Any Real Number
Let r be any real number. Let r' be a nearest neighbor to r.
\begin{equation} r = r' - \epsilon \hspace{.5 cm} (\epsilon \geq 0) \end{equation}If $\epsilon > 0$, then $r$ and $r'$ are discrete, not continuous, and do not exist as real numbers. If $\epsilon = 0$, then $r = r'$ and neither $r$ nor $r'$ exist as distinguishable real numbers.
Therefore, no real number can be proven to be a distinguishable, continuous real number.
The Set of Real Numbers
Does the set of real numbers exist? If members of a set are discrete, then a set of continuous numbers cannot exist. If members of a set need not be discrete, the set of real numbers can be represented as follows:
\begin{equation} \mathbb{R} = \lim_{n\to\infty}[r-n\epsilon,...,r-2\epsilon,r-\epsilon,r,r +\epsilon, r+2\epsilon, ..., r+ n\epsilon] \end{equation}Once again, if $\epsilon > 0$, the members of $\mathbb{R}$ are discrete. If $\epsilon = 0$, then $\mathbb{R}$ has only $r$ as its one and only member, or, if it is also true that $n=\infty$, $\mathbb{R}$ has member $r$ and two additional undefined members. Therefore, the set of real numbers $\mathbb{R}$ does not exist as we understand it.
The False Premise
Premise: The set of real numbers exists.
Any proof based on the above premise is unsound.
