# Understanding a Proof: Open connected sets and differentiable paths

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. 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.

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.

• "you can replace this path by a smooth path from a to z. " But this is the crux of the matter. Obviously, since balls are convex, you can get a continuous path from $a$ to $z$ by attaching a segment. But the question is how do you pass to a smooth map? The rest of the proof follows easily from this. – Matematleta Oct 3 '18 at 23:22