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I have two questions which pertain to differentiability, connectivity and path connectivity. Ocasionally, I will encounter an author who defines connectivity in the following way:

An open subset $U$ of $\mathbb{R}^n$ is said to be connected if and only if given two points $a$ and $b$ of $U$ there exists a differentiable mapping $\phi: \mathbb{R} \rightarrow U$ such that $\phi(0) = a$ and $\phi(1) = b$.

This particular example is from Edward's Advanced Calculus of Several Variables p 84. Now, this is obviously not the standard definition we learn from topology which has nothing to do with differentiability but rather whether there exists two nonempty open subsets that comprise a separation. It also seems to me that what the author is really defining what it means for a space to be "smoothly path connected", which of course implies connectivity and, it seems to me, considerably more.

My first question is: Is "smoothly path connected", as defined above actually equivalent to "connected", in the topological sense, in $\mathbb{R}^n$?

Next, in Vector Calculus by Baxandall and Liebeck on p 150 the authors state the existence of a continuous path $\alpha$ from the closed interval $[0,1]$ to an open subset $D$ of $\mathbb{R}^n$ where $\alpha (0) = a \in D$ and $\alpha(1) = b \in D$ guarantees the existence of a differentiable path with the same properties. This claim is stated without proof.

My second question is: Can someone provide a reference to a proof of the above claim or explain why it is so?

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For the second question, is the codomain supposed to be, say, an open subset $U$ of $\mathbb{R}^n$? –  Qiaochu Yuan Jun 17 '11 at 20:41
    
Author doesn't define it that way; the path is from the closed interval $[0, 1]$ to $\mathbb{R}^n$ –  ItsNotObvious Jun 17 '11 at 20:49
    
@Qiaochu Scratch that last comment; you are correct about the codomain. I will fix the post –  ItsNotObvious Jun 17 '11 at 20:59

3 Answers 3

up vote 7 down vote accepted

Proposition: Let $X$ be a locally path-connected space (in particular, any open subset of $\mathbb{R}^n$ has this property). Then every path component of $X$ is a connected component of $X$. In particular, $X$ is connected if and only if it is path-connected.

Proof. Let $U$ be a path component of $X$. If $x \in U$, then there is an open neighborhood $V$ containing $x$ which is path-connected, hence $V \subseteq U$. It follows that $U$ is open. If $x \in \bar{U}$, again choose an open neighborhood $V$ containing $x$ which is path-connected. By assumption, this neighborhood intersects $U$, so it follows that $V \subseteq U$. Hence $U$ is closed. It follows that $U$ and its complement in its connected component $C$ are disjoint open sets whose union is $C$, hence that $U = C$.

Proposition: Let $X$ be an open subset of $\mathbb{R}^n$ (or a smooth manifold). If two points $a, b$ are connected by a path in $X$, then they are connected by a smooth path in $X$.

Proof. Let $\alpha : [0, 1] \to X$ be such a path. Choose for each point $\alpha(t)$ an open ball $U_t$ containing $\alpha(t)$ and contained in $X$. By compactness, the $U_t$ have a finite subcover $U_{t_1}, ... U_{t_n}$. Now it is not hard to explicitly write down a smooth path from $a$ to $b$ going through the balls $U_{t_i}$. (For example, it is trivial to write down a piecewise-linear path with this property, and then one just has to deform this path slightly in a neighborhood of each of its points of nondifferentiability using a smooth bump function.)

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1

Yes, smooth-path-connectedness is the same thing as path-connectedness.

For example, suppose $\gamma:[0,1]\to U$ is a path in an open set $U\subseteq\mathbb R^n$ from $a=\gamma(0)$ to $b=\gamma(1)$. Then you can show that there is a map $\eta:[0,1]\to U$, also from $a$ to $b$, which is in fact piecewise linear and whose linear segments are parallel to the coordinate axes. Next, show that such a map can be deformed very slightly into a smooth one.

2

Path-connectedness and connectedness is the same thing for open subsets of $\mathbb R^n$.

That the first implies the second is more or less immediate. The converse implication is a consequence of the fact that open subsets of $\mathbb R^n$ are locally path-connected: each point has an open neighborhood which is path connected: namely, a sufficiently small open ball centered at it.

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I edited the question to try to make it more clear, but not really sure why a path with domain $[0, 1]$ and codomain $\mathbb{R}^n$ doesn't make sense. –  ItsNotObvious Jun 17 '11 at 20:51
1  
@·Sphere: if the domain is $\mathbb R^n$, there is no need of any hypothesis (like the existence of a continuous map... etc) to guarantee there is a smooth map from $a$ to $b$: just take the line passing through the points! –  Mariano Suárez-Alvarez Jun 17 '11 at 20:53
    
Note that for 1, the hypothesis that $U$ is open is essential. A non-open subset of $\mathbb{R}^n$ may be connected without being path connected (e.g. the topologist's sine curve), or path connected without being smoothly path connected (e.g. the graph of $y = |x|$ in $\mathbb{R}^2$). –  Nate Eldredge Jun 17 '11 at 20:57
    
Yes, your right, will fix the post. The path is in an open subset of $\mathbb{R}^n$ –  ItsNotObvious Jun 17 '11 at 20:58

I will try to give a detailed proof of the second question by ItsNotObvious above. My justification for answering a year-old, already answered question, which is probably a very well-known fact as well, is that I couldn't find a detailed proof of it anywhere in the literature.

Proposition: Let $M$ be a connected smooth manifold. Then for every two points $p,q \in M$ there exists a smooth (i.e. smooth at all of $(0,1)$) path $$\gamma: [0,1] \rightarrow M$$ such that $\gamma(0)=p$ and $\gamma(1)=q$.

