A neighborhood $Y$ of a set $X$ such that the line segment connecting any point in $Y$ and its projection to $X$ is contained in $Y$

Question

A direct line from a point $p$ to a set $X$ is a line segment with one endpoint at $p$ and one endpoint in $X$, which is as short as any other line segment from $p$ to $X$. Given a closed set $X$ and open set $Z$ with $X \subset Z \subset \mathbb{R}^n$, must there exist an open set $Y$ in between $X$ and $Z$ which includes a direct line to $X$ from each point in $Y$?

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My thoughts:

(1) My guess is the answer is yes, and in fact I think the following stronger statement is true: there is a $Y$ such that all direct lines to $X$ from each point in $Y$ are included in $Y$.

(2) If $X$ is compact, then we know the answer is yes because in this case there exists a $\tau > 0$ such that $Y := \{y : d(X, y) < \tau\} \subset Z$. (Of course, we define $d(X, y) = \min_{x \in X} d(y, x)$.)

Originally posted on MSE without any responses.

Answer 1

This is to confirm the stronger claim mentioned in (1).

Set $X_r=X\cap \overline{B_r(0)}$ and $X^r=X\setminus B_r(0)$ (both are closed, for any $t$). Define $$ f(t)=\min\left\{1,\sup\left\{r\colon U_r(X_{t})\subseteq Z\right\}\right\}, $$ where $U_r(A)$ is the (open) $r$-neighborhood of $A$.

Define $$ Y_t=U_{f(t)}(X_t)\cap\{y\colon d(y,X_t)<d(y,X^t). $$ Notice that $X\cap B_t(0)\subseteq Y_t$, the set $Y_t$ is open and is contained in $Z$. Hence the set $$ Y=\bigcup_{t>0} Y_t $$ is also open, is contained in $Z$ and contains $X$. It remains to check the principal condition.

Consider any $y\in Y$ (say, $y\in Y_t$), and let $x\in X$ be a closest to $y$ point in $X$. By definition of $Y_t$, we have $x\in X\setminus X^t$. Now, if $y'\in(x,y)$, then $y'\in U_{f(t)}(X_t)$. Moreover, $y'$ has a unique closest point in $X$ --- namely, $x$. Hence $y'\in Y_t\subseteq Y$.