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Homework Statement
Let X by a completely regular space and let A and B be closed, disjoint subsets of X. Prove that if A is compact, then there is a continuous function f : X --> [0,1] such that f(A) = {0} and f(B) = {1}.
The attempt at a solution
Let {U} be an open covering of A, U_1, ..., U_n be the corresponding finite open covering afforded by compactness of A. Let a_i in U_i and let f_i : X --> [0,1] be the continuous function separating a_i and B afforded by the complete regularity of X. Define f by f_1 * ... * f_n. This is all I have at the moment. Unfortunately, my f is only 0 at a_i and I don't know how to extend it to be 0 on all of A without breaking continuity.
I believe that the open covering I choose for A must somehow play a role. For example, if the open covering is just {A}, then, proceeding as I did above, compactness plays no role. Any tips?
Let X by a completely regular space and let A and B be closed, disjoint subsets of X. Prove that if A is compact, then there is a continuous function f : X --> [0,1] such that f(A) = {0} and f(B) = {1}.
The attempt at a solution
Let {U} be an open covering of A, U_1, ..., U_n be the corresponding finite open covering afforded by compactness of A. Let a_i in U_i and let f_i : X --> [0,1] be the continuous function separating a_i and B afforded by the complete regularity of X. Define f by f_1 * ... * f_n. This is all I have at the moment. Unfortunately, my f is only 0 at a_i and I don't know how to extend it to be 0 on all of A without breaking continuity.
I believe that the open covering I choose for A must somehow play a role. For example, if the open covering is just {A}, then, proceeding as I did above, compactness plays no role. Any tips?