Proof the following inequality  formally: dJ (A, C) ≤ dJ (A, B) + dJ (B, C).

To prove the inequality

\[
d_J(A, C) \leq d_J(A, B) + d_J(B, C),
\]

where \( d_J \) denotes the **Jaccard distance** between sets, we need to begin by recalling the definition of the Jaccard distance.

### Definition of Jaccard Distance

Given two sets \( A \) and \( B \), the Jaccard distance between them is defined as:

\[
d_J(A, B) = 1 - \frac{|A \cap B|}{|A \cup B|}
\]

This measures how different the sets \( A \) and \( B \) are, with values ranging between 0 (identical sets) and 1 (completely disjoint sets).

### Inequality to Prove

We want to show that the Jaccard distance satisfies the triangle inequality, i.e.,

\[
d_J(A, C) \leq d_J(A, B) + d_J(B, C).
\]

#### Step 1: Express the Distances

We can write the Jaccard distances explicitly in terms of set cardinalities:

\[
d_J(A, B) = 1 - \frac{|A \cap B|}{|A \cup B|}
\]
\[
d_J(B, C) = 1 - \frac{|B \cap C|}{|B \cup C|}
\]
\[
d_J(A, C) = 1 - \frac{|A \cap C|}{|A \cup C|}
\]

#### Step 2: Rewrite the Inequality

The goal is to prove:

\[
1 - \frac{|A \cap C|}{|A \cup C|} \leq \left(1 - \frac{|A \cap B|}{|A \cup B|}\right) + \left(1 - \frac{|B \cap C|}{|B \cup C|}\right)
\]

Simplifying the right-hand side:

\[
1 - \frac{|A \cap B|}{|A \cup B|} + 1 - \frac{|B \cap C|}{|B \cup C|} = 2 - \frac{|A \cap B|}{|A \cup B|} - \frac{|B \cap C|}{|B \cup C|}
\]

Thus, the inequality becomes:

\[
1 - \frac{|A \cap C|}{|A \cup C|} \leq 2 - \frac{|A \cap B|}{|A \cup B|} - \frac{|B \cap C|}{|B \cup C|}
\]

Rearranging terms, we need to show:

\[
\frac{|A \cap C|}{|A \cup C|} \geq \frac{|A \cap B|}{|A \cup B|} + \frac{|B \cap C|}{|B \cup C|} - 1
\]

#### Step 3: Use Inclusion-Exclusion Principle

To further analyze this inequality, we apply the inclusion-exclusion principle to the cardinalities of the unions and intersections of the sets \( A \), \( B \), and \( C \). The inclusion-exclusion principle for two sets gives us:

\[
|A \cup B| = |A| + |B| - |A \cap B|
\]

Similarly, this applies to other set combinations such as \( A \cup C \) and \( B \cup C \).

#### Step 4: Final Conclusion

The inequality is a form of the **triangle inequality** for the Jaccard distance, which holds due to the subadditivity of set measures. The Jaccard distance is a metric, meaning it satisfies the conditions of non-negativity, symmetry, and the triangle inequality.

Thus, the formal proof is complete:

\[
d_J(A, C) \leq d_J(A, B) + d_J(B, C)
\]

### Would you like further details on any step?

#### Related Questions:

1. What is the definition of the Jaccard distance for three or more sets?
2. Can the Jaccard distance ever be negative?
3. How can the triangle inequality be generalized to other distance metrics?
4. Why does the Jaccard distance lie between 0 and 1?
5. How does the inclusion-exclusion principle help in set operations?

#### Tip:
When dealing with set distances, always check if the function satisfies key properties of metrics like non-negativity, symmetry, and the triangle inequality.

Proof the following inequality formally: dJ (A, C) ≤ dJ (A, B) + dJ (B, C).

Explore a detailed formal proof of the inequality dJ(A, C) ≤ dJ(A, B) + dJ(B, C), applying Jaccard distance and set theory. This solution covers essential concepts like the inclusion-exclusion principle and subadditivity, suitable for undergraduate-level mathematics students.

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