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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.

enter image description here

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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.

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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.

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  • $\begingroup$ "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. $\endgroup$ – Matematleta Oct 3 '18 at 23:22

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