1

I have seen the diagonal proof that implies such a construction is impossible, but I do not understand what property the reals possess that prevents this construction from happening.

I am asking because I'm trying to understand what it really means for there to be multiple sizes of infinity.

EDIT: The answers I'm getting are not quite the kind I am looking for. I'm looking for a more abstract answer. If we have two collections of things, one being the size of the natural numbers (call it A) and one being the size of the real numbers (call it B), what property does B possess that A does not that (intuitively) prevents a bijection from being constructed between the two.

IgnorantCuriosity
  • 1,423
  • Simple, not all inverses are covered; in other words, how would you map something to $\sqrt{2}$ only using a function involving the natural numbers? – JohnColtraneisJC Sep 01 '17 at 20:42
  • 1
    maybe the different types of infinity might be a start. –  Sep 01 '17 at 20:44
  • @BenjaminMoss You could define the function$ f(2)= \sqrt{2}$ – Quality Sep 01 '17 at 20:44
  • 1
    @RoddyMacPhee is right, the reals are uncountably infinite, natural numbers countably infinite. – JohnColtraneisJC Sep 01 '17 at 20:45
  • 1
    @Quality but $f(a)= \sqrt(a)$ doesn't map all reals to natural number. –  Sep 01 '17 at 20:45
  • 3
    It isn't anything algebraic about the reals. It's just a factor of the size of the set of real numbers. – Thomas Andrews Sep 01 '17 at 20:46
  • 1
    @benjamin, the algebraic numbers are countable! See https://math.stackexchange.com/questions/50655/proving-the-countability-of-algebraic-numbers – Mark S. Sep 01 '17 at 20:46
  • 1
    @RoddyMacPhee Presumably, that was in reply to the comment: "how would you map something to $\sqrt 2$ only using a function involving the natural numbers." That (the original comment above) was what I'd call at best an uninformative response. – Thomas Andrews Sep 01 '17 at 20:47
  • @RoddyMacPhee I never implied that it did. – Quality Sep 01 '17 at 20:47
  • @Quality ah but you did imply a point of a function but can that function be bijective ? if not then there's not a 1 to 1 correspondence which would imply the same cardinality. –  Sep 01 '17 at 20:51
  • https://math.stackexchange.com/questions/130788/isnt-there-a-bijection-between-real-numbers-and-natural-numbers What I said originally was in line with a few comments in this question, I'm not sure how my response was uninformative – JohnColtraneisJC Sep 01 '17 at 20:52
  • sorry correction: there's not a bijection which would be the indicator of equal cardinality. –  Sep 01 '17 at 20:57
  • 1
    How are you defining the real numbers - definitions differ, though isomorphism/uniqueness of the reals is provable in various contexts. But if you want to know which part of the definition makes the difference, you need to specify all the components. – Mark Bennet Sep 01 '17 at 21:04
  • 7
    Uncountability. The property of being too big. – BrianO Sep 01 '17 at 21:05
  • @BrianO: Isn't this a circular argement? You effectively say, the reals uncountable because they are uncountable. – gammatester Sep 01 '17 at 21:23
  • 3
    @gammatester: but that's the point! The OP is asking a question of the form: if we have too things and one is bigger than the other, then what property of the bigger thing makes it bigger? It isn't circular to say that the property that distinguishes between a red door and a blue door is colour. Likewise it isn't circular to say that the property that distinguishes between a set and a bigger set is size. – Rob Arthan Sep 01 '17 at 21:56
  • It would be helpful to consider a simpler example. E.g., consider the set $X = {0}$ with one element and the set $Y = {0, 1}$ with two elements. What abstract properties of $X$ and $Y$ prevent us constructing a bijection between $X$ and $Y$? I can't think of any answer other than the difference in cardinality, i.e., the non-existence of a bijection between $X$ and $Y$. What would you like to see in a more "intuitive" explanation why $1 \neq 2$? – Rob Arthan Sep 01 '17 at 22:12
  • Perhaps someone can give the kind of answer you're looking for if you demonstrate it for finite sets. I have one set being the same size as the set of planets (call it P) and one being the size of the set of fingers on my hand (call is A). What property does B possess that A does not that (intuitively) prevents a bijection from being constructed between the two? – Jim H Sep 02 '17 at 04:31
  • @gammatester It's a joke. I don't think it's helpful to ask, "What property of set $A$ makes it bigger than set $B$?" Being bigger than is a relation, and the only "property" of $A$ that could account for its being bigger than $B$ smuggles in $B$: the property of not being bijectable with B$. – BrianO Sep 12 '17 at 18:09

3 Answers3

1

It would be the supremum axiom: The fact that for every bounded above set there exists a least upper bound, i.e. supremum. This implies the existence of irrational numbers within the reals.

For example, $\sqrt{2}$ would be the supremum of the bounded above set $[0,\sqrt{2})$.

For you to have an intuition of what is going on, you can think of the natural/rational number line as a number line full of holes whereas the real number line is absolutely continuous. The holes in the natural number line are obvious (for example between $1$ and $2$ we have $1.1, 1.5,\ldots$) and those in the rational number line are the ones left by the irrational numbers.

Samuel González-Castillo
  • 103
  • Why does this prevent the existence of the bijection/injection? The hole argument seems to be the same for algebraic numbers instead of reals. – gammatester Sep 01 '17 at 21:01
  • 2
    @gammatester If you try to do this with the algebraics in place of the reals, how do you know that the resulting "antidiagonal" real will be algebraic? (It won't.) – Noah Schweber Sep 01 '17 at 21:13
1

When you construct an infinite decimal in the diagonlization argument, the least upper bound axiom of $\mathbb{R}$ guarantees that the infinite decimal you've constructed is actually a real number; if you tried to perform the diagonlization argument with $\mathbb{Q}$, that would be your stumbling block.

Duncan Ramage
  • 6,928
  • 1
  • 20
  • 38
1

The underlying truth is the following:

Given any set $X$ there is no surjective map $f: \>X\to{\cal P}(X)$.

This means that ${\cal P}(X)$ has "essentially more elements" than $X$, say $2^n$ compared to the $n$ elements of $X$, if $X$ is finite.

Proof. Consider an arbitrary map $f: \>X\to{\cal P}(X)$. This map assigns to each $x\in X$ a subset $A_x\subset X$. Now look at the special set $$Q:=\{x\in X\,|\, x\notin A_x\}\quad \in{\cal P}(X)\ .$$ I claim that $Q\notin f(X)$; hence $f$ is not surjective. Assume to the contrary that $Q=f(x_*)$ for some $x_*\in X$. We now argue about the membership of $x_*$ in $Q$. By definition of $Q$ and of $x_*$ one has $$x_*\in Q\quad\Leftrightarrow\quad x_*\notin f(x_*)\quad\Leftrightarrow \quad x_*\notin Q\ ,$$ which is absurd.$\quad\square$

Christian Blatter
  • 226,825