Let (X,d) be a metric space, and suppose that f_n: X -> X are continuous and have dense orbit.?

Question

If f_n: X -> X converges uniformly to f. Does f have a dense orbit, if not when?

Answer 1

By dense orbit, do you mean that the images f_n(X) are dense in X? If so, then this is true.

Fix some ε>0.  Now choose an N such that

d( f_n(x), f(x) ) < ε/2  for all n ≥ N (by uniform convergence).

Since f_N(X) is dense in X, for any y∈X, the ε/2-ball centered at y contains some point of the image, so we can say for some x∈X, f_N(x)∈ B(y, ε/2)., which implies

d( f_N(x), y) < ε/2.  Thus,

d( f(x), y) ≤ d(f_N(x), f(x)) + d( f_N(x), y) < ε,

so since ε and y were arbitrary, f(X) intersects every open ball at any point, and is thus dense.

Steve