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In my calc class we constructed the real numbers using the following 5 axioms:

  1. The set is nonempty.
  2. The set has an ordering.
  3. The set has no first/last point.
  4. The set is connected.
  5. The set contains a countable dense subset.

We then proved that for any set that fulfills these axioms, there exists an order preserving bijection from this set to the set of Dedekind cuts, with the order on the latter set being the standard strict subset ordering of cuts.

I've been studying in preparation for next year's analysis course and all material I've studied outlines the construction of the reals as an ordered field equipped with a Least Upper Bound property, i.e. Dedekind completeness.

Are these definitions equivalent? Are there any nuances I should be aware of comparing and contrasting these definitions?

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  • $\begingroup$ Does this help? $\endgroup$ Jun 28, 2022 at 16:55
  • $\begingroup$ @quanticbolt Not really. I already looked through it. I was hoping to elucidate the connection between density and connectedness and completeness. $\endgroup$
    – Andrew Lys
    Jun 28, 2022 at 17:15

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Here’s the way I would think about this. Let $L$ be a set endowed with a linear order relation $\leq$. In particular, $\leq$ is “reflexive” (for all $x \in L$, we have $x \leq x$), “anti symmetric” (for all $x,y \in L$ if $x \leq y$ and $y \leq x$, then $x = y$), and “transitive” (for all $x,y,z \in L$, if $x \leq y$ and $y \leq z$, then $x \leq z$.

From an order, we may endow $L$ with a topology generated by the open rays. That is, sets of the form $\{x \in L : x < a\}$ and of the form $\{x \in L : a < x \}$ for a fixed $a \in L$. Here “$x<Y$” just means that both $x \leq y$ and $x \neq y$ hold. This is called the “order topology”, and it generates the “standard” way we think of points being “close together” on the reals. This order topology is more general though.

Here’s a nice theorem about the order topology:

Theorem: Let $L$ be an ordered set endowed with the order topology. Then $L$ is connected if and only if $L$ is dense and complete as an order.

Here, to be “dense” just means that if you give me two distinct points $x < y$ in $L$, then we may find a point in between them $z$ such that $x < z < y$.

“Complete” means that if you give me a bounded subset $A \subseteq L$, then $A$ has a greatest lower bound and a least upper bound.

Now here are two different theorems that completely characterize the real numbers:

Theorem 1: Let $L$ be an ordered set without endpoints that is dense, complete, and has a countable dense subset. Then $L$ is order-isomorphic to the reals. That is, there is an order preserving bijection from $L$ to $\mathbb{R}$.

This is exactly the theorem you were given in class.

Theorem 2: Let $\mathbb{F}$ be an ordered field with the least upper bound property. Then $\mathbb{F}$ is isomorphic to $\mathbb{R}$ as an ordered field.

This theorem does not immediately follow from theorem 1. The equivalence between these two sets of conditions reveals some non-trivial properties of linear orderings. Indeed, what do these two sets of conditions have in common? Every ordered field has characteristic 0, so we necessarily don’t have endpoints. Every ordered field is also dense, and has the “rationals” as a dense sub-field. Thus we always have a countable dense subset.

Thus the “common hypotheses” are being separable (countable dense subset), lacking endpoints, and being dense as an order.

So suppose we have some ordered set $L$ satisfying those hypotheses. Theorems 1 and 2 tell us that if we also have that $L$ is connected in the order topology ($L$ is complete as an order), then we may put field operations on $L$ and turn it into an isomorphic copy of $\mathbb{R}$. That’s crazy! $L$ was just an ordered set. There was no algebraic structure at all, and we can now bring one into existence.

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  • $\begingroup$ Great stuff. Do you know where I can learn more about order topology? I'd like to see the proof for the first theorem and read more about this $\endgroup$
    – Andrew Lys
    Jun 28, 2022 at 17:15
  • $\begingroup$ The Wikipedia page en.wikipedia.org/wiki/Order_topology has a good explanation of the order topology. For the equivalence between “connected” and “dense + complete”, I think that’s a proof you could do on your own! Indeed, if we try to think about two “prototypical” examples of disconnected ordered sets, we might think of $[0,1] \cup [2,3]$ or $(-\infty, 0) \cup (0, \infty)$ as subsets of $\mathbb{R}$. The first is not dense as an order (there is no point between $1$ and $2$. the second is not complete as an order $(-\infty, 0)$ has no least upper bound. $\endgroup$
    – Joe
    Jun 28, 2022 at 17:20
  • $\begingroup$ Try to prove it yourself, and if you run into any trouble let me know. I can write up a proof if it’s needed. $\endgroup$
    – Joe
    Jun 28, 2022 at 17:21

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