4
$\begingroup$

In Ahlfors' Complex Analysis text, pages 289-290, the author discusses analytic continuation along arcs. Among other things, he proves the equivalence with analytic continuation using a chain of function elements $(f_k ,\Omega_k)$.

The fact that two function elements being related by a chain of direct analytic continuations implies that an appropriate analytic continuation along an arc is possible is clear to me. However, in page 290 Ahlfors is trying to prove the converse. Namely, that the arc from the analytic continuation can be "followed" by a chain of direct analytic continuations, from the initial point of the arc to its endpoint.

His words are:

Conversely, if $\bar{\gamma}$ is given, we can find a chain of direct analytic continuations which follows the arc $\gamma$ in the same way as in the preceding construction. In fact, by Heine-Borel's lemma the parametric interval $[a,b]$ can be subdivided into $[a,t_1], [t_1,t_2], \dots , [t_{n-1},b]$ such that $\bar{\gamma}(t)$ = $(f_k,\gamma(t))$ in $[t_{k-1},t_k]$ for suitably chosen function elements $(f_k,\Omega_k)$. Although $(f_k,\Omega_k)$ and $(f_{k+1},\Omega_{k+1})$ need not be direct analytic continuations of each other, they are at least direct continuations of their common restrictions to a neighborhood of $\gamma(t_k)$.

I'm having a hard time understanding this. As far as I know the Heine Borel lemma characterises compact subsets of Euclidean space alone as bounded and compact ones. The only set I see here which lies in Euclidean space is the parameter interval $[a,b]$ of the arc. But, even if that's where the lemma is applied, the outcome should be a finite open subcover of $[a,b]$, while the intervals $\{[t_{k-1},t_k]\}$ are only closed.

Can anyone please help me understand this proof? Thanks!

P.S. Here $\bar{\gamma}:[a,b] \to \mathfrak{S_0}(\mathbf{f})$ takes values on some Riemann surface, and $\gamma= \pi \circ \bar{\gamma}:[a,b] \to \mathbb C$ is its projection.

$\endgroup$
3
$\begingroup$

Correct,he is talking about the interval, and yes, Heine-Borel applies to open covers, but once you have a finite open cover, it is a triviality to construct a decomposition into closed intervals (take any two two intersecting sets of your cover, they will give rise to three closed/half-closed intervals. Use those. In the end, throw out redundant intervals.

$\endgroup$
3
$\begingroup$

This could be called "partition form of the Heine-Borel theorem", just as there is a "separation form of the Hahn-Banach theorem". All the ingredients are the same, they are just packaged into a different shape, which is more convenient for certain uses. In the paper Applications of full covers in real analysis, the [undergraduate] authors give the partition form a separate name, "Thomson's lemma". I don't think this is going to catch on, seeing as the paper by Thomson is dated by 1980 rather than 19th century. As the authors say,

Thomson’s lemma is similar in concept and execution to the version of the Heine–Borel theorem

Then they go on to reprove some theorems in real analysis using the partition form.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.