I have read that $\mathbb{S}^{2} \subset \mathbb{R}^{3}$ does not have boundary.
However, given a set $X \subset \mathbb{R}^{n}$ its boundary is the set $\partial X = \overline{X}\cap (\overline{\mathbb{R}^{n} - X})$. We have that $\overline{\mathbb{S}^{2}} = \mathbb{S}^{2}$ since it is closed. And $\overline{\mathbb{R}^{n} - X} = \mathbb{R}^{n}$. Therefore $\partial \mathbb{S}^{2} = \mathbb{S}^{2}$.
What is wrong with my argument?
Thank you!
Being without boundary as a manifold means that locally it always looks like the full Euclidean space of lower dimension, once we "unfold" it. A manifold with boundary some parts of it only look like a half space. For example, if you look at the set $$\{(x,y,z)|x^2+y^2=1, 0\leq z\leq 1\}$$, you will see that any point where $z\in (0,1)$ we can take a small enough open set around any point and we can map it to an open set in $\mathbb{R^2}$ via a homeomorphism. However, at the boundaries $z=0$ and $z=1$, we can only map open sets around such points to open sets in the half space where we only take from the $x$ axis up (or down).
A sphere, on the other hand, there's a small neighborhood around every point that maps to an open set in $\mathbb{R}^2$ via a homeomorphism, so there is no boundary as a manifold