Prove that if $F$ is a nonempty closed subset of $\Bbb R$ and $x \notin F$, then $\exists y \in F$ s.t. $|z-x| \geq |y-x|$ for all $z \in F$. Prove that if $F$ is a nonempty closed subset of $\Bbb R$ and $x \notin F$, then there exists at least one point $y \in F$ such that $|z-x| \geq |y-x|$ for all $z \in F$.
Any help would be appreciated!
 A: Outline: For $R>0$ call $F_R=F\cap \bar{B}_R(x)$. Since $F$ is non-empty we have that $F_R\neq \emptyset$ for some $R>0$. Clearly any candidate $y$ must be in this $F_R$. Now define $f(z)=|x-z|$ for $z\in F_R$. Since $F_R$ is compact $f$ attains its minimum at some $y$, this $y$ does the trick.
A: I think this might be a good start, but as you will see once you get to the end, this solution is not complete, because I am having trouble proving a final statement.
Since $F$ is nonempty and closed, that means it contains its limit points.  Let $x \in \Bbb R - F$ (i.e., $x \in \Bbb R$ and $x \not \in F$).  
Suppose by contradiction that for each $y \in F$, there is some $z \in F$ such that $|y - x| > |z - x|$.  (What we are saying is that for each $y \in F$, we can find a $z \in F$ that is closer to $x$ than $y$ is).
In particular, if we pick some $y_{1} \in F$ and look at $|y_{1} - x|$, then we can find some $y_{2} \in F$ such that $|y_{2} - x| < |y_{1} - x|$.  But then we can find $y_{3} \in F$ such that $|y_{3} - x| < |y_{2} - x|$.  So, we are a sequence of elements of $F$ that is getting closer and closer to $x$.  If $\epsilon_{n} = |y_{n} - x|$, then $\epsilon_{n}$ represents the distance of the $n$-th term of the sequence from $x$.
Clearly, {$\epsilon_{n}$} is a sequence of real numbers that is monotonically decreasing ($\epsilon_{n} > \epsilon_{n + 1}$) and is bounded from below by $0$, and so it converges.  The only problem now is: does this sequence converge to $0$?  If so, then the sequence $y_{n}$ is getting arbitrarily close to $x$, which would make $x$ a limit point of $F$, and this would be a contradiction since we assumed $x \not \in F$, but we also assumed $F$ is closed (i.e., it contains its limit points).  So we would have $x \in F$ and $x \not \in F$.
I'm not entirely sure yet how to prove $\epsilon_{n}$ does not converge to some positive number.  When I figure it out, I will update this answer.
