In my complex analysis class there was an example presented following this definition:
Definition - Pointwise, Uniformly: Let $f_n: D \rightarrow \mathbb{C}$ be a collection of functions, we say that $f_n \rightarrow f$ pointwise if $\forall x \in D, f_n(x) \rightarrow f(x)$ and $f_n \rightarrow f$ uniformly if $\sup_{x \in D} |f_n(x) - f(x)| \rightarrow 0.$
The example was that for $D = [0, 1], f_n(x) = x^n,$ then $f_n(x) \rightarrow 0$ if $x < 1$ and $f_n(x) \rightarrow 1$ if $x = 1$. My professor said that this example converges pointwise but not uniformly. However, doesn't it converge uniformly to $\lfloor x \rfloor$? I believe that for $x < 1$, $|f_n(x) - f(x)| \rightarrow |0 - \lfloor x \rfloor| = 0.$ When $x = 1$, $|f_n(x) - f(x)| \rightarrow |1 - \lfloor 1 \rfloor| = 0,$ so doesn't this satisfy the definition for $f_n \rightarrow f$ uniformly?