Prove |X| = |Y| When X\Y and Y\X are Equipotent Sets

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Discussion Overview

The discussion revolves around proving that if the sets \(X \setminus Y\) and \(Y \setminus X\) are equipotent, then the cardinalities of sets \(X\) and \(Y\) are equal. The scope includes mathematical reasoning and exploration of set theory concepts.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about how to begin the proof, mentioning a failed attempt to establish a bijection between \(X \setminus Y\) and \(Y \setminus X\).
  • Another participant notes the decomposition of sets \(X\) and \(Y\) into disjoint unions, suggesting a structural approach to the problem.
  • A third participant reiterates the decomposition of sets but admits uncertainty about its application in the proof.
  • A later reply introduces a practical example involving married couples and children, proposing that a one-to-one correspondence can be constructed between sets \(X\) and \(Y\) based on this scenario.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on how to approach the proof, with multiple viewpoints and methods suggested but no agreement on a definitive solution.

Contextual Notes

There are unresolved assumptions regarding the nature of the sets and the conditions under which the bijections are to be established. The discussion reflects varying levels of understanding and approaches to the problem.

Who May Find This Useful

Readers interested in set theory, mathematical proofs, and the concept of equipotent sets may find this discussion relevant.

Guest2
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Prove that if X\Y and Y\X are equipotent sets then |X| = |Y|.

The problem is that I've no clue where to start...

(Futile) attempt: There is bijection $f: X\backslash Y \to Y\backslash X$. For every $r_1 \in X\backslash Y$ there exists $r_2$ s.t. $r_2 \in Y\backslash X$. That's $r_1 \in X$ and $r_2 \in Y$. So there's a bijection $f: X \to Y$. :confused:
 
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Note that $X=(X\setminus Y)\sqcup (X\cap Y)$ where $\sqcup$ denotes the union of disjoint sets, and similarly $Y=(Y\setminus X)\sqcup (X\cap Y)$.
 
Evgeny.Makarov said:
Note that $X=(X\setminus Y)\sqcup (X\cap Y)$ where $\sqcup$ denotes the union of disjoint sets, and similarly $Y=(Y\setminus X)\sqcup (X\cap Y)$.
I'm not really sure how to use that.
 
Suppose that you have several married couples with kids. Let $X$ be the set of husbands and children and $Y$ be the set of wives and children. Then, obviously, $X\setminus Y$ (the set of husbands) is in natural one-to-one correspondent with $Y\setminus X$ (the set of wives). Can't you construct a one-to-one correspondence between $X$ and $Y$?
 

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