# Intuitive explanation of tangent space of an open set of $\mathbb{R^n}$

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

• A disc in $\Bbb R^2$ is 2D, whereas a spherical hemisphere in $\Bbb R^3$ is not 3D. If you pick an open ball in $\Bbb R^3$, then all of the tangent spaces may be identified with $\Bbb R^3$, just as with an open disc in $\Bbb R^2$. Similarly, if you let $\Omega$ be a semicircle in $\Bbb R^2$, the tangent spaces will just be 1D lines.
– anon
Commented Dec 24, 2020 at 19:15
• @runway44 Ok , but then the tangent spaces may be identified with $\mathbb{R^2}$ as intuition suggests, then how come the definition or property is that $T_{P}\Omega=\mathbb{R^n}$, for $\Omega$ an open set of $\mathbb{R^n}$? Commented Dec 24, 2020 at 19:19
• If $\Omega$ is an open subset of $\Bbb R^n$ then its tangent spaces may be identified with $\Bbb R^n$. Note that for $\Omega$ to be an open subset of $\Bbb R^n$, it must have the same dimension $n$. In general, the tangent spaces of any manifold have the same dimension as the manifold itself.
– anon
Commented Dec 24, 2020 at 19:20
• @runway44 I don't see how taking an open ball in $\mathbb{R}^3$ , the tangent space to a point which I think like the plane tangent to the sphere could possibly be $\mathbb{R}^3$ Commented Dec 24, 2020 at 19:24
• I said ball, not sphere. Do you know the difference? What if someone, say the 2D version of you yourself, told you they don't understand how the tangent space to a point in the unit disk is all of $\Bbb R^2$ and not a tangent line to the circle?
– anon
Commented Dec 24, 2020 at 19:25

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.
• How can I understand intuitively that the tangent space of an open set $\Omega$ of $R^3$ is $R^3$ and not $R^2$?. Taking an open ball as $\Omega$, I can't help thinking that the tangent space is for a point in the surface still a plane, and for a point inside I can't think of a tangent space. Commented Dec 24, 2020 at 23:54
• My professor has defined the tangent space of a surface at a point as the set of all tangent vectors to the curves contained in the surface at the given point so that for a subset of $R^2$ the tangent space coincides with $R^2$ itself, but I fail to do the same for a subset of $R^3$ Commented Dec 25, 2020 at 0:01
• If the tangent space is an approximation of the underlying open set, I don't see how all of $R^3$ can be an approximation of an open ball, For one thing $R^3$ is infinite while, the sphere is not, so they don't resemble each other. Furthermore, I am not sure in what sense are you talking about an approximation. Commented Dec 25, 2020 at 0:55