Limit of Sequence to Zero Distance Point

Theorem
Let $S$ be a non-empty subset of $\R$.

Suppose the distance $d \left({\xi, S}\right) = 0$ for some $\xi \in \R$.

Then there exists a sequence $\left \langle {x_n} \right \rangle$ in $S$ such that $\displaystyle \lim_{n \to \infty} x_n = \xi$.

Corollary
If $S$ is bounded above, then there exists a sequence $\left \langle {x_n} \right \rangle$ in $S$ such that $\displaystyle \lim_{n \to \infty} x_n = \sup S$.

If $S$ is unbounded above, then there exists a sequence $\left \langle {x_n} \right \rangle$ in $S$ such that $\displaystyle x_n \to +\infty$ as $n \to \infty$.

Proof
First we show that $\forall n \in \N^*: \exists x_n \in S: \left|{\xi - x_n}\right| < \dfrac 1 n$.

Suppose the contrary: that $\exists n \in \N^*: \not \exists x \in S: \left|{\xi - x}\right| < \dfrac 1 n$.

Then $\dfrac 1 n$ is a lower bound of the set $T = \left\{{\left|{\xi - x}\right|: x \in S}\right\}$.

This contradicts the assertion that $d \left({\xi, S}\right) = 0$.

We have from Power of Reciprocal that $\displaystyle \lim_{n \to \infty} \dfrac 1 n = 0$.

So as $\left|{\xi - x}\right| < \dfrac 1 n$ it follows from the Squeeze Theorem that $\displaystyle \lim_{n \to \infty} x_n = \xi$.

Proof of Corollary
If $\xi = \sup S$, then from Distance from Subset of Real Numbers, $d \left({\xi, S}\right) = 0$ and the result follows directly from the main result.

Note that the terms of this sequence do not necessarily have to be distinct.

If $S$ is unbounded above, then $\forall n \in \N^*: \exists x_n \in S: x_n > n$.

Hence $x_n \to +\infty$ as $n \to \infty$.