Suppose we have a function $ f: X \to Y $ such that $ d_1 (f (x), y_0) <A_1 $ whenever $ d_0(x, x_0) <A_0 $ and so such a function $ g:Y \to W $ such that $ d_2 (g (y), w_0) <A_3 $ if $ d_1 (y, y_0) <A_4 $. $d_0,d_1,d_2$ are the metrics. It´s not true that the limit of $h$ can be calculated simply as the limit of g if and tends to $ y_0 $ (only this is not necessarily equal to $f$ and thus compose want to know what is wrong with the show) if poorly explain my doubt, see the example I put where it is shown
The question is what is wrong in the next proof, that try to demonstrate this false property?
Taking $ A_3 = A_2 $ exists $ A_0 $ such that: $ d_2 (g (y), w_0) <A_2 $ whenever $ d_1 (y, y_0) <A_1 $ but using the fact that $ d_1 (f (x), y_0) <A_1 $ whenever $ d_0(x, x_0) <A_0 $ we have that those $f (x)$ exist in the ball of $ d_1 (y, y_0) <A_0 $ therefore evaluating $g$ in those points will also be at a distance of less than $ A_2 $ of the point $ w_0 $ Therefore $ h (x) \to w_0 $ as $ x \to x_0 $ But this proof is bad! but i dont know why! counterexample
$ f(x) = 0 $ for x $ \in [0,1] $
$ g(0) = 0 $
$ g(x) =1 $ for $x \in (0,1] $
$ h(x) = g(f (x)) $ clearly $g(x) \to 1$ as $x \to 0$ but $g(f(x)) \to 0$ as $x \to 0$.