# An Incomplete Metric Space

I am reading Kolmogorov's "Introductory Real Analysis" and I have come across this example of an incomplete metric space ($$C^2_{[a,b]}$$) on page 59: If $$\phi_n(t)= \left\{ \begin{array}{lcc} -1 & -1 \leq t \leq -\frac{1}{n} \\ nt & -\frac{1}{n}\leq t \leq \frac{1}{n} \\ 1 & \frac{1}{n}\leq t \leq 1 \end{array} \right.$$ then {$$\phi_n(t)$$} is a fundamental sequence in $$C^2_{[-1,1]}$$, since $$\\$$ $$\int^1_{-1}[\phi_n(t)-\phi_{n^{'}}(t)]^2dt \leq \frac{2}{{min\{n,n^{'}\}}}$$, where $$C^2_{[a,b]}$$ is the metric space on the set of all functions continuous on the interval [a,b], equipped with the distance function $$\rho(x,y)=(\int^a_{b}[x(t)-y(t)]^2dt)^{1/2}$$. It is worth noting that 'fundamental sequence' and 'Cauchy sequence' are used interchangeably here. The problem I have is that I don't understand there this upper bound comes from and why it is chosen here. From desmos:https://www.desmos.com/calculator/mrd9tjyrup, the max distance between 2 elements only ever reaches $$\frac{2}{3}$$, but the given bound $$\frac{2}{{min\{n,n^{'}\}}}=2$$ for $${min\{n,n^{'}\}}=1$$. Also, it is my understanding that $$\int^1_{-1}[\phi_n(t)-\phi_{n^{'}}(t)]^2dt = \frac{2}{{min\{n,n^{'}\}}}$$ is a consequence of the limit taken as $$n^{'} \rightarrow \infty.$$ Any help would be much appreciated.

• Not that of course the limit function is discontinuous, which is why this metric space is not complete. – Math1000 Dec 23 '19 at 4:45

Let $$min(n,n')=m$$ then notice $$\phi_n(t)$$ and $$\phi_{n^{'}}(t)$$ agree off of $$[-\frac{1}{m},\frac{1}{m}]$$ so, $$\int^1_{-1}[\phi_n(t)-\phi_{n^{'}}(t)]^2dt = \int^{\frac{1}{m}}_{-\frac{1}{m}}[\phi_n(t)-\phi_{n^{'}}(t)]^2dt$$. Now notice on this interval, there difference is bounded by 1 as both functions are always of the same sign and have modulus at most one. Also the length of the interval is $$\frac{2}{m}$$ so by the max length bound we get $$\int^{\frac{1}{m}}_{-\frac{1}{m}}[\phi_n(t)-\phi_{n^{'}}(t)]^2dt\leq \frac{2}{m}$$