Give an example of a sequence of continuous functions which converges on a compact set to a function that has an infinite number of discontinuities.

Question

Give an example of a sequence of continuous functions which converges on a compact set to a function that has an infinite number of discontinuities.

Analysis is something that is very difficult for me, and I am not fully sure what I am supposed to give an example of, but this is my attempt (hopefully some of it is right):

Let C be the Cantor Set (since it is compact and has an infinite number of discontinuities).

Let C1, C2,... be subsets of C and the union of all Cn's = C.

Let F = (fn) where for all n, fn: [0,1] -> Cn

Let fn = |x|

Then the midpoints of all Cn will converge to 0 and the endpoints of each Cn will converge to 1.

Any help or corrections is greatly appreciated!

Answer 1

Here is a variation of the hint given by David Mitra. Let $C\subset [0,1]$ be a closed set with empty interior (could be the Cantor set, if you wish). Recall the definition of distance function $$d(x,C)=\inf_{y\in C}|x-y|$$ The triangle inequality implies $|d(x_1,C)-d(x_2,C)|\le |x_1-x_2|$. In particular, $d$ is continuous. Define $$f_n(x) = \max(0,1-n\,d(x,C))$$ which is a continuous function, being built from $d$ in a way that preserves continuity. If $x\in C$, then $f_n(x)=1$ for all $n$. If $x\notin C$, then $d(x,C)>0$ and consequently $f_n(x)=0$ for all large enough $n$.

Conclusion: the limit of $f_n$ is the characteristic function of $C$. The latter is discontinuous at every point of $C$.

Bonus features: you can get an uncountable set of discontinuities in this way; moreover, if $C$ is a fat Cantor set, the limit function is not even Riemann integrable (because the set of discontinuities has positive measure).