Prove that $d(X,Y) = |X\setminus Y| + |Y\setminus X|$ is a distance

Question

I was trying to prove that, for $S$ the power set of $\{1,2,\dots,n\}$, the following function

$$d(X,Y) = |X\setminus Y| + |Y\setminus X|$$

is a distance function. I managed to prove positivity and symmetry easily, but I am stuck in how to prove the triangle inequality $d(X,Z) \leq d(X,Y) + d(Y,Z)$. I have tried to look at particular cases, like $Y \subseteq Z$, but I can't seem to get to a solution.

Any hints would be appreciated.

As mentioned in one of the answers to the linked question, one could try to prove triangle inequality using the observation that $d(X,Y)=\sum\limits_{a\in S} \lvert\chi_X(a)-\chi_Y(a)\rvert$. — Martin Sleziak, Sep 24 '23 at 10:12

score 2 · Accepted Answer · answered Sep 22 '23 at 10:44

Hint: If $X^c$ denotes the complement of $X$ then:

$|X\setminus Z|=|X\cap Z^c|=|X\cap Z^c\cap Y|+|X\cap Z^c\cap Y^c|.$

And you can similarly decompose $|Z\setminus X|$. Can you find an upper bound for the right hand side? For example, note that $X\cap Z^c\cap Y\subseteq Y\setminus Z$.

Prove that $d(X,Y) = |X\setminus Y| + |Y\setminus X|$ is a distance

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