Proof. We will establish the proof by proving two lemmas, which when combined imply the proposition. First we need to facts.

Fact 1: Every n-dimensional (not necessarily smooth) manifold $M$ admits a covering by coordinate charts $(U_{i}, \varphi_{i})$, where each of the $U_{i}$ is homeomorphically mapped by $\varphi_{i}$ to an open ball around the origin in $\mathbb{R}^{n}$. In particular $M$ is locally path-connected.

In fact the above covering can be chosen to be countable, but we won't need that fact.

Fact 2: A locally path-connected topological space is connected, if and only if it is path-connected. Hence connected manifolds are path-connected.

More or less detailed proofs of those facts can be found in J.M. Lee “Introduction to Smooth Manifolds” Lemma 1.6. (Fact 1) and in “Introduction to Topological Manifolds” Proposition 4.26 e) (Fact 2) by the same author.

Lemma 1: Let $M$ be a connected smooth manifold. Then for every two points $p,q \in M$ there exists a path, which is non-smooth at only finitely many points in $(0,1)$, and such that for all $t \in [0,1]$ all left and right derivatives of any order exist (i.e. $\gamma$ is piecewise smooth).

Proof. Let $\gamma: [0,1] \rightarrow M$ be a path joining $p$ and $q$. By Fact 1 there is a cover $\{U_{i}\}_{i \in I}$ of $\gamma([0,1])$ consisting of charts $(U_{i},\varphi_{i})$, such that each of the open sets $U_{i}$ is homeomorphic to an open ball around the origin via $\varphi_{i}$. Now the preimages $\gamma^{-1}(U_{i})$ form an open cover of $[0,1]$. By the Lebesgue Lemma there is an $n \in \mathbb{N}$, such that the finitely many intervals $$[\frac{k}{n},\frac{k+1}{n}] \qquad k=0,…,n-1$$ cover $[0,1]$ and such that for every $k$ there exists a $j_{k} \in I$ with $\gamma([\frac{k}{n},\frac{k+1}{n}]) \subset U_{j_{k}}$.

Let $\alpha_{k}: [\frac{k}{n},\frac{k+1}{n}] \rightarrow \varphi_{j_{k}}(U_{j_{k}})$ be a straight line, parametrized by $[\frac{k}{n},\frac{k+1}{n}]$ connecting $\varphi_{j_{k}}(\gamma(\frac{k}{n}))$ with $\varphi_{j_{k}}(\gamma(\frac{k+1}{n}))$, which is possible as $\varphi_{j_{k}}(U_{j_{k}})$ is an open ball and thus convex. Now define a new path $$\tilde{\gamma}: [0,1] \rightarrow M$$ piecewise by $$\tilde{\gamma}(t) = \varphi_{j_{k}}^{-1}(\alpha_{k}(t)), \qquad \text{if} \quad t \in [\frac{k}{n},\frac{k+1}{n}]$$ It is clear from the definition that $\tilde{\gamma}$ is a piecewise smooth path joining $p$ and $q$.

Lemma 2: Let $\gamma_{1}, \gamma_{2}: [0,1] \rightarrow M$ be two smooth curves with $\gamma_{1}(1)=\gamma_{2}(0)$. There is a smooth curve $$\gamma: [0,1] \rightarrow M$$ such that $\gamma(0)=\gamma_{1}(0)$ and $\gamma(1)=\gamma_{2}(1)$. In particular any two points $p,q \in M$, which can be connected by a piecewise smooth curve can be connected by a smooth curve.

Proof. Define $$\sigma_{1}: [0,1] \rightarrow [0,1], \qquad t \mapsto 1-e^{-\frac{1}{1-t}+1}$$ $$\sigma_{2}: [0,1] \rightarrow [0,1], \qquad t \mapsto e^{-\frac{1}{t}+1}$$ Both maps are homeomorphisms of $[0,1]$ onto itself. Consider the path $$\gamma:=(\gamma_{1} \circ \sigma_{1}) \star (\gamma_{2} \circ \sigma_{2})$$ where $\star$ denotes the composition of two paths. It is clear that $\gamma$ is smooth everywhere on $(0,1)$, except for possibly $t=\frac{1}{2}$.

Denote by $\partial_{+}$ the right derivative and by $\partial_{-}$ the left derivative. By applying the chain rule for semi-differentiable functions it is easy to see that $$\partial_{+}\gamma(\frac{1}{2})=\partial_{+}(\gamma_{1} \circ \sigma_{1})(1)=\partial_{+}\gamma_{1}(\sigma_{1}(1)) \partial_{+}{\sigma_{1}(1)}= \partial_{+}\gamma_{1}(\sigma_{1}(1)) \cdot 0=0$$ and likewise $\partial_{-}\gamma(\frac{1}{2})=0$. This shows in particular that the left and the right derivative of $\gamma$ in $\frac{1}{2}$ agree, and a similar calculation shows that all higher left and right derivatives agree at $\frac{1}{2}$ as well. Hence $\gamma$ is smooth

Putting the two facts and the two lemmas together gives the proof of the proposition.

Edit: One could alternatively prove Lemma 1 by showing that for a given point $p \in M$ the set of points which can be connected by a piecewise smooth path with $p$ is non-empty and both open and closed in $M$ and hence all of $M$, as $M$ is connected. The proof is almost the same as the proof of Fact 2. Roughly speaking, in the proof of Fact 2 (sketched by Qiaochu in his answer) you replace "path" by "piecewise smooth path" everywhere. I just realized this after posting my answer.

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