Let $X$ be $\{(x,\sin{\frac{1}{x}})\mid x\in(0,1]\}$ together with the vertical set of accumulation points, the vertical line segment at $0$. This is closed subspace of $\mathbb{R}^2$. Is there a triangulation of this space? If there isn't, how would one go about proving this fact? I saw examples, for example in Hocking and Young, of defining a similar simplicial complex, but they omit discussing an explicit homeomorphism.
To be more concrete, let $Y$ be a complex constructed as a collection of $1$-simplices such that there is an "oscillation" on every interval from $1/(2\pi(n+1))$ to $1/(2\pi n)$, together with a vertical set of accumulation points, the vertical $1$-simplex at $0$. It's topology is made by declaring subsets $A\subset Y$ closed if $A \cap \sigma$ is closed in $\sigma$ for all simplices $\sigma$.
Someone tried to explain to me that in the above example, there can't be a homeomorphism, it went as follows. Vertical segment and the "body" of $Y$ are both closed, and they are complements, so they are both open. So $Y$ is open in this topology as union of two open sets. So we are trying to find a homeomorphism that will take an open set to a closed set, which I guess is a contradiction? I am a bit lost in this argument, I would appreciate if someone could verify this is fine. If it is fine, can we construct some complex that will work? Or how could we generalize this contradiction to "arbitrary" complexes? Thanks!
The topologist's sine curve $X$ is connected. Assume it has triangulation $K$. This means that the polytope (geometric carrier) $\lvert K \rvert$ is homeomorphic to $X$. But a polytope is connected iff it is pathwise connected. Thus $X$ should be pathwise connected, which is not true.