Check whether it's a submanifold of $\Bbb R^2$

Question

I was asked to prove the following:

Consider $A:=\left\{\frac1n: n=1,2,3,...\right\}$ and $B:=A\cup\{0\}$. Then, $B\times \Bbb R$ is not a smooth sub-manifold of $\Bbb R^2$, but $A\times \Bbb R$ is a smooth submanifold of $\Bbb R^2$.

I have the following reasonings:

$(1)$ There doesn't exist an open subset $U$ of $\Bbb R^2$ containing $(0,0)$ such that $U\cap B$ homeomorphic to an open subset of $\Bbb R$: The set $U\cap B$ consists of a sequence of vertical open intervals converging to another vertical interval. Also, any open subset of $\Bbb R$ is a collection of pairwise disjoint open intervals. So, $B$ is not a sub-manifold of $\Bbb R^2$.

$(2)$ Note that each vertical slice of $A\times \Bbb R$ is isolated from any other vertical slices, maintaining a fixed distance(though it may be small) at least from other slices.

Are these reasonings correct?

Thanks in advance.

Answer 1

Your argument is convincing, though not rigorous enough.

For (1) maybe the cleanest way is to observe that $(0,0)$ doesn't have a connected neighborhood in $B\times\Bbb R$, whereas in $\Bbb R^n$ every point has.

For (2) every point has a vertical open segment as neighborhood.