Is the set $A = \{ (x,\sin(\frac{1}{x}) ) | x \in (0,1)\}$ a manifold with and without the origin.
So without the origin $(0,0)$ I'm pretty sure that it is a manifold, since in each point the set is a graph of a smooth function. we can also present $A$ with the implicit function $F(x,y) = y - sin(\frac{1}{x})$.
However, with the origin I am pretty sure that $A$ is not a manifold, but I don't know how to prove it formally. The way I know to prove a set is not a manifold, is to show that at any open environment of the origin $(0,0)$, $A$ is not a graph of a function. I haven't learned anything about manifolds and connectedness.
Any help would be appreciated.
If $A$ is a manifold then any neighborhood of $(0,0)$ is homeomorphic of open subset of $\mathbb{R}^n$. But open subset of $\mathbb{R}^n$ are path connected while every neighborhood of $(0,0)$ in $A$ are not path-connected. Hence $A$ can not be manifold if $(0,0)$ is included because path-connectedness property is preserved by homeomorphism.