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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.

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  • $\begingroup$ Sphere = surface or also the interior of the ball? $\endgroup$ – user370967 Nov 13 '18 at 21:43
  • $\begingroup$ If it is the latter: show two easy things: (1) sphere is convex (2) convex implies path connected $\endgroup$ – user370967 Nov 13 '18 at 21:45
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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. $\Box$

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  • $\begingroup$ Thank you for your complete answer but I have revise the last part of the question to be clear what I expect as the answer. I am not allowed to use the fact that you have mentioned. We just have that definition of arcwise connected and have been asked to prove that. How can we do that? $\endgroup$ – Saeed Nov 13 '18 at 21:26
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    $\begingroup$ @Saeed I'm not really sure what exactly you are not allowed to use? There are few details missing in my answer (like showing that each function is continuous) but both facts + lemma have a proof here. $\endgroup$ – freakish Nov 13 '18 at 21:28
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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.

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  • $\begingroup$ Could you tell me how to do stereographic projection? It is not clear to me. I have read wiki on this but does not help. $\endgroup$ – Saeed Nov 13 '18 at 20:33
  • $\begingroup$ @Saeed You can read about it on standard topology text, such as Lee’s Topological Manifold page 56. The point is that stereographic projection is a homeomorphism so you can pass your path in $R^2$ to the sphere minus a point. $\endgroup$ – Sou Nov 13 '18 at 20:42
  • $\begingroup$ My problem is I am not a math student and I have taken real analysis as my graduate course and this question is a homework problem. My professor has said nothing about your answer. So there should be some solution which does not depend on stereographic projection. $\endgroup$ – Saeed Nov 13 '18 at 20:55

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