I am referring to an example in Reed & Simon (1980) Methods of Modern Mathematical Physics I: Functional Analysis, p. 12 et seq.
We know that the space $C[a,b]$ is complete in the metric induced by the $L^{\infty}$-norm, but not by the one induced by the $L^1$-norm. As an example of the lack of completeness of $(C[a,b],\lVert \cdot \rVert_1)$, consider a sequence of functions in $C[0,1]$, $f_n$, given by the graph in the book (see here). This sequence is Cauchy in $\lVert \cdot \rVert_1$, but does not converge to any function in $C[0,1]$ as it, intuitively, converges to the (non-continuous) indicator function $I_{[1/4,3/4]}$. All of this is clear, but how does this change when we consider this in the supremum-norm instead? The only explanation I have come up with is that $f_n$ is not Cauchy in $\lVert \cdot \rVert_{\infty}$, because $$\sup_{x \in [0,1]} |f_n(x) - I_{[1/4,3/4]}(x)|$$ is constant equal to 0.5 for all $f_n \in C[0,1]$ and only becomes 0, intuitively, for $n = \infty$, i.e. for a function outside of $C[0,1]$. Is this argument mathematically correct (in principle)? I am only starting to wrap my head around functional analysis, so please apologise any inaccuracies, inconsistencies and outright stupidities...
