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Let $F(S) $ be the set of all finite subsets of a set $S$. For all $A, B ∈ F(S)$, let $∆ (A, B)=(A\setminus B) ∪ (B\setminus A) $ be the symmetric difference between A and B. Let $d(A, B ) $ be the cardinality of $∆ (A, B).$ Is $d$ a metric?
I am able to prove positivity, definiteness, symmetry. For any three sets $A, B, C$ I have to show $|(A\Delta B)| \le |(A \Delta C) |+| (C\Delta B) |$. My intuition suggests triangle inequality may also hold, but I can't make it rigorous, please elaborate.

Pat
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SoG
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2 Answers2

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It's quite intuitive to show this from a slighly higher view:

For any set $S$ and the space $C_c(S, \mathbb R)$ (arbitrary functions with finite support), $\|f\| = \sum_{x\in S} |f(x)|$ is the well-know $\mathcal l^1$ norm. In particular, $d(f, g) = \|f-g\|$ is a metric over $C_c(S, \mathbb R)$.

Now we can embed the space of finite sets of $S$ to $C_c(S, \mathbb R)$ by mapping each finite subset $A$ to its characteristic function $\mathbb 1_A$. Then $|A\Delta B|= \|\mathbb 1_A-\mathbb 1_B\|$, hence as the latter is a metric, so is the former.

Just a user
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Since symmetric difference is commutative and associative, we have $$A\Delta B\ =\ (A\Delta C)\,\Delta\, (C\Delta B)\ \subseteq (A\Delta C)\cup(C\Delta B)$$ and the latter has at most $|A\Delta C|+|C\Delta B|$ elements.

Berci
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