(metric spaces) - how to prove f has a continuous extension?

Question

Let X and Y be metric spaces and let A be a subset of X. Suppose f:A-->Y is continuous on A and has a limit at every limit point of A. Show that f has a unique continuous extension to the closure of A.

Thank you

Answer 1

It's not so immediate like in the previous answer.

Let dx be the metric defined in X and dy be metric defined in Y.

Let A' be the closure of A and let f* be the extension of f to A' defined by

f*(x) = f(x) for x in A
f*(x) = lim ( y -> x) f(y) for x in A' - A (x in A' but not in A, which implies x is a limit point of A)

Any other definition for x in A' - A would make f* discontinuous at x. So, this definition is the only candidate to a possible continuous extension from A to A'.

Now, we have to prove f* is continuous on A'. Let x belong to A' and let (x_n) be any sequence in A' that converges to x. Each x_n is in A or is a limit point of A (or both). Since f is continuous at x_n if x_n is in A or has limit f*(x_n) at x_n if x_n is a limit point of A (no matter if, in this case, x_n is in A or not), we can find y_n in A satisfying:

dx(x_n, y_n) < 1/n (1) and dy(f*(x_n), f*(y_n)) < 1/n

If x_n is in A, then the first inequalty is possible because we can put y_n = x_n, for example. If x_n is not in A, then x_n is a limit point of A and we can find elements in A arbitrarily close to x_n. Therefore, in this case the 1st inequality is always possible, too

The second inequality is consequence of the definition of f* and of the continuity of f at x_n (if x_n is in A) or of lim ( y -> x_n) f(y) = f*(x_n), if x_n is a limit point of A.

Therefore, for each n we can find y_n satisfying (1) and(2). All we have to do is choose y_n arbitrarily close (possibly equal) to x_n.

(1) implies that dx(x_n, y_n) --> 0. Therefore, since x_n --> x, do does y_n. Since (y_n) is a sequence in A and f is continuous at x_n or has limit f*(x_n) at x_n if x_n is not in A, it follows that lim f(y_n) = lim f*(y_n) = f*(x). On the other hand, (2) implies that dy(f*(x_n), f*(y_n)) --> 0. And since (f*(y_n) = f*(x), it follows that

lim f*(x_n) = f*(x), valid for every sequence in A' that converges to x. Therefore, f* is continuous at every x of A' and actually is the unique continuous extension from A to A'.

The proof is now complete.

Answer 2

let c be a point in closure(A) - A. Then c is a limit point of A. So define f(c) = lim(f(x) for x in A, x approaches c.

Let e > 0. by def of limit, there exists d > 0 such that D(x,c) < d implies D(f(x), f(c)) < d where D is the metric. So the extended f is continuous over A closure.

Answer 3

holdm, I think you meant to write "f(c)) < e".