I'm reading Carother's Real Analysis and currently on the chapter about functions of bounded variations on the interval $[a,b]$, that is $BV([a,b])$. I came across the statement below:
"... Thus, $C([a,b])$ is contained in the closure of $BV([a,b])$ under uniform convergence but not in $BV([a,b])$ itself. That is, $BV([a,b])$ is evidently not closed under uniform convergence (and hence is not complete under uniform convergence)."
I am assuming the author is working with the Sup Norm.
I don't understand the bolded portion of the statement. Is uniform convergence necessary for completeness in functional spaces? If so, can someone point me in the direction for a proof? I have tried looking online. Thanks for the assistance.
If $f$ is a function in $C[a,b]$ which is not of bounded variation then there exists a sequence of functions in $BV[a,b]$ ( in fact a sequence of polynomials) converging uniformly to $f$. This sequence in $BV[a,b]$ is a Cauchy sequence. If it converges to some element of $BV[a,b]$ then this element has to be $f$ but $f$ is not bounded variation. Hence $BV[a,b]$ is not complete.