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 $$\lim_{n \to \infty} x_n = \xi$$.

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

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

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

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

Then $$\frac 1 n$$ 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 $$\lim_{n \to \infty} \frac 1 n = 0$$.

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

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 $$n \to \infty$$.