Tangle-Independent Set Gives Connected Set¶
Claim/Theorem¶
Let \(\mathcal T\) be a tangle of order \(k\) in a matroid \(M\), and let \(X\subseteq E(M)\) be independent in the tangle matroid \(M(\mathcal T)\) with
Then \(X\) is a cardinality-based connected set in \(M\):
for every \(A\subseteq E(M)\).
Equivalently, \(X\) is an (|X|+1)-connected set in the sense used by [[tangle-breadth-gives-k-connected-set.md]].
Proof sketch:
- Every subset \(Y\subseteq X\) is also independent in \(M(\mathcal T)\).
- By Lemma 15(iii) from the source paper, since \(|Y|<k-1\), this means:
- \(Y\) is \(\mathcal T\)-small, and
- there is no \(\mathcal T\)-small set \(B\supseteq Y\) with
- Now let \(A\subseteq E(M)\) and set
If \(s=0\) the claim is trivial. Assume \(s>0\) and, for contradiction, that
Then in particular \(\lambda_M(A)\le k-2\), so by the tangle axiom exactly one of \(A\) or \(E(M)-A\) is \(\mathcal T\)-small. 4. Whichever side is \(\mathcal T\)-small contains a subset \(Y\subseteq X\) of size \(s\). But then this \(\mathcal T\)-small side is a \(\mathcal T\)-small superset of \(Y\) with connectivity less than \(|Y|=s\), contradicting Step 2.
Hence
So any sufficiently small tangle-independent set is already a genuine connected set, even without any breadth hypothesis.
Dependencies¶
- [[tangle-breadth-gives-k-connected-set.md]]
Conflicts/Gaps¶
- This yields a connected set whose size is at most the tangle order minus
2, not a dense set spanning a constant fraction of the ground set. - The theorem is strongest when one has a source of large tangle-independent sets, such as [[high-tangle-order-gives-large-tangle-independent-set.md]].
- It does not by itself solve the separator-alignment problem for Conjecture 3, because a connected set of size
Omega(n/log n)is still far from the dense linear regime needed by [[dense-k-connected-set-forces-balanced-cut-rank.md]].
Sources¶
10.37236/12467