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..
 A: 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.
A: Hint: Path-connected implies connected. (A proof of this is at ProofWiki.)
A: 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.
A: 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 $n \geq 1$).
A: *

*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 $(x_1,...,x_n, -\sqrt{1-x_1^2-...-x_n^2})$.


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

*$S^n$ 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.

*$S^n$ 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.

*$S^n$ 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 $y$.
A: 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).
