Extension of a continuous function

[radou]

Homework Statement

Let f : A --> Y be a continuous function, where A is a subset of X and Y is Haussdorf. Show that, if f can be extended to a continuous function g : Cl(A) --> Y, then g is uniquely determined by f.

The Attempt at a Solution

I think I can solve this on my own, but I got a bit stuck, perhaps I didn't understand the problem right, so let me check.

I assume an extension of a function in this case means that for any x in A, f(x) = g(x), and further on, I assume that g is uniquely determined by f if, for any x in Cl(A) , f(x) = g(x), too. Am I right on this one?

 

[Office_Shredder]

f(x) doesn't exist for points in Cl(A) necessarily. When they say that g is uniquely determined, it means that suppose g(x) and h(x) are continuous functions on Cl(A) such that f(x)=g(x)=h(x) on A. Then g(x)=h(x) for all points in Cl(A)

 

[radou]

f(x) doesn't exist for points in Cl(A) necessarily. When they say that g is uniquely determined, it means that suppose g(x) and h(x) are continuous functions on Cl(A) such that f(x)=g(x)=h(x) on A. Then g(x)=h(x) for all points in Cl(A)

OK, thanks. I'll try to do it right away.

 

[radou]

OK, here's the solution.

Let g and h extensions of f, and assume g =/= h. Let x be a point in Cl(A). Then, since Y is Haussdorf, take two disjoint neighborhoods of g(x) and h(x), V1 and V2. Since g and h are continuous, for V1 and V2 there exist neighborhoods U1 and U2 of x such that g(U1) is contained in V1 and g(U2) in V2. Since x is in the closure of A, both U1 and U2 intersect A. Take some x' from the intersection. Since x' is in A, g(x) = f(x) is contained in V1, and h(x) = f(x) is contained in V2. But then V1 and V2 are disjoint neighborhoods of f(x), which is impossible.

 

[radou]

Since x' is in A, g(x) = f(x) is contained in V1, and h(x) = f(x) is contained in V2. But then V1 and V2 are disjoint neighborhoods of f(x), which is impossible.

Just a correction - since x' is in A, g(x') = f(x'), etc. replace all x with x'.