8 ms·

Maybe I misunderstood the article but if the set of real numbers is finite then it should be countable. But I can easily prove that the set of real numbers or

Maybe I misunderstood the article but if the set of real numbers is finite then it should be countable. But I can easily prove that the set of real numbers or any subset of real numbers is not countable. Been a really long time since I’ve thought about this but wondering what I’m missing.

Sharlin 5y ago
No, what the article is talking about is the question whether or not the cardinality of real numbers is the smallest uncountable infinity or some other, larger uncountable infinity. The only countable infinity is aleph-0, the cardinality of natural numbers, and Cantor showed that aleph-0 is too small to hold all reals. So reals must be uncountable, but there is an infinite hierarchy of uncountable infinities, and it is not known which one is the cardinality of reals (although in practice it's suspected to be either aleph-1, which is what the Continuum Hypothesis states, or aleph-2).
- keymone 5y ago
  unrelated: how do we know there are no alephs between 0 and 1?
  - Sharlin 5y ago
    If you assume Axiom of Choice, you can define cardinalities in terms of ordinal numbers (an extension of naturals that generalizes the notion of "counting", or "indexing"). And ordinal numbers are well-ordered, ie. every element has a unique "successor" element.
  - elcomet 5y ago
    That's the definition of aleph-1. The continuum hypothesis is that 2^aleph-0 (number of sets of integers or number of reals) = aleph-1.
- orangecat 5y ago
  (although in practice it's suspected to be either aleph-1, which is what the Continuum Hypothesis states, or aleph-2) There are some arguments that it's "really" much larger, as in greater than aleph-n for any finite n. See https://risingentropy.com/the-continuum-hypothesis-is-false/ https://risingentropy.com/the-continuum-hypothesis-is-false/ (incidentally an excellent blog if you're interested in these sorts of things).
TameAntelope 5y ago
How can you easily prove that the set of real numbers is not countable? I don't think it's as easy as you claim, but I'm kind of a dummy so it's quite probably I'm wrong.
- twelfthnight 5y ago
  Here's some background on the proof [1]. Here's a video explaining it little better [2]. [1] https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument [2] https://www.youtube.com/watch?v=elvOZm0d4H0 https://www.youtube.com/watch?v=elvOZm0d4H0
  - TameAntelope 5y ago
    I guess I was focusing on the word "easy", heh.
    - mmmBacon 5y ago
      You do this in sophomore level Real Analysis. So as far as pure math goes, easy.
      - TameAntelope 5y ago
        > sophomore level Real Analysis Hmm?
        rpearl 5y ago
        We did this first semester freshman year at my school in a single lecture in a 100 level "concepts of mathematics" course. I get that there's a certain amount of "it's hard to wrap your head around infinity" but really this proof is pretty well trodden by students not far into learning mathematics.
        TameAntelope 5y ago

Sharlin 5y ago

No, what the article is talking about is the question whether or not the cardinality of real numbers is the smallest uncountable infinity or some other, larger uncountable infinity. The only countable infinity is aleph-0, the cardinality of natural numbers, and Cantor showed that aleph-0 is too small to hold all reals. So reals must be uncountable, but there is an infinite hierarchy of uncountable infinities, and it is not known which one is the cardinality of reals (although in practice it's suspected to be either aleph-1, which is what the Continuum Hypothesis states, or aleph-2).

keymone 5y ago
unrelated: how do we know there are no alephs between 0 and 1?
- Sharlin 5y ago
  If you assume Axiom of Choice, you can define cardinalities in terms of ordinal numbers (an extension of naturals that generalizes the notion of "counting", or "indexing"). And ordinal numbers are well-ordered, ie. every element has a unique "successor" element.
- elcomet 5y ago
  That's the definition of aleph-1. The continuum hypothesis is that 2^aleph-0 (number of sets of integers or number of reals) = aleph-1.
orangecat 5y ago
(although in practice it's suspected to be either aleph-1, which is what the Continuum Hypothesis states, or aleph-2) There are some arguments that it's "really" much larger, as in greater than aleph-n for any finite n. See https://risingentropy.com/the-continuum-hypothesis-is-false/ https://risingentropy.com/the-continuum-hypothesis-is-false/ (incidentally an excellent blog if you're interested in these sorts of things).

TameAntelope 5y ago

How can you easily prove that the set of real numbers is not countable? I don't think it's as easy as you claim, but I'm kind of a dummy so it's quite probably I'm wrong.

twelfthnight 5y ago
Here's some background on the proof [1]. Here's a video explaining it little better [2]. [1] https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument [2] https://www.youtube.com/watch?v=elvOZm0d4H0 https://www.youtube.com/watch?v=elvOZm0d4H0
- TameAntelope 5y ago
  I guess I was focusing on the word "easy", heh.
  - mmmBacon 5y ago
    You do this in sophomore level Real Analysis. So as far as pure math goes, easy.
    - TameAntelope 5y ago
      > sophomore level Real Analysis Hmm?
      - rpearl 5y ago
        We did this first semester freshman year at my school in a single lecture in a 100 level "concepts of mathematics" course. I get that there's a certain amount of "it's hard to wrap your head around infinity" but really this proof is pretty well trodden by students not far into learning mathematics.
        TameAntelope 5y ago