Problem: Let $S=\{(x,y):0\le y\le 1, x=0\;\, \text{or}\;\, x=1/n \quad\text{for}\; n=1,2,\cdots\}\cup ([0,1]\times {0}$), Show that $S$ is not triangulable.
Note: my definition for triangulation is that a set $X$ is triangulabe if there exists simplicial complex $K$ such that $|K|$ is homeomorphic to $X$
My attempt: So first thing I spotted was that $S$ is clearly closed, and hence compact in $\mathbb{R}^2$. Therefore, if a simplicial complex $|K|$ was homeomorphic to it, then it must also be compact, meaning that $|K|$ is finite. But here is one point of my concern: I'm not supposed to know that all compact simplicial complexes are finite, simply because I haven't yet covered it. Is there other way more elementary arguement to show that $|K|$ is finite, other than referring to the theorem?
Now there is explicit hint given in the problem:
show that for any finite simplicial complex for any $x\in |K|$ and open set $U$ containing $x$ there is a connected open set $V$ such that $x\in V\subseteq U$
Apart from the fact that I cannot see how this might help me solving problem, I also am a bit puzzled that it seems too much obvious for me to prove it: since $|K|$ can be embedded into finite dimensional $\mathbb{R}^n$ space, with standard metric on it we can always take $V$ to be a open ball around $x$, and we are done. Is that right?
I also see that $S$ is clearly path-connected, so my $|K|$ must be edge-connected and connected. But I seem to be stuck at this stage.
It would be really great if someone could explain how one would do this with given hint, if possible.
Thanks in advance,
This space (the comb space) is not locally connected: any small enough neighborhood of a point like $(0,1/2)$ splits into several components (try discs, for instance). The hint you are given is that simplicial complexes are locally connected: each point has a open nbhd basis of connected sets. Hence the comb cannot be simplicial.