Show that the unit sphere in $\mathbb{R}^3$ is pathwise(=arcwise) connected?

Question

I know this question has been already asked but no one does not address the definition of pathwise(=arcwise) connectedness as the following:

A set $C$ in metric space $(M,d)$ is pathwise(=arcwise) connected if any two points of $C$ can be joined by an arc that is entirely lies in $C$.

Formally,

$$ \forall p,q \in C\,\,\, \exists \,\,\,\, \phi: [a,b] \rightarrow C \,\,\, \text{s.t.} \phi(a)=p, \phi(b)=q $$ where $\phi$ is a continuous function.

Now to show sphere is pathwise(=arcwise) in $\mathbb{R}^3$ we need such a $\phi$ that maps a closed interval in $\mathbb{R}$ to the sphere.

What is that function $\phi$ and that interval?

This is my homework problem and what we have learned in our real analysis course is the above so the professor expects us to solve the problem from the above point of view.

Please answer in this direction.

Answer 1

You can try to use stereographic projection $ \sigma : \Bbb{S}^2\smallsetminus \{*\} \to \Bbb{R}^2$ with appropriate “north pole” each time you want to construct a path between two points $p$ and $q$ in $\Bbb{S}^2$.

The key fact is that the stereographic projection is a homeomorphism, so we can pass the path in the plane to the sphere.

For any two points $p,q \in \Bbb{S}^2\smallsetminus \{*\} $ you have the corresponding points $\sigma(p),\sigma(q)\in \Bbb{R}^2$. Choose a path $\gamma : I \to \Bbb{R}^2$ joining those points, and then compose it with $\sigma^{-1}$. So you have path $\sigma ^{-1} \circ \gamma : I \to \Bbb{S}^2\smallsetminus \{*\}$ joining $p$ and $q$. If you want to be strict, just composing it one more time with inclusion map $ i : \Bbb{S}^2\smallsetminus \{*\} \hookrightarrow \Bbb{S}^2$. Then you are done.

Alternatively (without constructing paths), note that $\Bbb{S}^2$ is the image of the continous map $ \pi : \Bbb{R}^3 \smallsetminus \{0\} \to \Bbb{R}^3$ defined by $x \mapsto \frac{x}{||x||}$. Since the image of any path-connected space under a continous map is path-connected, therefore $\Bbb{S}^2$ is path-connected.

Answer 2

So there are many ways to show that a space is path connected. Spherical coordinates may be more difficult then the approach I propose. We start with this useful fact:

Lemma. If $f:X\to Y$ is a continous function and $X$ is path connected then so is the image $f(X)$.

The proof is quite simple: you take two points $a,b\in f(X)$, you connect points from their preimages and then compose that newly created path with $f$. $\Box$

With that you have a new tool to show that some space $Y$ is path connected. You first show that a different space $X$ is path connected (which might be easier) and then you show that $Y$ is a continuous image of $X$. So let's have a look at a simplier space:

Fact. $\mathbb{R}^n\backslash\{0\}$ is path connected for $n>1$.

Proof. Indeed, let $v,w\in\mathbb{R}^n$, $v,w\neq 0$. We will consider two cases:

(1) if $v, w$ are linearly independent then the path we are looking for is given by $$\varphi:[0,1]\to\mathbb{R}^n\backslash\{0\}$$ $$\varphi(t)=tv+(1-t)w$$ You can easily verify that this is a well defined continous function. The assumption that $v,w$ are linearly independent obviously means that $\varphi(t)\neq 0$ for any $t$.

(2) for $v,w$ linearly dependent take any point $z\in\mathbb{R}^n$ such that $z$ does not lie on the line connecting $v, w$ (or in other words such that $v,z$ are not linearly dependent). Such point always exists when $n>1$, you gain it by taking $v=(v_1,\ldots, v_n)$, now taking its first non-zero coordinate say $v_m$ and reversing its sign into $-v_m$.

In that case we can apply (1) to both $(v,z)$ and $(z,w)$ pairs in order to produce two paths. You join those two paths to complete the proof. $\Box$

Fact. The $n$-dimensional sphere $\mathbb{S}^n=\{v\in\mathbb{R}^{n+1}\ |\ \lVert v\rVert=1\}$ is path connected for $n\geq 1$.

Proof. Let

$$N:\mathbb{R}^{n+1}\backslash\{0\}\to \mathbb{S}^n$$ $$N(v)=\frac{v}{\lVert v\rVert}$$

and note that this is a surjective, continuous function. Both Lemma and previous Fact apply. $\Box$