Constructing $\mathbb{R}$ from $\mathbb{Z}$?

Question

I have been told that the real number line $\mathbb{R}$ can be constructed from the cartesian product $\mathbb{Z} \times [0,1)$.

How exactly is that true? Surely, the cartesian product $\mathbb{Z} \times [0,1)$ would give a set of ordered pairs of numbers? How is this equivalent to $\mathbb{R}$?

Answer 1

There is a bijection $f((x,r)) = x + r$, so $x$ is the integer part of the real number (before the digital point) and $r$ is the part after the digital point. Of course this assumes we write $1$ as corresponding to $(1,0)$ etc.

This is a bijection but not a homeomorphism (as the product is disconnected as a topological space).

Answer 2

Define $$ f:\mathbb{Z} \times [0,1)\to\mathbb R $$ as follows: for any $(m,x)\in \mathbb{Z} \times [0,1)$, $ f(m,x)=m+x$. Then $f$ is a bijective map. I don't know if that is what you mean with 'contructed'

Answer 3

If you had $\mathbb{R}$, a natural map from the set you have to $\mathbb{R}$ would be $$\mathbb{Z} \times [0,1) \to \mathbb{R}$$ $$(x,y) \mapsto x+y.$$ This is a bijection, as you can verify. However, since we "do not have" $\mathbb{R}$, we must "define":

$$\mathbb{R}:=\mathbb{Z} \times [0,1).$$

But... I think that by any reasonable meaning of "construct", you are entailing at least the operations and ordering of $\mathbb{R}$. And I see no easy way to do this in the set above without using explicitly (or at least "implicitly explicitly") the operations and ordering in $\mathbb{R}$, which would turn out to be circular.

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If you admit as given the "structure" inherited by $[0,1)$ (although this is unclear, since it is not a good algebraic structure. It is an ordered set in a natural way, but the operations are not defined on all elements), we can proceed as follows to get the operations:

We use the lexicographic order on "$\mathbb{R}$". From the order, we already have the topology.

To do addition, if $x+y$ is defined, then $$(n,x)+(m,y):=(n+m,x+y).$$ Now, before defining addition in the whole set, we define what it means to multiply by two in $[0,1)$, a unary operation which we will call $d$.

If $x<1/2$, we define $d(x)=2x$. If $x \geq 1/2$, we define $d(x)=x-(1-x)$*. Now we can define addition everywhere as follows: if $x+y$ is not defined, then $$(n,x)+(m,y):=\left(n+m+1,d\left(\frac{x}{2}+\frac{y}{2}\right)\right).$$

Having defined addition, we have a natural way to define multiplication by integers (we can verify that, indeed, our set with addition is an Abelian group). Now, to define multiplication, we do $$(n,x)\cdot(m,y):=(nm,0)+n(0,y)+m(0,x)+(0,xy).$$

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*- $(1-x)$ can be defined as $\sup\{y \in [0,1) \mid x+y \text{ is defined}\}$ (with regards to what "is defined" means if it is not clear, see Ian's comment below). For $x \neq 0$ this is well-defined. Hence, it is well-defined for all $x \geq 1/2$, as is needed for our definition.

Answer 4

I have been told that the real number line $\mathbb{R}$ can be constructed from the cartesian product $\mathbb{Z} \times [0,1)$.

"constructed from" is perhaps relatively vaguely defined, but I assume you mean "has the same cardinality as" or, equivalently, "a bijection exists to".

$$ f((n, r)) = n + r $$ is exactly such a bijection.

How exactly is that true? Surely, the cartesian product $\mathbb{Z} \times [0,1)$ would give a set of ordered pairs of numbers? How is this equivalent to $\mathbb{R}$?

"Equivalent" makes no sense in this case. One could define equivalence in many distinct ways. When you say "constructed from" I assume you are searching for a bijection.