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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?

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  • $\begingroup$ The property is not correctly stated. Onw should say that for every real number $a$, there is a natural number $n$ such that $n\gt a$. One can use quantifiers to say this, though on the whole it is better not to. $\endgroup$ – André Nicolas Aug 24 '13 at 0:11
  • $\begingroup$ The Archimedean property actually says that for each $x\in\Bbb R$ there is an $n\in\Bbb N$ such that $n>x$ or, equivalently, that for each real $x>0$ there is an $n\in\Bbb Z^+$ such that $0<\frac1n<x$. But even after those corrections are made, I’m not sure just what you’re asking. $\endgroup$ – Brian M. Scott Aug 24 '13 at 0:13
  • $\begingroup$ Fixed; and what I am asking is if we can create infinitely many disjoint open balls $S_r(x)$ @BrianM.Scott $\endgroup$ – Don Larynx Aug 24 '13 at 0:27
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    $\begingroup$ No, I’m afraid that it’s not fixed. Your first version is now ambiguous: it could mean $\forall a\in\Bbb R\exists n\in\Bbb N(a<n)$, which is correct, but because you wrote and, it could also mean $\exists n\in\Bbb N\forall a\in\Bbb R(a<n)$, which is clearly false. And the second version is uninterpretable, because your notation is non-standard. // The balls $S_1(2n)$ for $n\in\Bbb Z$ are pairwise disjoint. $\endgroup$ – Brian M. Scott Aug 24 '13 at 0:32
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    $\begingroup$ If $n$ is a positive integer, let $A_n=(\frac{1}{n+1}, \frac{1}{n})$. The sets $A_n$ are pairwise disjoint. If we think of the reals as Dedekind cuts, then there certainly are Dedekind cuts (reals) $a$, $b$ such that $a\lt x$, for every $x\in \bigcup A_n$ and $x\lt b$ for every $x\in \bigcup A_n$. $\endgroup$ – André Nicolas Aug 24 '13 at 0:54
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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.

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