1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

In need of proof checking

  1. Nov 6, 2014 #1
    Hello everyone
    I was given a question in a homework (this is not a homework thread though as I have submitted it)
    is was to :
    Show that there is no infinite set A such that |A| < |Z+| =ℵ0.
    I thought of it and tried to work my way out and came up with those proofs , which I am not quiet sure about

    Please tell me what are the mistakes I made and if my assumptions had been proven before now or not

    Long answer :

    Let A be the infinite set of real numbers

    Knowing that from an infinite set we can have a countable infinite set that is a subset of the uncountable infinite set .

    I will assume that we can express the cardinality of the

    infinite set A as : |a1| +|a2| +|a3| +|a4| ... (the cardinality of infinite countable-infinite sets)

    [My acheles kneel]
    where |a1| , |a2| , |a3| are the cardinalities of countable infinite sets

    with the condition that (a1,a2,a3...) → A is a one-to-one correspondance

    |a1| +|a2| +|a3| +|a4| ... < ℵ0

    one of these countable infinite sets can be a set of natural numbers and it's cardinality is ℵ0

    if that was the case :

    ℵ0+|a2| +|a3| +|a4| ... < ℵ0

    which can only be true if the cardinalities of the rest of the countable infinite subsets

    |a2| +|a3| +|a4| ... , is negative which is impossible

    Short Answer :

    Since A is an infinite set , we can have a countable infinite set that is a subset of it

    this set can be the set of natural numbers (which have the cardinality ℵ0)

    if that was the case the inequalitiy can't be true .
  2. jcsd
  3. Nov 6, 2014 #2
    There is an inherent problem in making this assumption. Let A1 and A2 be two disjoint countably infinite sets. Show that [itex]A_1\cup A_2[/itex] is also countably infinite, and thus [itex]|A_1| + |A_2| = \aleph_0[/itex]. Notice how you can then use induction to show that the countably infinite union of countably infinite sets must also be countably infinite, thus your sum above cannot represent the cardinality of an uncountably infinite set.

    A more pressing problem with the "Long Form" above is that you start out by assuming that A is a very specific set (the set of all real numbers). The original conjecture only requires A to be an infinite set. Thus, the only assumption you may make when starting a constructive proof is that A is infinite. You might then break it down into cases where A is a certain type of infinite set in each, but then you must make sure you cover every type of infinite set.
    An alternative is to start a proof by contradiction, of course. In this type of proof, you would start by assuming all the precepts are true, but the conclusion is false. Then show that this leads to an internal contradiction: a statement that does not have a consistent truth value. This method would start with you assuming that there is at least one infinite set A such that |A| < |Z+|. But note that we still start with A being an arbitrary infinite set, not a particular one.

    The Short Form is a promising outline of a proof. It just needs to be made more formal. :-)
    Last edited: Nov 6, 2014
  4. Nov 8, 2014 #3
    Thanks you :) !
    I should really be more careful about my assumptions.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted