Let $D$ be an open subset in $\mathbb{R}^n$. Let $\alpha$ be a path in $D$ from $x$ to $y$, and set $d=\inf\{|\alpha(s)-w|:w\in \partial D, 0\le s\le 1\}$. Show that if $\beta$ is any path in $D$ from $x$ to $y$ such that $|\beta(s)-\alpha(s)|<d,0\le s\le 1$, then $\beta$ is homotopic to $\alpha$ with endpoints fixed.
I get the picture. If $D$ is a convex set then this is automatically true. But even if $D$ is not, it wouldn't be a problem because $\beta(s)$ would not leave the set $D$ since the "distance" between $\beta$ and $\alpha$ are restricted. But I'm having a hard time formalizing this idea. Any idea? Thanks.
We have $$ \inf\{|\alpha(s)-w|:w\in \partial D, 0\le s\le 1\}=\inf\{|\alpha(s)-w|:w\notin D, 0\le s\le 1\}. $$ Indeed, the inequality $\ge$ follows from the inclusion $\partial D\subset\mathbb R^n\setminus D$. For $\le$ consider for every $\alpha(s)\in D$ and $w\notin D$ the segment $[\alpha(s),w]$. This segment is connected, hence it must meet $\partial D$: otherwise the open sets $D, \mathbb R^n\setminus\overline D$ would separate the segment. Consequently, there is some $w'\in\partial D$ closer to $\alpha(s)$ than $w$, and we get the $\le$ inequality.
After this, set $H_t(s)=(1-t)\alpha(s)+t\beta(s)$. Then $H_t(s)\in D$ for $0\le s,t\le 1$ because $$ |\alpha(s)-H_t(s)|=t|\alpha(s)-\beta(s)|<d=\inf\{|\alpha(s)-w|:w\notin D, 0\le s\le 1\}. $$