Suppose $a\in\mathbb{R}^n$ is a point and $Y\subset\mathbb{R}^n$ is a closed set. Prove that $\exists\:y\in Y$ s.t. $d(a,Y)=|a-y|$

Question

$\bf Definition:$ If $X,Y\in\mathbb{R}^n$ are two sets, we say the distance $d(X,Y)$ between them is the greatest lower bound of $|x-y|$ where $x\in X$ and $y\in Y.$

My idea is as follows: Let $B$ be a closed ball centered at $a.$ Let $f:B\cap Y\rightarrow \mathbb{R}$ defined by $f(x)=|x-a|.$ Since $B\cap Y$ is compact, every sequence of points has subsequences that converge to a point in the set which means there is a "minimizing" point. From here I'm not sure how to conclude that $\exists\:y\in Y$ s.t. $d(a,Y)=|a-y|.$ Any thoughts?

Also, another confusion on my part is the statement in the definition "greatest lower bound of $|x-y|$." Does this mean that $d(X,Y)= \text{inf}\:|x-y|$?

Answer 1

Last part: yes, greatest lower bound is same as inf. Proof of existence of $y$: there iexisrs $\{y_n\} \subset Y$ such that $|a-y_n| \to d(a,Y)$.I t follows that $|y_n| \leq |y_n-a|+|a|<1+|a|$ for $n$ large enough , so $\{y_n\}$ is a bounded sequence. By Bolzano _ Weirstrass theorem there is a subsquence $\{y_{n_k}\}$ which converges to some point $y$. Since $Y$ is closed, $y \in Y$. Now $d(a,y)=\lim d(a, y_{n_k})=d(a,Y)$.

Answer 2

First Part: The question of existance of minimum: As in the question if $B$ is a closed ball of radius $R$ with $B \cap Y$ not empty then all points in $Y$ not in $B\cap Y$ have distnces from $x$ greater than $R$ and all the points in $B \cap Y$ have distances atmost $R$. So the point of minimum distance in $B \cap Y$ also is the point of minimum distance in $Y$