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Let $G$ be an open-connected subset of $\mathbb{C}$.

Let $a,b$ be two distinct points in $G$.

How do I prove that there exists a $C^1$-curve $\alpha:[0,1]\rightarrow G$ such that $\alpha(0)=a$ and $\alpha(1)=b$?

Here's how I tried:

I have proven that there exists a polygonal path joining $a,b$ just like below.

enter image description here

Then, this curve is $C^1$-curve except for the "edges" of the curve.

Now let's focus on an edge.

enter image description here

Since the image is lying in an open set $G$, we can have an open neighborhood $N$ of an edge. And if we transform the curve in $N$ to a dotted line, then it would be a $C^1$ cirve around an edge.

However, I have a trouble with formalizing this idea.

How do I formally show that a curve-image around an edge can be transformed into a dotted line?

Rubertos
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2 Answers2

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The big gap in your proof is that you've merely drawn one open set and one pair of points and shown by picture how to join the points by a path within the open set.

In order to get a formal proof you need a strategy that depends on an open set $G$ and a pair of points $a$ and $b$ in $G$. Here's the usual approach:

Let $U$ denote the set of points $x$ in $G$ such that there exists a $C^{1}$ curve $\alpha:[0, 1] \to G$ such that $\alpha(0) = a$ and $\alpha(1) = x$. Show that $U$ is non-empty (easy), open (along the lines of your "smoothing" argument for polygonal paths, see below), and closed (by the preceding step, the complement of $U$ in $G$ is open).

The primary technical step is the following lemma: If $x \in U$, and if $r > 0$ is a real number such that the disk $D_{r}(x)$ of radius $r$ centered at $x$ is contained in $G$, then $D_{r}(x) \subset U$.

The idea of the proof is: By hypothesis, you can join $a$ to $x$ by a $C^{1}$ curve. Moreover, if $y \in D_{r}(x)$, then $x$ and $y$ can obviously be joined by a $C^{1}$ curve (e.g., a radial segment). Now perform the "rounding off" argument to join $a$ to $y$ be a $C^{1}$ curve.

Andrew D. Hwang
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Another approach: A corollary of the Weierstrass approximation theorem says that if $f:[0,1]\to \mathbb {R}$ is continuous, then there is a sequence of polynomials $p_n$ converging uniformly to $f$ on $[0,1]$ such that $p_n(0) = f(0),p_n(1) = f(1)$ for all $n.$

Now let $\alpha :[0,1]\to G$ be a continuous map (like your polygonal map say) such that $\alpha (0)=a, \alpha (1) =b.$ Write $\alpha = u+iv.$ Then there are polynomials $p_n\to u, q_n\to v$ as in the Weierstrass corollary. The maps $p_n+iq_n$ are polynomial maps into $\mathbb {C}$ connecting $a$ to $b$ for all $n.$ The range of $p_n+iq_n$ lies in $G$ for large $n$ because of the uniform convergence, and we're done.

zhw.
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