What ordinals can be embedded in ℚ and ℝ?

Last time we talked a little bit about some properties of the order type of ℚ. I want to go into more detail about these properties, and actually prove them to you. The proofs are nice and succinct, and ultimately rest heavily on the density of ℚ.

Every Countable Ordinal Can Be Embedded Into ℚ

Take any countable well-ordered set (X, ≺). Its order type corresponds to some countable ordinal. Since X is countable, we can enumerate all of its elements (the order in which we enumerate the elements might not line up with the well-order ≺). Let’s give this enumeration a name: (x1, x2, x3, …).

Now we’ll inductively define an order-preserving bijection from X into ℚ. We’ll call this function f. First, let f(x1) be any rational number. Now, assume that we’ve already defined f(x1) through f(xn-1) in such a way as to preserve the original order ≺. All we need to do to complete the proof is to assign to f(xn) a rational number such that the ≺ is still preserved.

Here’s how to do that. Split up the elements of X that we’ve already constructed maps for as follows: A = {xi | xi ≺ xn} and B = {xi | xi > xn}. In other words, A is the subset of {x1, x2, …, xn-1} consisting of elements less than x_n and B is the subset consisting of elements greater than xn. Every element of B is strictly larger than every element of A. So we can use the density of the rationals to find some rational number q in between A and B! We define f(xn) to be this rational q. This way of defining f(xn) preserves the usual order, because by construction, f(xn) < f(xi) for any i less than n exactly in the case that xn < xi.

By induction, then, we’ve guaranteed that f maps X to ℚ in such a way as to preserve the original order! And all we assumed about X was that it was countable and well-ordered. This means that any countable and well-ordered set can be found within ℚ!

No Uncountable Ordinals Can Be Embedded Into ℝ

In a well-ordered set X, every non-maximal element of X has an immediate successor (i.e. a least element that’s greater than it.) Proof: Take any non-maximal x ∈ X. Consider the subset of X consisting of all elements greater than x: {y ∈ X | x < y}. This set is not empty because α is not maximal. Any non-empty subset of a well-ordered set has a least element, so this subset has a least element. I.e, there’s a least element greater than x. Call this element S(x), for “the successor of x”,

Now, take any well-ordered subset X ⊆ ℝ (with the usual order). Since it’s well-ordered, every element has an immediate successor (by the previous paragraph). We will construct a bijection that maps X to ℚ, using the fact that ℚ is dense in ℝ (i.e. that there’s a rational between any two reals). Call this function f. To each element x ∈ X, f(x) will be any rational such that x < f(x) < S(x). This maps every non-maximal element of x to a rational number. To complete this, just map the maximal element of x to any rational of your choice. There we go, we’ve constructed a bijection from X to ℚ!

The implication of this is that every well-ordered subset of the reals is only countably large. In other words, even though ℝ is uncountably large, we can’t embed uncountable ordinals inside it! The set of ordinals we can embed within ℝ is exactly the set of ordinals we can embed within ℚ! (This set or ordinals is exactly ω1: the set of all countable ordinals).

Final Note

Notice that the previous proof relied on the fact that between any two reals you can find a rational. So this same proof would NOT go through for the hyper-reals! There’s no rational number (or real number, at that!) in between 1 and 1+ϵ. And in fact, you CAN embed ω1 into the hyperreals! This is especially interesting because the hyperreals have the same cardinality as the reals! So the embeddability of ω1 here is really a consequence of the order type of the hyperreals being much larger than the reals. And if we want to take a step towards even crazier extensions of ℝ, EVERY SINGLE ordinal can be embedded within the surreal numbers!

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