Let $X$ be a compact manifold. Is any triangulation of $X$ by simplices already a finite triangulation? Can you give a proof or counter example?
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I think that this is more of a definition question than a proof/counter-example question. You can certainly find a covering by infinitely many triangles (with disjoint interiors), the question is whether that is defined to be a triangulation. – Michael Burr Nov 28 '16 at 14:19
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7A triangulation is a homeomorphism from a simplicial complex to $X$. An infinite simplicial complex is not compact. (It contains an infinite discrete closed set - the centers of all the simplices.) – Nov 28 '16 at 14:26
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@MikeMiller: thanks, this solves the issue. – shuhalo Nov 30 '16 at 02:14