4
$\begingroup$

The question is simple one, how do we know that there is a bijection between the points in a line and an uncountable set (such as the power set of natural numbers)? Why don’t countably infinite sets that are dense (such as the rational numbers) correspond 1-1 with all the points on a line? The simplest possible explanation that assumes minimal background knowledge of the inquirer would be greatly appreciated.

-Thank you

$\endgroup$
6
  • 3
    $\begingroup$ Do you know why the set of real numbers is uncountable? $\endgroup$
    – Thomas Andrews
    Commented Aug 10 at 5:14
  • $\begingroup$ @ThomasAndrews Yes, it is because it contains more elements than can be corresponded to by the set of natural numbers, this is shown by Cantor’s diagonal argument. $\endgroup$
    – Max Maxman
    Commented Aug 10 at 5:18
  • 2
    $\begingroup$ Without essentially knowing the line is in one-to-one correspondence with the real numbers, you actually can't show the line is uncountable. Euclid's postulates hve a model with a countable line, and a countable plane. $\endgroup$
    – Thomas Andrews
    Commented Aug 10 at 5:18
  • 1
    $\begingroup$ I mean, there exist rather simple models of plane geometry - for example, the set of pairs of points $(x,y)$ with $x,y$ algebraic numbers satisfies Euclid's postulates. $\endgroup$
    – Thomas Andrews
    Commented Aug 10 at 5:24
  • 1
    $\begingroup$ But to prove it, you have to define geometry in set theory. There is no such thing as the functions, the natural numbers, etc in formal geometry. $\endgroup$
    – Thomas Andrews
    Commented Aug 10 at 5:27

1 Answer 1

Reset to default
6
$\begingroup$

This is due to Cantor–Dedekind axiom. In rigorous formulations of Euclidean Geometry, this statement is taken as an axiom or is a theorem. This statement is actually independent of Euclid's original axioms and is assumed implicitly in Elements.

$\endgroup$
9
  • 2
    $\begingroup$ its a theorem in rigorous formulations of Euclidean Geometry (eg. Hilbert, Artin etc). In Euclid's original work, with appropriate interpretation, there are models which don't satisfy this statement. $\endgroup$
    – Ajin Shaji Jose
    Commented Aug 10 at 5:30
  • 1
    $\begingroup$ Lucid and concise, bless your soul. Thank you so much! $\endgroup$
    – Max Maxman
    Commented Aug 10 at 5:33
  • 1
    $\begingroup$ @MaxMaxman if you are interested, read books by George Martin or Borsuk on Foundations of Geometry $\endgroup$
    – Ajin Shaji Jose
    Commented Aug 10 at 5:56
  • 1
    $\begingroup$ I think this answer still misses one piece of the story. The Cantor-Dedekind axiom is indeed a theorem of Euclidean geometry if you add a still simpler axiom added. In the last edition of Hilbert's Grundlugen this was the axiom of line completeness. $\endgroup$
    – Lee Mosher
    Commented Aug 10 at 15:11
  • 1
    $\begingroup$ @MaxMaxman i am not aware of any Euclidean Geometry axiom system that have negation of Cantor-Dedekind as its consequence. However, I know of a model of plane that satisfies Euclid's axioms but not Cantor-Dedekind. Consider all ordered pair of algebraic numbers and we can define other stuff similar to standard Cartesian plane $\endgroup$
    – Ajin Shaji Jose
    Commented Aug 11 at 3:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .