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Let the set $L$ be definded as $$L=\{\{x\mid x<q(i)\}\mid i\in\mathbb{N}\},$$

where $q(i)$ is some bijection from $\mathbb{N}$ to $\mathbb{Q}$.

Clearly, every member of $L$ is neither an empty set nor $\mathbb{Q}$.

Every member of $L$ is closed downward.

Every member of $L$ has no largest member.

So, all members of $L$ are Dedekind cuts.

If $L$ is the set of all possible Dedekind cuts, then real numbers would be countably infinite. So, my question is, why is $L$ not the set of every possible Dedekind cut?

Edit:

I'm aware that irrational numbers are not included in $L$.

So to clarify my question, how could $\{x\mid x^2<2\lor x<0\}$ be not in $L$ although $L$ has all possible cuts corresponding to every single rational number?

I'm looking for explanation, not a counter example.

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    $\begingroup$ The set $\{x : x\leq 0 \text{ or } x^2 \leq 2 \}$ is a Dedekind cut which is not in your set $L$ (it's the real number $\sqrt{2}$). $\endgroup$
    – Sambo
    Commented May 15 at 16:22
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    $\begingroup$ No irrational belongs to your set! $\endgroup$
    – Yathi
    Commented May 15 at 16:22
  • $\begingroup$ $q(i)$ is not a bijection, $q$ is the bijection. $\endgroup$
    – Thomas Andrews
    Commented May 15 at 16:23
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    $\begingroup$ In fact, $L$ is the canonical identification of $\Bbb Q\subset\Bbb R$. So really, any choice of irrational will do $\endgroup$
    – FShrike
    Commented May 15 at 16:53
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    $\begingroup$ Is your question really "How come my intuition tells me something wrong?" Because you are asking why you expect something false to be true, but will not accept a simple counterexample as evidence that what you "expect" is just not so. To quote Mark Twain: it's not what you don't know, it's what you "know" that just ain't so. $\endgroup$
    – Arturo Magidin
    Commented May 15 at 18:48

2 Answers 2

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Your definition of the family $L$ is unnecessarily complicated. Since we already know that $\mathbb Q$ is countable, we can simply define your countable family $L$ to be the family of sets $\{x\in\mathbb Q : x < i\}$ as $i$ ranges over $\mathbb Q$. This makes it more obvious that you have simply defined the Dedekind cuts corresponding to the rationals.

Taking your favorite irrational, say $\pi$, you get a Dedekind cut $\{x\in\mathbb Q : x<\pi\}$ not in $L$.

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This might be an approach to consider.

If I understand your definition of $L$, it appears to be equivalent to a "rational cut." That is, $L$ is associated with the rational number $q$, or perhaps call it the rational cut $L\left(q\right)$. In that case, $q \in \mathbb{Q}$ is the least upper bound of the set $L\left(q\right).$ Each set of form $L$ likewise has a least upper bound in $\mathbb{Q}.$

Now, in the second case $\{x | x^{2} \lt 2 \lor x \lt0 \}$, this set does not have a least upper bound in $\mathbb{Q}.$ This is shown in Rudin Principles of Mathematical Analysis (3rd Ed.) in Example 1.9 (discussing Example 1.1). So, it is different than the family of sets $L\left(q\right)$, but still a cut.

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  • $\begingroup$ Your answer made me think again about this particular cut. Let $H=\{q(i)\mid q^2(i)>2 \land q(i)>0\}$, now does every single member of $H$ has a corresponding cut in $L$? I think the answer is yes including the cut $\{x\mid x^2<2 \lor x<0\}$. $\endgroup$
    – Mohamed Mostafa
    Commented May 16 at 5:25

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