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Exercise 34 - Chapter 1

edited April 2018 in Exercises

Is the usual \(\leq\) ordering on the set \(\mathbb{R}\) of real numbers a total order?

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  • 1.
    edited April 2018
    1. Reflexivity holds
    2. For any \( a, b, c \in \tt{R} \) \( a \le b \) and \( b \le c \) implies \( a \le c \)
    3. For any \( a, b \in \tt{R} \), \( a \le b \) and \( b \le a \) implies \( a = b \)
    4. For any \( a, b \in \tt{R} \), we have either \( a \le b \) or \( b \le a \)

    So, yes.

    Comment Source:1. Reflexivity holds 2. For any \\( a, b, c \in \tt{R} \\) \\( a \le b \\) and \\( b \le c \\) implies \\( a \le c \\) 3. For any \\( a, b \in \tt{R} \\), \\( a \le b \\) and \\( b \le a \\) implies \\( a = b \\) 4. For any \\( a, b \in \\tt{R} \\), we have either \\( a \le b \\) or \\( b \le a \\) So, yes.
  • 2.

    Perhaps, due to our interest in things categorical, we can enjoy (instead of Cauchy sequence methods) to see the order of the (extended) real line as the Dedekind-MacNeille_completion of the rationals. Matthew has told us interesting things about it before.

    Hausdorff, on his part, in the book I mentioned here, says that any total order, dense, and without \( (\omega,\omega^*) \) gaps, has embedded the real line. I don't have a handy reference for an isomorphism instead of an embedding ("everywhere dense" just means dense here).

    Comment Source:Perhaps, due to our interest in things categorical, we can enjoy (instead of Cauchy sequence methods) to see the order of the (extended) real line as the [Dedekind-MacNeille_completion of the rationals](https://en.wikipedia.org/wiki/Dedekind%E2%80%93MacNeille_completion#Examples). Matthew has told us interesting things about it [before](https://forum.azimuthproject.org/discussion/comment/16714/#Comment_16714). Hausdorff, on his part, in the book I mentioned [here](https://forum.azimuthproject.org/discussion/comment/16154/#Comment_16154), [says](https://books.google.es/books?id=M_skkA3r-QAC&pg=PA85&dq=each+everywhere+dense+type&hl=en&sa=X&ved=0ahUKEwjLkJao-9DaAhWD2SwKHVrkBcIQ6AEIKTAA#v=onepage&q=each%20everywhere%20dense%20type&f=false) that any total order, dense, and without \\( (\omega,\omega^*) \\) [gaps](https://en.wikipedia.org/wiki/Hausdorff_gap), has embedded the real line. I don't have a handy reference for an isomorphism instead of an embedding ("everywhere dense" just means dense here).
  • 3.

    I believe the hyperreal numbers give an example of a dense total order that embeds the reals without being isomorphic to it. (I can’t speak to the gaps condition though, and it’s just plausible that they’re isomorphic at the level of mere posets rather than ordered fields.)

    Comment Source:I believe the [hyperreal numbers](https://en.wikipedia.org/wiki/Hyperreal_number) give an example of a dense total order that embeds the reals without being isomorphic to it. (I can’t speak to the gaps condition though, and it’s just plausible that they’re isomorphic at the level of mere posets rather than ordered fields.)
  • 4.

    That's an interesting question, Jonathan.

    Comment Source:That's an interesting question, Jonathan.
  • 5.
    edited April 2018

    Jonathan Castello

    I believe the hyperreal numbers give an example of a dense total order that embeds the reals without being isomorphic to it. (I can’t speak to the gaps condition though, and it’s just plausible that they’re isomorphic at the level of mere posets rather than ordered fields.)

    In fact, while they are not isomorphic as lattices, they are in fact isomorphic as mere posets as you intuited.

    First, we can observe that \(|\mathbb{R}| = |^\ast \mathbb{R}|\). This is because \(^\ast \mathbb{R}\) embeds \(\mathbb{R}\) and is constructed from countably infinitely many copies of \(\mathbb{R}\) and taking a quotient algebra modulo a free ultra-filter. We have been talking about quotient algebras and filters in a couple other threads.

    Next, observe that all unbounded dense linear orders of cardinality \(\aleph_0\) are isomorphic. This is due to a rather old theorem credited to George Cantor. Next, apply the Morley categoricity theorem. From this we have that all unbounded dense linear orders with cardinality \(\kappa \geq \aleph_0\) are isomorphic. This is referred to in model theory as \(\kappa\)-categoricity.

    Since the hypperreals and the reals have the same cardinality, they are isomorphic as unbounded dense linear orders.

    Puzzle MD 1: Prove Cantor's theorem that all countable unbounded dense linear orders are isomorphic.

    Comment Source:[Jonathan Castello](https://forum.azimuthproject.org/profile/2316/Jonathan%20Castello) > I believe the hyperreal numbers give an example of a dense total order that embeds the reals without being isomorphic to it. (I can’t speak to the gaps condition though, and it’s just plausible that they’re isomorphic at the level of mere posets rather than ordered fields.) In fact, while they are not isomorphic as lattices, they are in fact isomorphic as mere posets as you intuited. First, we can observe that \\(|\mathbb{R}| = |^\ast \mathbb{R}|\\). This is because \\(^\ast \mathbb{R}\\) embeds \\(\mathbb{R}\\) and is constructed from countably infinitely many copies of \\(\mathbb{R}\\) and taking a [quotient algebra](https://en.wikipedia.org/wiki/Quotient_algebra) modulo a free ultra-filter. We have been talking about quotient algebras and filters in a couple other threads. Next, observe that all [unbounded dense linear orders](https://en.wikipedia.org/wiki/Dense_order) of cardinality \\(\aleph_0\\) are isomorphic. This is due to a rather old theorem credited to George Cantor. Next, apply the [Morley categoricity theorem](https://en.wikipedia.org/wiki/Morley%27s_categoricity_theorem). From this we have that all unbounded dense linear orders with cardinality \\(\kappa \geq \aleph_0\\) are isomorphic. This is referred to in model theory as *\\(\kappa\\)-categoricity*. Since the hypperreals and the reals have the same cardinality, they are isomorphic as unbounded dense linear orders. **Puzzle MD 1:** Prove Cantor's theorem that all countable unbounded dense linear orders are isomorphic.
  • 6.

