Suppose $S$ is a subset of $\mathbb R$ and $x \in \mathbb R$. Show that there are at most two nearest points of $S$ to $x$

Question

Suppose $S$ is a subset of $\mathbb R$ and $x \in \mathbb R$. Show that there are at most two nearest points of $S$ to $x$

Attempt:

Suppose there are three nearest points $a,b,c$ of $S$ to $x$. Let $dist(x,S) =t= \inf \{~~ d(x,s) ~|~s\in S \} $ represent the distance of $x$ from $S$.

Now, by the given assumption : $d(a,x)=d(b,x)=d(c,x)=t=dist(x,S)$

Strategy $1$: If we bring a contradiction that when there are $3$ or more nearest points to $S$, one of them, say, $d(a,x)>t$, then, we can prove that there can be atmost $2$ nearest points.

Strategy $2$: If we prove that $a=b$ and $b \ne c$ by proving that $d(a,b)=0,$ then also, we can prove it.

The second way of going seems somehow better to me.

I know this result $dist(x,S) \le |x - \sup S|$ , with equality when $x > \sup S$

Could anyone please tell me how to proceed ahead.

Thank you very much for your help in this regard.

Answer 1

Hint: For a fixed real number $x$ and a number $t \geq 0$, how many real numbers are at a distance $t$ from $x$?

Answer 2

I will follow your notation.

If $t=0$, there is nothing to prove. Let's suppose $t>0$.

Now, none of $a,b,c$ is equal to $x$; thus, without loss of generality, we can suppose $a,b<x$. Since $t=x-a=x-b$, we have $a=b$.