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.
Best Answer
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 $D^n$