I know that it's a theorem by Moise that every compact 3-manifold admits a finite triangulation but to me the astounding part of that statement is the existence part instead of the finite one. So I was wondering: Is there an easier (possibly more modern) way to show that every compact, triagulable 3-manifold already admits a finite triangulation? All attempts I've made so far end up with me choosing a metric but I'd rather avoid that.

  • $\begingroup$ What is your definition of "triangulation"? If in particular any triangulation gives your space the structure of a CW complex, it is a standard theorem that any infinite CW complex is not compact. $\endgroup$ – Eric Wofsey Feb 25 '16 at 20:28
  • $\begingroup$ By a triangulation of a $3$-manifold $M$ I mean that there exists a simplicial complex, say $\sigma$, with geometric realization $\vert \sigma \vert$ homeomorphic to M. The finiteness condition then means that $\sigma$ contains only finitely many $3$-simplicies. I should have made that clear. $\endgroup$ – Nephry Feb 25 '16 at 20:30

Suppose you have a compact space $X$ and a triangulation (or more generally, a CW complex structure) of $X$. Choose one point in the interior of each simplex, and let $S\subseteq X$ be the set of all these points. Then $S$ is closed, because its intersection with each closed simplex is finite and hence closed (it contains only one point in the interior of each face of the simplex). Thus $S$ must be compact. But the exact same argument shows that any subset $A\subset S$ is also closed, so $S$ has the discrete topology. Thus $S$ must be finite, so the triangulation of $X$ can have only finitely many simplices.

  • $\begingroup$ Woderful, thank you very much! $\endgroup$ – Nephry Feb 25 '16 at 20:36

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