I have a question about pointwise convergence of a sequence of functions. The question is very specific and simple, and is at the last section below.
Here is my understanding of the concept (based on definition in Spivak's Calculus)
Given a function $f(x)$, if we can say that $f(x)=\lim\limits_{n\to\infty} f_n(x)$ for each $x$ in an interval $A$ (but not for all of the points simultaneously), that is, we can say that
$$\forall \epsilon>0, \forall x, x\in A \implies [\exists N, \forall n, n>N\implies |f(x)-f(x_n)|<\epsilon]$$
then we say that $\{f_n\}$ converges pointwise to $f$ on $A$.
Note that there is a stronger concept, that of uniform convergence of the sequence $\{f_n\}$, which occurs when we can say
$$\forall \epsilon>0, \exists N, \forall n, n>N \implies [\forall x\in A \implies |f(x)-f(x_n)|<\epsilon]$$
This essentially means that by choosing $N$ large enough, the graph of the function $f_n$ with $n>N$ is arbitrarily close to the graph of $f$ at all points in an interval at the same time.
Consider the sequence of functions $\{f_n(x)\}=\{\sqrt[n]{x}\}$.
A graphical depiction of this sequence of functions, and also of the sequence of numbers $\{f_n(x_1)\}$ is shown below
Let $$f(x)=\lim\limits_{n\to\infty} f_n(x)=\begin{cases} 1, x\in (0,1] \\ 0, x=0 \end{cases}$$
The question
Let $x \in (0,1]$.
$$\sqrt[n]{x}>1-\epsilon \implies \frac{1}{n}\log{x}>\log{(1-\epsilon)}$$
$$\implies n>\frac{\log{x}}{\log{1-\epsilon}}$$
That is, for any $x \in (0,1]$, $\exists N, \forall n, n>N\implies |f(x)-f_n(x)|<\epsilon$.
Therefore we say that $\{f_n\}$ converges pointwise to $f$ on $(0,1]$.
I just want to make sure of the following: we can't say that it is pointwise convergent on $[0,1]$, right?