A power series of finite radius has a singular point on its circle of convergence

Suppose toward a contradiction that every point of the circle $C:=\{z:\mathrm{\mid }z-a\mathrm{\mid }=R\}$ is regular. By [L1], each $\zeta \in C$ then has a disc $B(\zeta ,r_{\zeta })$ and a holomorphic extension $F_{\zeta }$ on that disc agreeing with the original series on $B(\zeta ,r_{\zeta })\cap D(a,R)$. The circle $C$ is closed and bounded in ${\mathbb{R}}^2$, hence compact by [L4], so finitely many of these discs cover $C$.

The union of those finitely many extension discs is an open neighbourhood of $C$, so some $\epsilon >0$ satisfies $\{z:R-\epsilon <\mathrm{\mid }z-a\mathrm{\mid }<R+\epsilon \}$ inside that union. Hence $D(a,R+\epsilon )$ is covered by $D(a,R)$ together with the finitely many extension discs. On overlaps, each extension agrees with the original series on a nonempty open subset of $D(a,R)$, so [L5] makes all the local definitions agree on overlaps. Therefore they glue to one holomorphic function $F$ on $D(a,R+\epsilon )$ that agrees with the original series on $D(a,R)$.

Because $F$ is holomorphic on $D(a,R+\epsilon )$, [L3] gives a Taylor expansion $F(z)={\sum }_{n\ge 0}\frac{F^{(n)}(a)}{n!}(z-a{)}^n(\mathrm{\mid }z-a\mathrm{\mid }<R+\epsilon )\mathrm{.}$ On $D(a,R)$ the original series already represents $F$, so [L6] gives $c_n=F^{(n)}(a)\mathrm{/}n!$ for every $n$. Thus the original series itself converges on $D(a,R+\epsilon )$, contradicting [L2] because its radius was $R$.

Therefore the assumption of step 1.1 is false, and some point of $\mathrm{\mid }z-a\mathrm{\mid }=R$ is singular.