A contradiction to Dini's theorem?

Question

Dini's theorem - Let $\{ f_n \} _n$ be a sequence of continuous functions on a compact set $K$. Suppose $f_{n+1} \geq f_n$ and $\{ f_n \} _n$ converges pointwise to a continuous function $f$ on $E$. Then $\{ f_n \} _n$ converges uniformly to $f$ on $E$.

This looks like a counterexample - $f_n(x)=x^n, 0 \leq x \leq 1$. $\{f _n \}_n$ is defined on $[0,1]$.

Because it converges pointwise to $0$, a continuous function, each of the functions are continuous on $[0,1]$, the sequence decreases monotonically but they don't converge uniformly to $0$.

Proof : We need to show $\exists \varepsilon >0 \ \ \forall N \in \mathbb N \ \ \exists n \geq N \ \ \exists x \in [0,1] \\ |f_n(x)|\geq \varepsilon $

We take $\varepsilon =\frac{1}{2}$, choose $N $ to be abitrary and set $n = N$ and $x= (\frac{1}{2})^{\frac{1}{n}}$. Then $f_n(x)=\frac{1}{2}$ hence we have proved that these functions do not converge uniformly to $0$.

Can someone point out the error? Thank you.

Answer 1

The limit function is not continuous, it is the function $f$ such that $f(x)=0$ if $x <1$ and $f(1)=1$.

Answer 2

No, it does not converge pointwise to the null function. You have$$\lim_{n\to\infty}f_n(x)=\begin{cases}0&\text{ if }x\in[0,1)\\1&\text{ otherwise.}\end{cases}$$So, you get a discontinuous function, and therefore you cannot apply Dini's theorem.