User:Alecscooper/Complete and Close Packed Metric Space is Dense-in-Itself

Theorem
Let $\left({S, \preceq}\right)$ be a close packed ordered set containing at least two distinct elements.

Let $M = \left({S, d}\right)$ be a complete metric space.

Let $\tau$ denote the topology on $A$ induced by $d$.

Let $T = \left({S, \tau}\right)$ be a topological space.

Then $T$ is dense-in-itself.

Proof
Assume, for contradiction, that $T$ is not dense-in-itself.

Then there exists some isolated point $x \in S$.

Since $S$ contains at least two distinct elements, there is at least one other point $y \in S$.

Without loss of generality, let $x \preceq y$.

As $\left({S, \preceq}\right)$ is close packed, there must exist some point $a_1 \in S$ with $x \prec a_1 \prec y$.

We can repeat this process for $i > 1$ to get a point $a_i \in S$ with $x \prec a_i \prec a_{i - 1}$.

Thus we get a sequence $\left \langle {a_n} \right \rangle$ in $S$.

This should be a Cauchy sequence, and it must converge since the space is complete. I'm just not sure where to go from there since the limit isn't necessarily $x$... I know I had a better idea of how to approach this a while ago, and promptly forgot it.