Possible condition on locally Euclidean subsets of Euclidean space to be embedded submanifold

Question

Consider a (locally closed if needed) locally Euclidean subset $X\subset \mathbb R^n$. Given a point $p\in X$, let $\dim_pX$ denote the dimension of $X$ at $p$ and consider the following condition:

• There are differentiable/smooth germs of paths in $X$ based at $p$ whose tangents at $p$ span a space of dimension $\dim_pX$.

I am hopeful this condition may express that $X\subset\mathbb R^n$ is smooth at $p$ by asserting it has a tangent space with the correct dimension.

If all points of a (locally closed if needed) locally Euclidean subset $X\subset\mathbb R^n$ satisfy the above condition, is $X\subset\mathbb R^n$ an embedded submanifold? Does this characterize submanifolds embedded in Euclidean space?

The question is motivated by thinking of manifolds as "things with tangent spaces".

Answer 1

Here's a counterexample. Define $f:\mathbb{R}^2\to\mathbb{R}$ by $f(r\cos\theta,r\sin\theta)=\sqrt[3]{r}\sin 2\theta$ and let $X\subset\mathbb{R}^3$ be the graph of $f$. Since $f$ is continuous, $X$ is closed in $\mathbb{R}^3$ and a topological surface. Since $f$ is smooth away from the origin, $X$ is smooth away from the origin. Moreover, the smooth paths $t\mapsto(t,0,0)$ and $t\mapsto(0,t,0)$ are contained in $X$ and have linearly independent tangents at the origin. However, $X$ is not smooth at the origin because $f$ is not smooth at the origin.

In fact, you can even arrange for there to be smooth curves whose set of tangents at the origin is an entire plane, rather than just spanning one. For instance, if you take $f(x,y)=\frac{x^3y}{x^4+y^2}$ (and $f(0,0)=0$), then $f$ is smooth away from the origin, is smooth when restricted to each line through the origin, and all its directional derivatives at the origin are $0$, but $f$ is not differentiable at the origin. The graph of $f$ will then not be a smooth submanifold of $\mathbb{R}^3$, but at the origin it will have an entire "tangent plane" worth of smooth paths through the origin, namely all the lines through the origin. (This example is taken from this answer.)