My differential geometry professor said that the tangent space of an open set $\Omega$ of $\mathbb{R^n}$ at a point $P$ coincides with $\mathbb{R^n}$ , that is
$T_P(\Omega)=\mathbb{R^n}$
He gave an intuitive example for n=2: if $\Omega$ is any region in $\mathbb{R^2}$, let'say a disc centerend in the origin, and $P$ any point in the disc, the tangent space if the set of all tangent vectors to the curves whose traietories pass by $P$ and are contained in the plane. cleary the $T_P(\Omega)$ consides with the plane $\mathbb{R^2}$ itself.
I tried to do my own example for $\mathbb{R^3}$, according to that is not true that $T_P(\Omega)=\mathbb{R^3}$. If $\Omega$ is half a spheric surface, and $P$ any point on the surface, It is absurd to say that the tangent plane to the sphere is $\mathbb{R^3}$, since is it a PLANE! as it is obvious form the picture, so $T_P(\Omega)=\mathbb{R^2}$
What I am getting wrong? Please refrain from talking about manifolds , varities or technical stuff. These are not part of my course
You have embedded a two dimensional object into a three dimensional space however when you move to a higher dimensional space open sets in the lower dimensional space may not be open anymore. A simple example would be the interval $(0,1)$ in $\mathbb{R}$ which is open as it contains all of its interior points, however a line segment is never open in $\mathbb{R}^2$ because it has no interior points.
In your example the open sets would be on the surface of the sphere which are not open in the ambient space. This is for the same reason, every neighborhood around every point will contain points not on the sphere and so it has no interior points, and is therefore not open. As you've correctly deduced this detail is non-trivial otherwise your get a mismatch in dimension.