The square is not a submanifold of $\mathbb{R}^2$

Question

Leb $X$ be the square in $\mathbb{R}^2$

$ X = \{(x,y) \in \mathbb{R}^2 : |x| + |y| = 1\} $

It's so easy to show that $X$ is a differentiable manifold of dimension one. But, it's not possible that $X$ be a regular submanifold of $\mathbb{R}^2$ and I dont know prove it.

Answer 1

Assume that $X$ is a submanifold of $\mathbb{R}^2$ and let $v$ be a vertex of the square $X$, then the tangent space of $X$ at $v$ should be one-dimensional$^1$, but $T_vX$ is seen to be zero-dimensional$^2$, whence a contradiction.

Another way to see it without (explicitely) using tangent spaces is that it cannot exists a regular parametrization of $X$ near $v$, such a parametrization must have a vanishing derivative at $v$, which prevents it to be an immersion, whence a contradiction.

_{$^1$ A zero-dimensional submanifold is discrete and a two-dimensional submanifold of $\mathbb{R}^2$ is locally open, but $X$ is none of them, so that it is one-dimensional.}

_{$^2$ A smooth curve drawn on $X$ passing through $v$ at $t=0$ must have a vanishing derivative at $t=0$.}

Answer 2

Expanding on C. Falcon's answer: For anyone wondering why the derivative at the vertex must be zero; let $c:I \to X$ be a curve such that $c(0)=v$ and $c'(0) \not = 0$. Let $f:I \to \mathbb{R}^2$ be such that $f(t) = \frac{c'(t)}{|c'(t)|}$. By hypothesis, $f$ is well defined and obviously continuous. Now, since $f(t)$ is the (unit) direction of the tangent vector to the square at the point $c(t)$ then, $f(t)$ is equal to the (unit) slope of the side of the square in which $c(t)$ lives. In particular, \begin{equation} \lim_{t \to 0^+} f(t) \not = \lim_{t \to 0^-} f(t) \end{equation} A contradiction since $f$ is continuous.

P.D. I'm still a newb so any corrections or comments will be greatly appreciated.