Show the usual metric on $C([0,1])$ does not give rise to a complete metric space. I know that I need to show there is a Cauchy sequence which converges to a point outwith $C[0,1]$ to show this. Is this the right way of going about it (I can only think of $f_n (x) = x_n = \frac{1}{n}$ as an example:

Let $x_n = \frac{1}{n}$. Then for $n \geq m &gt; 0$ $|x_n - x_m| = \frac{1}{n} - \frac{1}{m} $

$= \frac{m - n}{nm} &lt; \frac{m}{nm} = \frac{1}{n} &lt;\epsilon$
Set $N = \frac{1}{\epsilon}$ then for $n \geq m &gt; \epsilon$ $\frac{1}{n} \leq \frac{1}{m} &lt; \epsilon$ Hence $\frac{1}{n}$ is Cauchy.

Taking the limit as $n \to \infty$ $f_n (x) = 0$ for $0&lt;n&lt;1$ and $1 for n = 1$. Thus there is a discontinuity at 1, hence it cannot converge.

I feel this is the right track but think I've gone wrong somewhere, can someone spot where? Thanks
Start with a nice discontinuous function $f$ - something like the step function at $1/2$.  Find the Fourier series for this function.  The series should converge to $f$ in your metric (and not to any other continuous function, since the discontinuity is essential).  But the sequence of partial sums is made up of continuous functions, and so is contained in $C(0,1)$.

Or, if you don't like Fourier series, try approximating the same step function with a sequence of piecewise continuous (or even piecewise linear) functions.
Start with a nice discontinuous function $f$ - something like the step function at $1/2$.  Find the Fourier series for this function.  The series should converge to $f$ in your metric (and not to any other continuous function, since the discontinuity is essential).  But the sequence of partial sums is made up of continuous functions, and so is contained in $C(0,1)$.

Or, if you don't like Fourier series, try approximating the same step function with a sequence of piecewise continuous (or even piecewise linear) functions.