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Edge-connectivity of undirected and directed hypergraphs

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Section 3.4. Covering by uniform <strong>hypergraphs</strong> 63<br />

e = e ′ +w can be feasibly split <strong>of</strong>f if <strong>and</strong> only if e ∈ Q <strong>and</strong> there is no critical set containing<br />

w. Since e ′ was chosen so that e ′ + w ∈ Q for at least one w ∈ W , we can assume that<br />

there is at least one critical set in B2.<br />

Claim 3.25. Let X ∈ B2 be a critical set. If Y ∈ B1, then one <strong>of</strong> X ∩ Y , X − Y , <strong>and</strong><br />

Y − X is empty. If w ∈ W − X, then e ′ + w ∈ Q.<br />

Pro<strong>of</strong>. If X ∈ B2, Y ∈ B1, <strong>and</strong> X ∩ Y, X − Y, Y − X are non-empty, then p(X − Y ) ≤ p(X)<br />

<strong>and</strong> p(Y −X) < p(Y ) by the definition <strong>of</strong> B1 <strong>and</strong> B2, which contradicts (2.9). If w ∈ W −X,<br />

then e ′ +w ∈ Q unless there is a set Y ∈ B1 such that w ∈ Y <strong>and</strong> |e ′ ∩Y | = m(Y )−p(Y )+1.<br />

But then X ∩ Y = ∅, so X ⊆ Y , which contradicts the criticality <strong>of</strong> X.<br />

Suppose indirectly that every w ∈ W is in a critical set. Consider a family Z =<br />

{Z1, . . . , Zt} <strong>of</strong> maximal critical sets, such that every w ∈ W is in at least one <strong>of</strong> them, <strong>and</strong><br />

the family has minimal number <strong>of</strong> members. The following sequence <strong>of</strong> claims demonstrates<br />

that the existence <strong>of</strong> such a family leads to a contradiction. Let Z := ∩ t i=1Zi.<br />

Claim 3.26. If i = j, then m(Zi ∩ Zj) = m(Z) = ν − 1, <strong>and</strong> p(Zi) = m(Zi − Z) =<br />

m(Zi − Zj) = p(Zi − Zj).<br />

Pro<strong>of</strong>. We know that m(Zi ∩ Zj) ≥ m(Z) ≥ |e ′ ∩ Z| = ν − 1. Since Zi ∈ B2, we have<br />

p(Zi) = m(Zi) − (ν − 1), so (2.9) gives that 0 ≤ p(Zi − Zj) + p(Zj − Zi) − p(Zi) − p(Zj) ≤<br />

m(Zi − Zj) + m(Zj − Zi) − p(Zi) − p(Zj) = 2(ν − 1) − 2m(Zi ∩ Zj). This is possible only<br />

if equality holds throughout, so m(Zi ∩ Zj) = ν − 1, <strong>and</strong> m(Zi − Zj) = p(Zi − Zj).<br />

Clearly |Z| ≥ 2 <strong>and</strong><br />

m(V )<br />

ν<br />

≥ 2, since m(V − Zi) > 0 for every i. Suppose that |Z| = 2.<br />

Then p(Z1) = m(Z1 − Z2) = m(V − Z2) ≥ p(Z2) <strong>and</strong> vice versa, so p(Z1) = p(Z2) =<br />

m(Z1 − Z2) = m(Z2 − Z1) =<br />

m(V )<br />

ν<br />

m(V )−(ν−1)<br />

≤ 2<br />

m(V )−(ν−1)<br />

. This value can be integer only if ν ≥ 3, but then<br />

2<br />

= p(Z1), contradicting B3 = ∅. Therefore we may assume that |Z| ≥ 3.<br />

Claim 3.27. ∪ t i=1Zi = V , Zi − Z is a special singleton for every i, <strong>and</strong> p(Zi ∪ Zj − Z) > 0<br />

for every i, j.<br />

Pro<strong>of</strong>. For a set <strong>of</strong> indices I ⊆ {1, . . . , t}, let UI := ∪i∈IZi. We have seen that p(Zi) =<br />

m(Zi − Z) for every i. If X is a subset <strong>of</strong> V for which m(X) = |e ′ ∩ X| = ν − 1,<br />

then p(X) ≤ m(X) − |e ′ ∩ X| + 1 = 1. In particular, this holds for every Z ⊆ X ⊆<br />

∪i=j(Zi ∩ Zj). Thus we get from (3.7) that p(UI) ≥ m(UI − Z) − (|I| − 1) > 0 if UI = V .<br />

For I := {1, . . . , t} this implies that ∪ t i=1Zi = V , since p(∪ t i=1Zi) ≤ m(V − ∪ t i=1Zi) = 0.<br />

Now let I := {1, . . . , t} − {i0}, where i0 is chosen so that m(Zi0 − Z) is minimal. Then<br />

m(Zi0 − Z) ≥ p(UI) ≥ m(UI − Z) − (t − 2). Since |Z| ≥ 3, this is only possible if<br />

m(Zi − Z) = 1 for every i. Thus Zi − Zj is a special singleton for every i = j, which

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