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

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To prove that |X| = |Y| given that X\Y and Y\X are equipotent sets, one can start by establishing a bijection between the elements of X\Y and Y\X. The discussion highlights that both sets can be expressed as unions of disjoint subsets: X = (X\Y) ∪ (X∩Y) and Y = (Y\X) ∪ (X∩Y). By recognizing that the intersection X∩Y is common to both sets, a bijection can be constructed that includes these elements, leading to a conclusion that |X| equals |Y|. An analogy with married couples and their children illustrates the concept of establishing a one-to-one correspondence. This reasoning supports the assertion that the cardinalities of the two sets are indeed equal.
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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 $Y$?
 
Greetings, I am studying probability theory [non-measure theory] from a textbook. I stumbled to the topic stating that Cauchy Distribution has no moments. It was not proved, and I tried working it via direct calculation of the improper integral of E[X^n] for the case n=1. Anyhow, I wanted to generalize this without success. I stumbled upon this thread here: https://www.physicsforums.com/threads/how-to-prove-the-cauchy-distribution-has-no-moments.992416/ I really enjoyed the proof...

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