In a Functional Analisys assignment I was asked to check for convergence of the following sequence $$ n\in\mathbb{N}: g_n(x) = x^\frac{1}{n} $$ in $C[0,1]$ with respect to the following norm: $$ ||f||_\infty = \max{|f(x)|} $$ I am convinced that this does not converge and argued that is not a cauchy sequence. The grader just noted that the series is in fact convergent but did not note much more.
After having looked at my proof some times I am still unclear were the argument fails and would be greatful if somebody could help me figure this out.
My Proof
One can see that $\forall n$,since $x \in [0,1]$, we have that $$ \min(g_n(x)) = 0, \max(g_n(x)) = 1 $$ and $g_n(x)$ is continuos and monotonly growing.
Therefore, $\forall \epsilon \in (0,1)\exists x_0\in [0,1]$ so that $$ x_0^\frac{1}{n}+\epsilon <1 $$ Now we can see that $$ x_0^\frac{1}{m}-x_0^\frac{1}{n}>\epsilon \iff $$ $$ -\frac{1}{m}|ln(x_0)|>-|ln(\epsilon + x_0^\frac{1}{n})|\iff $$ $$ m>\frac{|ln(x_0)|}{|ln(\epsilon +x_0^\frac{1}{n})|} $$ one can pick $m>n$ that fullfills this inequality. Then we can se that $$ ||g_n(x)-g_m(x)||_\infty \geq x_0^\frac{1}{m} - x_0^\frac{1}{n}>\epsilon $$
And $g_n(x)$ is therefore not a cauchy sequence with respect to $||\cdot||_\infty$.
We have $$g_{2n}(x)-g_{n}(x)=x^{1/(2n)}-x^{1/n}=x^{1/(2n)}(1-x^{1/(2n)})$$ For $x_n=2^{-2n}$ we get $$g_{2n}(x_n)-g_{n}(x_n)={1\over 4}$$ which implies $\|g_{2n}-g_n\|_\infty \ge {1\over 4}.$ Hence the sequence $g_n$ does not satisfy the Cauchy condition.