In this last step 5-2 of our proof of \( \Pi(j) \Rightarrow \Pi(j+1) \), we double-check (10.3) we showed last time, making sure that it correctly implies all the conditions of \( \Pi(j+1) \). Please also open these links: 10/15/24 for \( \Pi (j) \); 12/2/24 for (8.3); 12/9/24 for (9.2).
We keep assuming \(m_* = q_{i, j+1} – |C_i \cap X_{j+1}|\) are nonzero for the three \( i \) since 10/21/24. Having (8.3) and (9.2) by this, we confirm (10.3) i)-v) and \( \Pi(j+1)\) i)-v) in the following.
We have verfied \( \Pi(j) \Rightarrow \Pi(j+1) \) when \( m_* >0 \) for the three \( i \). If \( m_* \) is zero for any \(i \) before Step 1, we just do nothing on \( \mathcal{F}_i \) and it satisfies the required five conditions: because of the simultaneous core growth and Theorem 6 on 7/29/24, any part of \( C_i \) has been excluded from the other two \( \mathcal{F}_{i’} \), so \( \Pi(j+1) \)-v) is already true for \( \mathcal{F}_i \) before Step 1. So are the other four conditions because \( C_i \) and \(\mathcal{F}_i \) remain the same.
Our induction step is complete now. We’ve proven the three-disjoint-set claim by this! Any family of \( m \)-sets satisfying the \( \Gamma( m^{\frac{1}{2} + \epsilon} ) \)-condition includes three mutually disjoint sets as long as \( m \) is larger enough than \(1/\epsilon \).
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