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Let $X = \{(x,y) \in \mathbb{R}^2 : 0 \le y \le 1 ,x=0 \text{ or }1/n \text{ for some } n \in \mathbb{N} \} \cup ([0,1] \times \{0\} ) $

I want to show $X$ cannot be triangulated. I have that given a finite simplicial complex $K$ , for any $x \in K $ and any open set $U$ containing $x$ there exists open connected $V$ such that: $$ x \in V \subset U $$

I can see, by taking $x = (0,1)$ and applying the above statement (no connected $V$ ), no finite simplicial complex triangulates $X$ .

Is this enough? In the definition of triangulable we do not require $K$ to be finite.

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up vote 2 down vote accepted

The finiteness is not necessary. Every simplicial complex (and more generally, every CW complex) is locally connected (more generally, it is locally contractible).

You can find a proof in Allen Hatcher's Algebraic Topology, in the Appendix, which deals with the topology of CW complexes. Each simplicial complex is a CW complex since a $\triangle^n$ is homeomorphic to the $n$-ball $D^n$

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I suppose this follows from being locally finite? do you know somewhere where i can find a proof of this statement? – JC784 Feb 24 '14 at 15:33
@JC784: Local finiteness isn't necessary either. The local finiteness of a CW complex (meaning each point has a neighborhood intersecting only a finite number of cells) is equivalent to local compactness, and also to first-countability. But the local contractibility is always there. – Stefan Hamcke Feb 24 '14 at 16:40
are infinite simplicial complexes homeomorphic to some $ \mathbb{R} ^n $ ? – JC784 Feb 24 '14 at 20:11
In general, no. Do you mean homeomorphic to a subset of $\Bbb R$? This isn't the case, either, as the infinite wedge of closed intervals ($[0,1]×\Bbb N/\{0\}×\Bbb N$) shows. That space is not first-countable, but every subset of $\Bbb R^n$ is. – Stefan Hamcke Feb 24 '14 at 20:28
sorry yes that's what i meant, thanks! – JC784 Feb 25 '14 at 14:02

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