.

6. Aug 24, 2010

Office_Shredder

Staff Emeritus
You don't know know that an intersection of open sets is open when there are infinite of them.

7. Aug 24, 2010