# 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 $x$

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 $$x > \sup S$$

Thank you very much for your help in this regard.

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