To remove paradoxes of naive set theory, We started with the axioms of Zermelo-Fraenkel and developed a set theory. Where we are building sets starting from a empty set. How to construct set of irrationals using such axioms?
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Once you settle on what are the rationals and what are the real numbers, it becomes particularly easy, since $\sf ZFC$ proves that if $A$ and $B$ are sets, then $A\setminus B$ is a set.
Why? Apply the axiom of separation for $\varphi(x,p_1)$ (where $p_1$ is a parameter) taking the formula to be $x\notin p_1$. Now the class $\{x\in A\mid x\notin B\}$ is a set.
But we have many possible ways to construct and understand what are the real numbers in the context of $\sf ZFC$. The reason is that our universe is composed only of sets, so the real numbers are not atomic objects, like we would normally think about them, but instead we encode them into sets and prove that there is a set $\Bbb R$ and there is a relation on that set which is $\leq$, and there are binary operators $+$ and $\cdot$, and so on.
One of the classical ways to do it is first build something that looks like $\Bbb N$, extend it to $\Bbb Z$ and then to $\Bbb Q$, and finally use Dedekind cuts to construct $\Bbb R$. This works, but then $\Bbb R\cap\Bbb Q=\varnothing$. This is perfectly acceptable mathematically, so there is no need to panic at this point.
While the sets we construct at each step are disjoint from one another, we do get very canonical ways to embed $\Bbb N$ into $\Bbb Z$, $\Bbb Q$ and $\Bbb R$; and we get very canonical ways to embed $\Bbb Z$ into $\Bbb Q$ and $\Bbb R$; and we get a very canonical way to embed $\Bbb Q$ into $\Bbb R$. So now a real number is rational if it is in the range of the canonical embedding of $\Bbb Q$, and the class of rational real numbers is a set again, this time because of the axiom of replacement (or you can define what it means being rational directly in $\Bbb R$ after you have $0,1,+,\cdot$ and you know the axioms of a field hold in $\Bbb R$).
So to sum up, yes, we can prove the irrational numbers form a set. But which set exactly would depend on how you choose to interpret the meaning of "real" and "rational" inside $\sf ZFC$. And if that bothers you, then just think about the many ways you can write a piece of code that implements an algorithm, or in how many ways you can walk from your house to the market. And remember this is not about how you do it, as much about the fact you can do it.
Some relevant links:
Asaf Karagila
- 393,674
Integers = Subset of $\mathbb{N} \times \mathbb{N}$ ;
Rationals = Subset of $\mathbb{Z} \times \mathbb{Z}$ ;
Reals = Subset of $2^{\mathbb{Q}}$ - See https://en.wikipedia.org/wiki/Dedekind_cut
And, of course, the irrationals are a subset of the real numbers.
PS: You got the name wrong.
– Aloizio Macedo Aug 02 '15 at 04:47Take $\mathcal{P}(\mathbb{Q})$. Now, consider the set $R$ of elements $A$ of $\mathcal{P}(\mathbb{Q})$ that satisfy:
(i) $A$ is neither empty nor the entire $\mathbb{Q}$ (ii) If $x \in A$, then $A$ contains all elements of $\mathbb{Q}$ that are smaller than $x$ (iii) If $x \in A$, then there exists a rational number greater than $x$ still in $A$.
This set $R$ is $\mathbb{R}$.
Now, for the irrationals, just take the subset of $\mathbb{R}$ consisting of elements which are not of the form ${x<r}$ for some rational $r$.
– Aloizio Macedo Aug 02 '15 at 05:09There is a way to define the real numbers with sequences of rational numbers (which are function from $\mathbb{N}$ to $\mathbb{Q}$, hence everything is fine). You take equivalence classes of cauchy sequences essentially.
– Aloizio Macedo Aug 02 '15 at 05:27