Proving the Equinumerosity of Infinite and Countable Sets

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Discussion Overview

The discussion centers on the cardinality of the sets of real numbers \(\mathbf{R}\) and the set of real numbers minus the rational numbers \(\mathbf{R} - \mathbf{Q}\). Participants explore methods to prove that these two sets have the same cardinality, with a focus on constructing a bijection and addressing the implications of the Continuum Hypothesis.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that if \(\mathbf{R} - \mathbf{Q}\) is not countable and has cardinality less than or equal to \(\mathbf{R}\), the Continuum Hypothesis could be invoked to conclude that its cardinality is aleph-1, the same as that of \(\mathbf{R}\).
  • Another participant points out that the union of \(\mathbf{Q}\) and \(\mathbf{R} - \mathbf{Q}\) equals \(\mathbf{R}\), and since \(|\mathbf{Q}| = \aleph_0\) and \(|\mathbf{R}| = c\), if \(|\mathbf{R} - \mathbf{Q}|\) were countable, it would lead to a contradiction.
  • A third participant agrees that while it proves \(|\mathbf{R} - \mathbf{Q}|\) is not countable, it does not establish that \(|\mathbf{R} - \mathbf{Q}| = | \mathbf{R}|\), noting the need for the Continuum Hypothesis to finalize that proof.
  • Another participant proposes a different approach, suggesting that if \(A\) is an infinite set and \(B\) is a countable set, then \(|A \cup B| = |A|\) without needing to invoke the Continuum Hypothesis.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of the Continuum Hypothesis for proving the cardinality of \(\mathbf{R} - \mathbf{Q}\) compared to \(\mathbf{R}\). There is no consensus on the best approach to establish this relationship.

Contextual Notes

Some assumptions regarding the properties of infinite sets and cardinality are not fully explored, and the discussion does not resolve the implications of the Continuum Hypothesis on the cardinality of these sets.

soumyashant
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Can you prove that [tex]\mathbf{R}[/tex] and [tex]\mathbf{R}-\mathbf{Q}[/tex] have same cardinality?

One way would be to say that [tex]\mathbf{R}-\mathbf{Q}[/tex] is not countable and must have cardinality <= [tex]\mathbf{R}[/tex] and invoke the Continuum Hypothesis to conclude that its cardinality is aleph-1 same as that of [tex]\mathbf{R}[/tex]..

Somehow this does not look appealing...

Can you explicitly construct a bijection and help me to visualise the situation better??

Thanks.
 
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Well the union of Q and R\Q is R right? |Q| = \aleph_0 and |R| = c so if |R\Q| was aleph_0 then you have a contradiction i.e. isn't the union of countable sets, countable?
 
Yes, that proves that R\Q is not countable. But it does not prove that card(R\Q)= card(R). As soumyashant said, You would need the contiuum hypothesis, that there is no set of cardinality strictly between that of R and that of Q, to finish that proof.
 
Try to prove the following fact: If A is an infinite set and B is a countable set, then [itex]|A \cup B| = |A|[/itex]. You shouldn't need to invoke the CH.
 

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