Archimedean property application The Archimedean property states that

for all $a \in \Bbb R$ and for some $n \in \Bbb N: a < n$.

Similarly,

for all $n \in \Bbb N | b \in \Bbb R | 0 < {1 \over n} < b$.

Thus, there can be found a rational number such that it is the boundary point of some open set.
In other words, there can be made Dedekind cuts $(A, B)$.
Is it true we can make infinitely many open sets - that is, does this imply that there can be infinitely many disjoint open sets $A, B$ - all bounded by some Dedekind cut?
 A: For any positive integer $n$, let $A_n=\left(\frac{1}{n+1},\frac{1}{n}\right)$.  Then the sets $A_n$ are pairwise disjoint open intervals.
If we think of the reals as Dedekind cuts, then there certainly are Dedekind cuts (reals) $a$ and b$ such that:
(i) $a\lt x$ for every $x\in \bigcup_{1}^\infty A_n$, and
(ii) $x\lt b$ for $x\in \bigcup_{1}^\infty A_n$.
If n is a positive integer, let An=(1n+1,1n). The sets An are pairwise disjoint. If we think of the reals as Dedekind cuts, then there certainly are Dedekind cuts (reals) a, b such that a

Remark: In stating the Archimedean property, we have to be very careful about the order of the quantifiers. It is often clearer to state the result without explicit use of logical symbols. For example, we could  say that for any fixed real number $x$, there is a natural number $n$  such that $n\gt x$.
In symbols, one could write $(\forall x\in \mathbb{R})(\exists n\in \mathbb{N}): n\gt x$.
The use of symbols has some disadvantages. It tends to encourage a symbol-manipulation approach to proving things. Ordinarily, for a statement with genuine mathematical content, the proof requires serious consideration of the meaning of the statement.
