Ring of Continuous Functions on a normal Space

[deluks917]

Homework Statement

Let (X,T) be a normal topological space. Let R be the ring of continuous real-valued functions (with respect to the given topology T) from X onto the real line. Prove that the that T is the coarsest Topology such that every function in R is continuous.

Homework Equations

Urshown's Lemma: If X is a normal Topological Space and A,B are closed subsets of X then there is a continuous real-valued function s.t. f(A) = 1, f(B) = 0 and 0<= f(x) <= 1.

Tietze Extension Theorem - If X is normal, A is a closed subset of X and f is a continuous real valued function on A then f can be extended to a continuous function on X (i.e. there exists a continuous function g on X s.t. the restriction of g to A is f). In fact we can have |g| <= max|f|.

The Attempt at a Solution

If X is a metric space we can consider the distance to the compliment of A. This function call it d is continuous and the inverse image of (0,inf) is A.

 

[micromass]

So ,what you can actually show is that every open set G of X, has the form G=f^{-1}(U), with U open in R.

To do this, apply the Urysohn lemma.

 

[deluks917]

I can't seem to do this though it seems like the right way to do the problem. Say we find a function that is zero on the compliment of A so that f^-1(0a,inf) = A. How do we get f to not equal zero on A itself? More generally how do we make f^-1(A) and f^-1(A^C) disjoint.

 

[micromass]

Well, to be honest, what I wrote in my first post is not completely true. But I wrote it that way to give an idea about how the proof should go.

The coarsest topology such that all continuous remains continuous is given by the following subbasis:

\{f^{-1}(U)~\vert~U~\text{open in R}, f\in \mathcal{C}(X,R)\}

So we must prove that the above set is a subbasis for the topology on X. For this you must show: for every open set G in X and for every x in G, there exist continuous functions fi and open set Ui in R, such that

x\in \bigcap_i{f_i^{-1}(U_i)}\subseteq G

Now, try to prove this...

 

[deluks917]

Thank you very much. We can just take f(x) = 0 and f(A^C) = 1 then f^-1((-inf,1)) is contained in A, contains x and is open.

 

[micromass]

Yes, that is good!