a continuous version of axiom of choice?

Question

Let $L_t(s):S^1 \rightarrow \mathbb{R}^2(t\in [0,1])$ be a Jordan curve, $O(t)$ be its interior and $H(t,s)=L_t(s)$. If $H$ is a homotopy from $L_0(s)$ to$L_1(s)$, does there exist a continuous function $f:[0,1]\rightarrow \mathbb{R}^2$ s.t. $\forall t \in[0,1],f(t)\in O(t)$?

Answer 1

By compactness one can replace the Jordan curves by a family of polygonal ones. The cut locus $C(t)$ of the Jordan curve $L_t$ is a rectifiable tree and one can choose the midpoint $m(t)$ of $C(t)$ is the value of $f$ at $t$, giving a continuous function with image in the interior of the Jordan curve for every $t$.

Answer 2

Here's a simple construction of the function $f$. It makes use of the Hausdorff metric on compact subsets of $\mathbb{R}^2$, defined by $$d_H(A,B) = \inf\bigl\{r \in [0,\infty) \,\bigm|\, A \subset N(B,r), \, B \subset N(A,r)\bigr\} $$ The image $L_t(S^1)$ is compact for each $t$. Continuity of the homotopy $H(t,s)$ implies continuity of the function $t \mapsto L_t(S^1)$ in the Hausdorff metric space on compact sets.

For each $t \in [0,1]$ make the following choices:

• pick $x_t \in O(t)$,
• then pick $r_t > 0$ so that $d(x_t,L_t(S^1)) > r_t$,
• then pick an open subset $I_t \subset [0,1]$ containing $t$ such that for all $s \in I_t$ we have $d_H(L_t(S^1),L_s(S^1)) < r_t$.

It follows from these choices that for each $t$ and each $s \in I_t$, the open ball $B(x_t,r_t)$ is a subset of $O(s)$. The proof is that $L_t(S^1)$ has winding number $1$ around $x_t$; but winding number of $L_s(S^1)$ around $x_t$ is continuous as a function $s \in I_t$, and so also equals $1$ for all $s \in I_t$.

The sets $I_t$ form an open covering of $[0,1]$. Choose a Lebesgue number $\lambda$ for this open covering. Let $$0 = t(0) < t(1) < \cdots < t(K) =1 $$ be a sequence of intervals such that $t(k) - t(k-1) < \lambda$ for $k=1,...,K$.

From the choices, it follows for each $k=1,...,K$ and each $s \in [t(k-1),t(k)]$ that the straight line segment $\overline{x_{t(k-1)},x_{t(k)}}$ is disjoint from $L_s(S^1)$. Thus by concatenating these segments from $k=1$ to $K$ we obtain a piecewise straight path $f(t)$ such that $f(t) \in O(t)$ for each $t \in [0,1]$.