Understanding a Proof: Open connected sets and differentiable paths

Question

I'm reading a book of complex analysis (Jerrold E. Marsden) and I came across a demonstration I can´t understand. I really want to understand it.

$\textbf{Proposition}:$ If $C$ is an open connected set and $a$ and $b$ are in $C$, then there is a differentiable path $\gamma:[0, 1]\rightarrow C$ with $\gamma(0)=a$ and $\gamma(1)=b$.

$\textbf{Proof}:$ Let $a\in C$. If $z_o\in C$, then since $C$ is open, there is an $\epsilon>0$ such that the disk $D(z_o; \epsilon)$ is contained in $C$.

$\textit{So far so good}$

By combining a path from $a$ to $z_o$ with one from $z_o$ to $z$ that stays in the disk, we see that:

$z_o$ can be connected to $a$ by a differentiable path if and only if the same is true for every point $z\in D(z_o;\epsilon)$

$\textit{I can't see how the above is true}$

This shows that both the sets $A=\{z\in\mathbb C | z\text{ can be connected to }a \text{ by a differentiable path}\}$ $B=\{z\in\mathbb C | z\text{ cannot be so connected to }a\}$

are open.

$\textit{Why they are open?}$

Since $C$ is connected, either $A$ or $B$ must be empty. Obviously it must be B.

$\text{I dont´have problems with the conclusion.}$

I thank everyone who helps me understand.

Answer 1

Let $c$ be a path from $a$ to $z_0$, the concatenation of $c$ and the seqment $[z_0,z]$ is a continuous path from $a$ to $z_0$ which is smooth everywhere but not always at $z_0$, you can replace this path by a smooth path from $a$ to $z$. Draw a picture.

The same method extend a path from $a$ to $z$ to a path from $a$ to $z_0$.

$A$ is open, because if $z_0\subset A, B(z_0,\epsilon)\in A$ where $B(z_0,\epsilon)\subset C$.

Suppose that $B$ is not open, there exists $z\in B$ such that for every integer $n>0$, there exists $x_n\in B(z,1/n)$ such that there exists a smooth path between $a$ and $x_n$. There exists an integer $N$ such that $B(z,1/N)\subset C$, since there exists a smooth path between $a$ and $x_N$ the first step implies the existence of a smooth path between $a$ and $z$. Contradiction.

Answer 2

First: C is open iff $\forall z\in C\ \exists \varepsilon >0: B(z,\varepsilon)\subset C$

Second: I'd like to give a slightly different and (hopefully) more accessible proof

Proposition 1 : If $C$ is an open connected set and $a$ and $b$ are in $C$, then there is a chain of line segments (i.e. a path which is linear up to finite many edges) $[z_0,\ldots,z_n]$ with $z_0=a$ and $z_n=b$.

Proposition 2 : If $C$ is an open set, a chain of line segments $[z_0,\ldots,z_n]\subset C$ admits a differentiable path $\gamma:[0, 1]\rightarrow C$ with $\gamma(0)=z_0$ and $\gamma(1)=z_n$.

Proof of 1 : Following the original proof we fix $a\in C$ and define $A_l=\{z\in C | z\text{ can be connected to }a \text{ by a chain of line segments}\}$ $B_l=\{z\in C | z\text{ cannot be so connected to }a\}$

For $z\in A_l$ choose a ball $B(z,\varepsilon)\subset C$. For $w\in B(z,\varepsilon)$ attach $[z,w]$ to the chain from $a$ to $z$. This shows $B(z,\varepsilon)\subset A_l$ and therefore $A_l$ open. The argument for $B_l$ is similar.

Proof of 2 : For each $z_k$ choose a small ball $B(z_k,\varepsilon_k)\subset C$ and replace the edge by an arc. Writing down the exact formulas is somewhat messy but much easier than for a generic smooth path. A picture should do for the necessary insight.

Answer 3

A part that might cause confusion is the differentiable trajectory. This simply means that the trajectory is smooth (contrary to having zigzags for example).

The proof might be rephrased as follows:

For a given fixed point $a \in C$, let's define two sets:

$A=\{z\in C | z\text{ can be connected to }a \text{ by a differentiable path}\}$

$B= C - A = \{z\in C | z\text{ can NOT be connected to }a \text{ by a differentiable path}\}$

Now let's prove that both sets are open relative to $C$. Note that if we prove that $B$ is open relative to $C$ then since $B$ is the complement of $A$ relative to $C$, $A$ would be closed relative to $C$. Thus $A$ would be closed and open relative to $C$ at the same time. The same logic would apply to $B$, and $B$ would be open and closed relative to $C$ at the same time. Since $C$ is connected, both $A$ and $B$ can only be either the empty set or $C$ itself, and because they are complements of each other one would have to be the empty set and the other $C$ itself.

To prove that the set $A$ is open relative to $C$, we have to prove that for any $z \in A$, there is an open disk $D(z; \epsilon)$, such that the $D(z; \epsilon) \cap C$ is a subset of $A$. So, given some $z \in A $, since $A \subset C$ and C is an open set we can find an open disk $D^*(z; \epsilon)$, such that it is wholly contained in C.

Now to show that $D^*(z; \epsilon) \cap C = D^*(z; \epsilon)$, is also contained in $A$, let's take some point $z_o \in D^*(z; \epsilon)$ and prove that there is a differentiable path from $a$ to $z_o$ and thus $z_o \in A$ and $A$ is an open set relative to C.

Since $z \in A$ we can find a differentiable path from $a$ to $z$, now we just need to connect this path to a differentiable path from $z$ to $z_o$, to get a differentiable path from $a$ to $z_o$. But this is easy to find just create a smooth arch from $z$ to $z_o$ (it is easy because we are creating this differentiable path inside a disk, which we were able to find because $C$ is open) and we have a differentiable path from $a$ to $z_o$. Thus we see that $A$ is open relative to $C$.

To see that $B$ is open relative to $C$, we follow a similar argument. Given a point, $z \in B$, since $A \subset C$ and $C$ is an open set we can find an open disk $D^{**}(z; \epsilon)$, such that it is wholly contained in $C$.

Let $z_o \in D^{**}(z; \epsilon)$, then $z_o$ can not belong to $A$ since if that were the case, we would be able to find a differentiable path from $a$ to $z_o$, and we could easily find a differentiable path from $a$ to $z$ by connecting the path from $a$ to $z_o$ and a smooth arch from $z_o$ to $z$ within the disk $D^{**}(z; \epsilon)$. Thus $z_o$ must belong to $B$, and thus $B$ is an open set relative to $C$.

Now we have proved that both $A$ and $B$ are open sets relative to $C$, and since $C$ is a connected set one must be the empty set and the other $C$ itself. But $A$ must have at least one point since we can take some point in $C$, let's call it $w$, get an open disk wholly contained in $C$ with the center at $w$ and then get another point $v$ in this disk, and draw a straight line between those 2. Thus at least $v$ and $w$ belong to $A$, thus $A$ is not the empty set, then $A$ must be $C$ and $B$ must be the empty set.