# Proving connectedness of the $n$-sphere

My homework assignment contains the following question:

Prove that $S^n = \{x \mid x \in \mathbb R^{n+1}, d(x,0)=1\}$ is connected.

Can you give me a hint please?

What I can do is nothing..

#### Solutions Collecting From Web of "Proving connectedness of the $n$-sphere"

Let $n\in \mathbb{N}$ now let us define a map $f: \mathbb{R}^{n+1}-\{0\}\rightarrow S^{n}$ by $f(X) = \frac{x}{||x\|}$. All that you need to show that this map is continuous. Now, since $\mathbb{R}^{n+1}-\{0\}$ is connected and the continuous image
of a connected space is connected, so $S^{n}$ is connected. Since our choice of $n$
was arbitrary, so we see that, for all $n \geq 1$, $S^{n}$ is connected.

Hint: Path-connected implies connected. (A proof of this is at ProofWiki.)

The union of two connected spaces that share at least one point is connected. You can use the fact that the sphere minus a point is homeomorphic to $\mathbb{R}^3$ and go from there.

Notice that $\mathbb{S}^n$ is the one-point compactification of $\mathbb{R}^n$, so there exists a continuous map $p : \mathbb{R}^n \to \mathbb{S}^n$ (think about the stereographic projection); in particular, $p(\mathbb{R}^n)= \mathbb{S}^n \backslash \{x_{\infty}\}$ for some $x_{\infty} \in \mathbb{S}^n$.

Let $f : \mathbb{S}^n \to \{0,1\}$ be a continuous map. Without loss of generality, you can suppose that there exists $x_0 \neq x_{\infty}$ in $\mathbb{S}^n$ such that $f(x_{\infty})=f(x_0)$. Then $f \circ p : \mathbb{R}^n \to \{0,1\}$ is also continuous and you just have to use the connectedness of $\mathbb{R}^n$ to conclude that $f$ is conStant.

Therefore, $\mathbb{S}^n$ is connected (for $n \geq 1$).

To prove that $S^n$ is path-connected, prove that through each two points $x, y \in S^n$ passes a unique big circle. Then prove that a circle is path-connected (represent it as the unit circle in $\mathbb C$ and use complex multiplication).

1. By taking $S^n$ to be the union of

$$A_1:=\{x=(x_1,…,x_{n+1}) \mid ||x||=1, ~ x_{n+1} \geq 0\},$$
$$A_2:=\{x=(x_1,…,x_{n+1}) \mid ||x||=1, ~x_{n+1} \leq 0\},$$

we have that $S^n$ is the union of two connected sets with one point in common (in fact, a lot of points in common), hence $S^n$ is connected. To check that $A_1$ and $A_2$ are connected, just notice that both are the image of a continuous function: $A_1=f(D^n)$, where $f: D^n \rightarrow S^n$ takes $(x_1,…,x_n)$ to $(x_1,…,x_n, \sqrt{1-x_1^2-…-x_n^2})$ and $A_2=g(D^n)$, where $g: D^n \rightarrow S^n$ takes $(x_1,…,x_n)$ to $(x_1,…,x_n, -\sqrt{1-x_1^2-…-x_n^2})$.

1. $S^n \cong D^n/ \sim$. Since $D^n$ is connected, so is $S^n$.

2. $S^n$ is the one-point compactification of a non-compact connected space (namely, $\mathbb{R}^n$). The result now follows due to the fact that the closure of a connected set is connected.

3. $S^n$ is (again) the union of two connected sets with one point in common (and yet again, a lot of points in common): $S^n$ without the north pole and $S^n$ without the south pole, since those two sets are homeomorphic to $\mathbb{R}^n$ by the stereographic projection.

4. $S^n$ is pathwise connected. This can be seen since any two antipodal points $x,y$ can be joined by the path $\displaystyle \frac{(1-t)x+ty}{||1-t)x+ty||}$. If $x,y$ are antipodal points, make a path from $x$ to another point $z \neq y$ as before, then join $z$ to $y$ and we have a path from $x$ to $y$.