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Im reading Chapter1 of Rudin's Principles of Mathematical Analysis, 3rd ed and a little confusing on his construction of $R$ from $Q$, enter image description here enter image description here

See the step3. "Define γ to be the union of all α∈A" means that γ is a subset of $Q$ which has no least upper bound property. So how could we write γ = sup A?

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  • $\begingroup$ No, it means that $\gamma$ is a subset of $\mathbb{Q}$, not that it belongs to $\mathbb{Q}$ $\endgroup$ – Brandon Jan 3 '14 at 4:22
  • $\begingroup$ @Brandon: Yes, you're right. I've edited it. $\endgroup$ – Bear and bunny Jan 3 '14 at 4:26
  • $\begingroup$ It is more an issue of notation than anything. People write "$x = f(y)$" and mean two separate things, firstly that "$f(y)$ is defined" and secondly that "$x = f(y)$". $\endgroup$ – user21820 Jan 3 '14 at 4:31
  • $\begingroup$ I don't understand your confusion. He goes on to prove $\gamma = \sup A$, where $\sup A$ is defined as usual for any ordered set. Intuitively, Dedekind cuts are just the observation that given a real number $x$, we can "recover" $x$ from $\alpha_x = \{q \in \mathbb{Q} : q < x\}$. The intuition behind Step 3 is that given a bunch of $\alpha_x$'s for $x \in A$, their union $\cup_{x \in A} \alpha_x$ is $\alpha_{\sup A}$. The proof is of course abstract since referencing the reals is circular, but this is the intuition, which is very straightforward. $\endgroup$ – J Swanson Jan 3 '14 at 4:35
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You misread the order. When we say $\sup A$, we don't mean in the rational numbers, we mean in the new world, $R$ whose ordering is $\subsetneq$.

The claim, if so, is that a union of a bounded set of cuts is a cut itself. And that cut is their supremum.

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  • $\begingroup$ Thanks for your reply. γ satisfies the three properties in definition of Cut but γ is a subset of $R$ or say γ can contain irrational numbers, is that right? $\endgroup$ – Bear and bunny Jan 3 '14 at 5:39
  • $\begingroup$ Now you're misreading the definition of $\gamma$. $\endgroup$ – Asaf Karagila Jan 3 '14 at 5:45
  • $\begingroup$ for example, A={p:1<p<$\sqrt2$, p∈$R$}. Then γ should be {p:p<$\sqrt2$,p∈$Q$}. So that is to say supA=$\sqrt2$ which belongs to γ? $\endgroup$ – Bear and bunny Jan 3 '14 at 16:47
  • $\begingroup$ @Frank, the elements of $A$ are subsets of $Q$. Their union is a subset of $Q$. $\sqrt2$, in this context, is defined as a subset of $Q$. $\endgroup$ – Asaf Karagila Jan 3 '14 at 16:55
  • $\begingroup$ In this context, $R$ is a subset of $Q$? $\endgroup$ – Bear and bunny Jan 3 '14 at 16:59
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$\gamma$ is a union of subsets of $Q$ and therefore is a subset of $Q$. In step 3, the author is showing that $R$ has the least upper bound property. To do so, he takes some subset $A \subset R$ and tries to show that $\gamma= \bigcup_{\alpha \in A} \alpha$ is both an element of $R$ and the desired upper bound. Does this make sense?

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    $\begingroup$ for example, A={p:1<p<$\sqrt2$,p∈$R$}. Then γ should be {p:p<$\sqrt2$,p∈$Q$}. So that is to say supA=$\sqrt2$ which belongs to γ? $\endgroup$ – Bear and bunny Jan 3 '14 at 16:45
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    $\begingroup$ Yes, that is correct. Keep in mind though that the element $\sqrt{2} \in R$ is the subset of $Q$ of all rationals less than $\sqrt{2}$. $\endgroup$ – Alexander Jan 4 '14 at 2:34
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$\alpha$ does not contain its least upper bound, but it has one. For example $$ \alpha={\mathbb Q}\cap (-\infty,0). $$ Then $$ \sup\alpha=0\quad\text{and}\quad 0\not\in\alpha. $$

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