In this article: Properties lost while passing from $\mathbb{R}$ to $\mathbb{C}$ , there is a discussion about the properties lost as we move towards a greater set, the complex numbers, from the set of real numbers.

I wonder if there other properties that we lose if we move from:

  1. $\mathbb{N}$ to $\mathbb{Z}$
  2. $\mathbb{Z}$ to $\mathbb{Q}$
  3. $\mathbb{Q}$ to $\mathbb{R}$
  • $\begingroup$ Depends on what kind of properties you're looking for. When moving from $\mathbb{N}$ to $\mathbb{Z}$ you can no longer say that every element is non-negative, for instance. $\endgroup$ – MisterRiemann Apr 26 '18 at 15:46
  • $\begingroup$ On the other hand, you gain some things, too: In going from $\mathbb{N}$ to $\mathbb{Z}$, each element now gains an additive inverse, which it didn't have before (except conceivably for $0$, if you put that in $\mathbb{N}$). And when you go from $\mathbb{R}$ to $\mathbb{C}$, you gain algebraic closure, which you didn't have before. (That is to say, there are polynomials with real coefficients that don't have real solutions.) $\endgroup$ – Brian Tung Apr 26 '18 at 16:09

Let me give a unifying example from the "structural," as opposed to algebraic, side: varieties of proof by induction.

The natural numbers are well-ordered: every (nonempty) subset has a least element. In other words, proof by induction works in $\mathbb{N}$.

When we go past $\mathbb{N}$, we still get versions of induction, but they become progressively more complicated:

$\mathbb{Z}$ still supports a version of proof by induction, but we have to work "in both directions:" "induction" in $\mathbb{Z}$ looks like

If $P$ is true of $0$, and $P(n)\implies P(n+1)$, and $P(n)\implies P(n-1)$, then $P$ is true of all integers.

$\mathbb{Q}$ also supports a kind of induction, only now it's even more complicated. The idea is to look at how rational numbers are "built up." The following is true:

Suppose $(i)$ $P$ is true of $0$, $(ii)$ $P(n)\implies P(n+1)$, $(iii)$ $P(n)\implies P(n-1)$, and $(iv)$ $P(n), P(m)\implies P({n\over m})$ (for $m\not=0$). Then $P$ is true of all rational numbers.

What about $\mathbb{R}$? Well, now we're in a pickle: $\mathbb{R}$ is uncountable, so there's not going to be a similar way to "build" the real numbers starting from $0$ and just using a few basic operations (that will only ever get countably many things - as long as "few" means "at most countably many"). However, something really cool happens:

Moving from $\mathbb{Q}$ to $\mathbb{R}$, we actually gained an important property: the least upper bound property, that every (nonempty) set of real numbers has a least upper bound. This is really the fundamental property of the real numbers, and is used for basically everything. One really interesting consequence is that the real numbers allow the following "real induction:"

Suppose $P$ is a property such that for all $X\subseteq [0, 1]$, if $P$ holds for every element of $X$ then $P$ holds for $\sup(X)$ *(or rather, the least upper bound in $[0, 1]$ of $X$ - the difference is that $\sup(\emptyset)=0$, giving our base case, rather than undefined)*. Then $P$ holds for every element of $[0, 1]$.

Note that this principle doesn't even make sense for $\mathbb{Q}$, since sets of rationals need not have rational least upper bounds. (I've stated real induction for the interval $[0, 1]$ above, since the most famous application of topological induction is a snappy proof of the Heine-Borel theorem, but you can easily rephrase it for the nonnegative reals, or for all reals (a la the $\mathbb{Z}$-example).) Proving real induction from the least upper bound property is a good exercise, and the principle is discussed in detail here.

Incidentally, I should admit that to the best of my knowledge real induction is really a curiosity, but it's a super cool one so I can't resist plugging it.

And in moving from $\mathbb{R}$ to $\mathbb{C}$, we lose the kind of "topological induction" above since $\mathbb{C}$ isn't linearly ordered anymore. We can still salvage it, though, via a kind of "double topological induction." However, I think it's a good idea to stop at this point.

On the other hand, at this point you can go through and cook up your own induction(-style) principles for various different mathematical structures. The most important one by far, incidentally, is induction on well-founded trees; but $\mathbb{N}$ is really the only "direct overlap" between the world of well-founded trees and the more algebraic world, so I'm not going to discuss it here.

  1. Every non-empty subset of $\mathbb N$ has a first element. This is false for $\mathbb Z$.
  2. Every bounded subset of $\mathbb Z$ is finite. This is false for $\mathbb Q$.
  3. There are bounded non-empty subsets of $\mathbb Q$ without a least upper bound in $\mathbb Q$. This is false for $\mathbb R$.
  • $\begingroup$ @NoahSchweber I've edited my answer. What do you think now? $\endgroup$ – José Carlos Santos Apr 26 '18 at 16:01
  • $\begingroup$ @NoahSchweber Sorry. I don't know what crossed my mind. $\endgroup$ – José Carlos Santos Apr 26 '18 at 16:03
  • $\begingroup$ My one remaining quibble is that it's not clear that 3 is really an example of a property that's lost going from $\mathbb{Q}$ to $\mathbb{R}$, but more a property that's gained going from $\mathbb{Q}$ to $\mathbb{R}$. Of course that's subjective, but that's my perspective. $\endgroup$ – Noah Schweber Apr 26 '18 at 16:05

Certainly there are many properties but to list a few:

$\quad(1)$ $\mathbb N$ to $\mathbb Z$

  • Every element is nonnegative
  • Every subset has a smallest element
  • Every element is uniquely determined by its distance to the origin $($to $0)$

$\quad(2)$ $\mathbb Z$ to $\mathbb Q$

  • Every bounded subset is finite
  • The set of powers of each element is either finite or unbounded

$\quad(3)$ $\mathbb Q$ to $\mathbb R$

  • Every element can be multiplied by a positive integer so as to become an integer
  • Every element has a finite or periodic decimal representation
  • Every element is the root of a degree $1$ polynomial with integer coefficients

If $a/b\ne c/d$ then $|ad-bc|\geq 1.$ This is true in $\Bbb Z$ , but not in $\Bbb Q.$

For each $x$ there exists $r\ne 0$ such that for all $a,b\in \Bbb Z$ with $b \ne 0,$ if $x\ne a/b$ then $|x-a/b|>|r/b|.$ This is true in $\Bbb Q$ but not in $\Bbb R.$

Prime decompositions in $\Bbb N$ or $\Bbb Z$: There are no primes in $\Bbb Q$ because $\Bbb Q$ is a field.

Topologically: With the induced order topology, $\Bbb N$ and $\Bbb Z$ are locally compact as they are discrete. $\Bbb Q$ is not locally compact. Also $\Bbb N$ and $\Bbb Z$ are (somewhat trivially) Baire spaces bur $\Bbb Q$ isn't. When we go up to $\Bbb R$ we recover the Baire property and local compactness.

$\Bbb Q$ is a universal space for linear spaces with the same or smaller cardinal: If $X$ is a linear space with $|X|\leq |\Bbb Q|$ there is an order-isomorphic homeomorphic embedding $f:X\to f(X)\subset \Bbb Q.$ This is not true if we replace $\Bbb Q$ with $\Bbb R.$


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