# The last step in Ahlfors' and Sario's proof of the triangulability of surfaces

In their book, Riemann Surfaces, Ahlfors and Sario write, at the bottom of pg. 109 to the top of pg. 110,

"Consider the sequences $$\{V_n\}$$ and $$\{W_n\}$$ introduced by Lemma 46B. We will show that there exist closed Jordan regions $$J_n$$, such that $$V_n \subset J_n \subset W_n$$, whose boundaries $$\gamma_n$$ have only a finite number of common points. They will then form a covering of finite character."

No proof is given for the last sentence. How does one prove it?

The precise content of Lemma 46B and the definition of "covering of finite character" will be given below, for convenience here. The essence of the question is then extracted.

Establishing the property of the cited last sentence constitutes the last step in the proof of triangulability of surfaces. The triangulation of a surface is constructed directly from $$\{J_n\}$$, based on the property in the cited last sentence. See the Remarks below, for further comments on the importance of the property in the cited last sentence.

Regarding the property (in the second cited sentence), "whose boundaries $$\gamma_n$$ have only a finite number of common points", the authors mean that for all $$m$$, $$n$$: $$\gamma_n \cap \gamma_m$$ consists of at most a finite number of points.

Here are the given properties of the open Jordan regions, $$V_n$$, $$W_n$$, and the desired and established properties of the closed Jordan regions, $$J_n$$, all in the connected surface (2-dim second countable manifold), $$F$$; the enumeration of the properties follows that of the authors. (An open Jordan region in $$F$$ is a subset of $$F$$ whose closure is homeomorphic to a closed disk in the Euclidean plane, in such a manner that the open region corresponds to the open disk. A closed Jordan region is the closure of an open Jordan region.)

Lemma 46B: There exist sequences $$V_n$$, $$W_n$$ ($$n = 1, 2, ...$$) of open Jordan regions in $$F$$ satisfying:

• (B1) $$\overline{V}_n \subset W_n$$.

• (B2) $$\bigcup_n V_n = F$$.

• (B3) No point of $$F$$ belongs to infinitely many $$\overline{W}_n$$.

Definition of 'covering of finite character': Closed Jordan regions $$J_n$$ ($$n = 1, 2, ...$$) in $$F$$, form a covering of finite character if

• (A0) $$\bigcup_n$$Int$$(J_n) = F$$, where Int$$(J_n)$$ denotes the interior of $$J_n$$.

• (A1) Each $$J_n$$ meets at most a finite number of others.

• (A2) For all $$m$$, $$n$$: $$\gamma_n \cap \gamma_m$$ consists of at most a finite number of points or arcs (possibly both), where $$\gamma_n := \partial J_n$$.

The constructed regions $${J_n}$$: The closed Jordan regions $$J_n$$, $$n = 1, 2, ...$$, in $$F$$ are constructed recursively in a manner such that for all $$n$$, $$V_n \subset J_n \subset W_n$$ and their boundaries $$\gamma_n := \partial J_n$$ satisfy: For all $$n$$, $$\gamma_n$$ meets $$\gamma_{n-1} \cap \cdots \cap \gamma_1$$ in at most a finite number of points.

The essential question: Does (A1) follow from (A0), (B3), and the $$\{\gamma_n\}$$'s property, for all $$m$$, $$n$$: $$\gamma_n \cap \gamma_m$$ consists of at most a finite number of points?

Remarks: If the sequences $$\{V_n\}$$ and $$\{W_n\}$$ are of finite length, then the whole business is trivial; so wlog the sequences are of infinite length. The result (A0) follows immediately from (B2) since Int$$(J_n) \supset V_n$$, for all $$n$$. The property (A2) is satisfied by construction. Thus only the verification of (A1) remains. From the purpose of triangulation, it would suffice to show that a subsequence of $$\{J_n\}$$ satisfies (A1). Thus if $$F$$ is compact, then we are done. The strength of the approach to triangulation that is adopted by the authors, is that it also handles noncompact surfaces (as well as surfaces with or without boundary).

Once $$\{J_n\}$$ have been constructed, the sequence $$\{V_n\}$$ has no further role to play, since (A0) can play the role of (B2). The same might be said of the sequence $$\{W_n\}$$ since it follows from (B3) that no point of $$F$$ belongs to infinitely many $$J_n$$. This leads to the following modified question.

Modified question: Does (A1) follow from (A0) if, in addition, no point of $$F$$ belongs to infinitely many $$J_n$$?

The last paragraph of the proof of Lemma 46B shows that the following result, which is stronger than (B3), holds: Each $$\overline{W}_n$$ meets at most a finite number of the other $$\overline{W}_m$$'s. Therefore, in particular, the $$J_n$$'s satisfy (A1) because, by construction, each $$J_n \subset W_n$$.
Whether or not the weaker condition, (B3), is sufficient for (A1) to hold (as in $$$$The Essential Question''), remains an open question; but an answer to the latter question is not required for the proof of triangulability of surfaces.