How can $\mathbb{R}^n$ has the same number of dimentions of $\mathbb{S}^n$?

Solution 1:

That is one way to define it. Another way is to define $S^n$ as the one-point compactification of $R^n$.

The idea is this: A circle ($S^1$) is like a line ($R^1$) with the ends glued together; or conversely a line is like a circle with one point deleted. There is an easy homeomorphism: Consider the circle with center at $(0, 1/2)$. For each point $P$ on the circle, draw a line through $P$ and $N = (0, 1)$.

This line intersects the $x$-axis in one point $P'$, and this is a homeomorphism between the $x$-axis and the circle minus the point $N$ itself.

One can do a similar mapping for the sphere onto a plane: for each point $P$ on the sphere, draw a line from the north pole $N$ of the sphere through $P$ to finds its intersection $P'$ on the plane. This is a one-to-one correspondence between the points of the plane and the points of the sphere except for the north pole. The mapping is called a stereographic projection.

Again the idea is that if you take a sphere and delete a point, you can stretch out the part around the deleted point and take it out to infinity, flatten it out, and what you get is the plane. Or you can do the same thing in reverse, adding a "point at infinity" that brings all the far-away parts of the plane together; this "point at infinity" is the "one point" in the "one-point compactification" I mentioned earlier.

The construction in higher dimensions is completely analogous.

Solution 2:

There are multiple definitions of $\mathbb{S}^n$ (although the topological spaces they define are all homeomorphic). It sounds like the one they're using is the one-point compactification of $\mathbb{R}^n$, which as a set is $\mathbb{R}^n\cup\{\infty\}$, and so literally contains $\mathbb{R}^n$. Think of gluing in one additional point at infinity to $\mathbb{R}^n$, so that it rolls up into a ball - this is easiest to visualise when $n$ is $1$ or $2$.

Solution 3:

You are correct. However, $\mathbb{S}^n - \{\mbox{one point}\}$ can be mapped onto $\mathbb{R}^n$ via stereographic projection. The reverse map $\mathbb{R}^n\hookrightarrow\mathbb{S}^n$ is an inclusion.

Another way of thinking about the inclusion is to see $\mathbb{S}^n$ as the one-point compactification of $\mathbb{R}^n$.

Solution 4:

$\mathbb{R}^n$ can be seen as homeomorphic to an $n$-sphere with a point removed via stereographic projection. When an author says 'include $\mathbb{R}^n$ into $\mathbb{S}^n$' he is refering to a homeomorphism $\mathbb{R}^n\rightarrow \mathbb{S}^n\setminus \{N\}$ where $N$ is the north pole of the sphere.