Confusion with limits with respect to a certain metric

Question

I was looking at some proofs that certain metric spaces are not complete, and I have been confused.

Here is a link to an example of a main source of my confusion:Show that this metric is not complete

Here is the line that drove my confusion:

No, there is a error in the proof. In fact the functions $f_n(x) = x^n$ converge to the zero function in the metric $\rho$ on $C([0,1]):$

$$ \sqrt {\int_0^1 |x^n-0|^2\,dx} = 1/\sqrt {2n+1} \to 0.$$

If I understand this correctly, if we have a metric space equipped with a certain metric $d$, all limits will be computed with respect to this metric. We can pretend no other metrics exist within the space, as we did with limits in $\mathbb{R}$ in the beginning. I do not understand the last line because it seems to me that the user is taking the limit with respect to the metric $d(x,y)=|x-y|$ in $\mathbb{R}$ when he writes $\frac{1}{\sqrt{2n+1}} \rightarrow 0$. I know I am really confused here, and would appreciate if someone could help clear this confusion.

Answer 1

If $f_n(x)=x^n$ and $\mathbf{0}$ is the constant $0$ function, then

$$\rho(f_n,\mathbf{0})=\sqrt{\int_0^1|x^n-0|^2\,dx}\,,$$

so

$$\rho(f_n,\mathbf{0})=\frac1{\sqrt{2n+1}}$$

for each $n$. This means that $\lim\limits_{n\to\infty}\rho(f_n,\mathbf{0})=0$, which in turn easily implies that the sequence $\langle f_n:n\in\Bbb N\rangle$ is $\rho$-Cauchy: given $\epsilon>0$, there is an $n_\epsilon\in\Bbb N$ such that $\rho(f_n,\mathbf{0})<\frac{\epsilon}2$ whenever $n\ge n_\epsilon$, so

$$\rho(f_m,f_n)\le\rho(f_m,\mathbf{0})+\rho(\mathbf{0},f_n)<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$$

whenever $m,n\ge n_\epsilon$.

Answer 2

As Brian showed, $f_n \to \mathbf{0}$ in $C([a,b])$ so it's not a counterexample to completeness of $C([a,b])$. You need a Cauchy sequence that is not convergent in $C([a,b])$ instead.

Note that this neglected answer gives a better example of $f_n$ that form a Cauchy sequence (and in fact a convergent sequence in the larger space $L^2[0,1]$) but without a continuous limit $f$ (the limit in $L^2[0,1]$ is the function that is $0$ on $[0,\frac12)$ and $1$ on $[\frac12,1]$, which has no continuous representative in its class, put technically), and this sequence does illustrate the non-completeness, in a similar way as a sequence of rationals converging to $\sqrt{2}$ does the incompleteness of $\Bbb Q$..