Are there path-connected but not polygonal-connected sets?

Question

This question came up in my mind.

In the scope of normed spaces, does there exist a path-connected but not polygonal-connected set?

I'd rather say no for open sets (my intuition is that they're 'locally convex').

Thanks for your suggestions.

Answer 1

The statement is not true for arbitrary path-connected sets. For example, the unit circle in $\mathbb{R}^2$ is path-connected, but no two points are polygonal-connected.

However, your intuition is correct for open, path-connected sets. To see this, let $U$ be open and path connected, and let $a,b\in U$ be any two points, with $\gamma:[0,1]\to U$ injective and continuous such that $\gamma(0)=a$ and $\gamma(1)=b$. Let $p_0=a$, $t_0=0$, and let $d_0=\mathrm{dist}(p_0,U^c)$. Then define $B_0$ to be the ball centered at $p_0$ with radius $d_0$. Then, inductively define $p_n$ to be the point $\gamma(t_n)$ where $t_n$ satisfies $1>t_n>t_{n-1}$ and is the smallest value $t$ in this region such that $\gamma(t)$ is on the boundary of $B_{n-1}$. If no such $t_n$ exists, then let $p_n=b$. Similarly, define $d_n=\mathrm{dist}(p_n,U^c)$ and $B_n$ to be the ball centered at $p_n$ of radius $d_n$.

If this process ever stops with $p_n=b$, then by the convexity of open balls, the polygonal path $p_0,p_1,\ldots,p_n$ is a polygonal path connecting $a$ and $b$ contained in $U$. So we must show that the process stops.

If not, then there are infinitely many $p_i$ which must have a limit point $p$ in $\mathrm{im}(\gamma)$. Let $t\in[0,1]$ be such that $\gamma(t)=p$. Let $B$ be centered at $p$ with radius $d=\mathrm{dist}(p,U^c)$. Notice by construction, we must have $t_n<t$ for all $n$. Since the $p_i$ approach $p$, we may find a $p_m$ such that $\mathrm{dist}(p_m,p)<d/2$. Since $p\notin B_m$ (else $t_{m+1}>t$), we have $d_m<d/2$. But we also have $d_m=\mathrm{dist}(p_m,U^c)>d/2$ because $B\subset U$. This is a contradiction.