    Hi Matthew, nice application of the categoricity theorem! One question if I may. You said:

    In fact, while they are not isomorphic as lattices, they are in fact isomorphic as mere posets as you intuited.

    But in my understanding the lattice and poset structure is inter-translatable as in here. Can two lattices be isomorphic and their associated posets not?

    Comment Source:Hi Matthew, nice application of the categoricity theorem! One question if I may. You said: > In fact, while they are not isomorphic as lattices, they are in fact isomorphic as mere posets as you intuited. But in my understanding the lattice and poset structure is inter-translatable as in [here](https://en.wikipedia.org/wiki/Lattice_(order)#Connection_between_the_two_definitions). Can two lattices be isomorphic and their associated posets not?
  • 7.
    edited April 2018

    (EDIT: I clearly have no idea what I'm saying and I should probably take a nap. Disregard this post.)

    Comment Source:(EDIT: I clearly have no idea what I'm saying and I should probably take a nap. Disregard this post.)
  • 8.
    edited April 2018

    Can two lattices be isomorphic and their associated posets not?

    If two lattices are isomorphic preserving infima and suprema, ie limits, then they are order isomorphic.

    The reals and hyperreals provide a rather confusing counter example to the converse. I am admittedly struggling with this myself, as it is highly non-constructive.

    From model theory we have two maps \(\phi : \mathbb{R} \to\, ^\ast \mathbb{R} \) and \(\psi :\, ^\ast\mathbb{R} \to \mathbb{R} \) such that:

    • if \(x \leq_{\mathbb{R}} y\) then \(\phi(x) \leq_{^\ast \mathbb{R}} \phi(y)\)
    • if \(p \leq_{^\ast \mathbb{R}} q\) then \(\psi(q) \leq_{\mathbb{R}} \psi(q)\)
    • \(\psi(\phi(x)) = x\) and \(\phi(\psi(p)) = p\)

    Now consider \(\{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \}\).

    The hyperreals famously violate the Archimedean property. Because of this \(\bigwedge_{^\ast \mathbb{R}} \{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \}\) does not exist.

    On the other than if we consider \( \bigwedge_{\mathbb{R}} \{ \psi(x) \, : \, x \in \mathbb{R} \text{ and } 0 < x \}\), that does exist by the completeness of the real numbers (as it is bounded below by \(\psi(0)\)).

    Hence

    $$ \bigwedge_{\mathbb{R}} \{ \psi(x) \, : \, x \in \mathbb{R} \text{ and } 0 < x \} \neq \psi\left(\bigwedge_{^\ast\mathbb{R}} \{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \}\right) $$ So \(\psi\) cannot be a complete lattice homomorphism, even though it is part of an order isomorphism.

    However, just to complicate matters, I believe that \(\phi\) and \(\psi\) are a mere lattice isomorphism, preserving finite meets and joints.

    Comment Source:> Can two lattices be isomorphic and their associated posets not? If two lattices are isomorphic preserving *infima* and *suprema*, ie *limits*, then they are order isomorphic. The reals and hyperreals provide a rather confusing counter example to the converse. I am admittedly struggling with this myself, as it is highly non-constructive. From model theory we have two maps \\(\phi : \mathbb{R} \to\, ^\ast \mathbb{R} \\) and \\(\psi :\, ^\ast\mathbb{R} \to \mathbb{R} \\) such that: - if \\(x \leq_{\mathbb{R}} y\\) then \\(\phi(x) \leq_{^\ast \mathbb{R}} \phi(y)\\) - if \\(p \leq_{^\ast \mathbb{R}} q\\) then \\(\psi(q) \leq_{\mathbb{R}} \psi(q)\\) - \\(\psi(\phi(x)) = x\\) and \\(\phi(\psi(p)) = p\\) Now consider \\(\\{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \\}\\). The hyperreals famously violate the [Archimedean property](https://en.wikipedia.org/wiki/Archimedean_property). Because of this \\(\bigwedge_{^\ast \mathbb{R}} \\{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \\}\\) does not exist. On the other than if we consider \\( \bigwedge_{\mathbb{R}} \\{ \psi(x) \, : \, x \in \mathbb{R} \text{ and } 0 < x \\}\\), that *does* exist by the completeness of the real numbers (as it is bounded below by \\(\psi(0)\\)). Hence $$ \bigwedge_{\mathbb{R}} \\{ \psi(x) \, : \, x \in \mathbb{R} \text{ and } 0 < x \\} \neq \psi\left(\bigwedge_{^\ast\mathbb{R}} \\{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \\}\right) $$ So \\(\psi\\) *cannot* be a complete lattice homomorphism, even though it is part of an order isomorphism. However, just to complicate matters, I believe that \\(\phi\\) and \\(\psi\\) are a mere *lattice* isomorphism, preserving finite meets and joints.
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