Return to Answer

It is a very nice question! The answer is yes, natural instances of the $\Delta$ system property, which hold under GCH, are in fact equivalent to the GCH.

Theorem. $\Delta(\omega_2,\omega_1)$ is equivalent to CH.

Proof: You've pointed out that CH implies the principle, since the hypothesis you mention for this case amounts to $\omega_1^{\lt\omega_1}<\omega_2$, which amounts to CH. So let us consider what happens when CH fails. Let $T=2^{\lt\omega}$ be the tree of all finite binary sequences, and label the nodes of $T$ with distinct natural numbers. Let $F$ be the subsets of $\omega$ arising as the sets of labels occuring on any of $\omega_2$ many branches through $T$. Thus, $F$ has size $\omega_2$, and any two elements of $F$ have finite intersection. I claim that this family of sets can have no $\Delta$-system of size $\omega_2$, and indeed, it can have no $\Delta$-system even with three elements. If $r$ is the root of $a$, $b$ and $c$ in $F$, then $r=a\cap b=a\cap c$, and so $a$ and $b$ branch out at the same node that $a$ and $c$ do, in which case $b$ and $c$ must agree one step longer, so $b\cap c\neq r$. QED

The same idea works for higher cardinals as follows:

Theorem. For any infinite cardinal $\delta$, we have $\Delta(\delta^{++},\delta^+)$ is equivalent to $2^\delta=\delta^+$.

Proof. If $2^\delta=\delta^+$, then your criterion, which amounts to $(\delta^+)^{\lt\delta^+}<\delta^{++}$, is fulfilled, and so the $\Delta$ property holds. Conversely, consider the tree $T=2^{\lt\delta}$, the binary sequences of length less than $\delta$. Let $F$ be a family of $\delta^{++}$ many branches through $T$, regarding each branch $b$ as a subset of $T$, the set of its initial segments. Each such branch has size $\delta$, since the tree has height $\delta$. But for the same reason as before, there can be no $\Delta$ system with even three elements, since the tree is merely binary branching, and so three distinct branches cannot have a common root. This contradicts $\Delta(\delta^{++},\delta^+)$, as desired. QED

Corollary. The full GCH is equivalent to the assertion that $\Delta(\delta^{++},\delta^+)$ for every infinite cardinal $\delta$.

Update. It seems that the same idea shows that the hypothesis you mention is optimal, for one can reverse the lemma from the conclusion to this hypothesis.

Theorem. The following are equivalent, for regular $\kappa$ and $\mu\lt\kappa$:

1. $\Delta(\kappa,\mu)$
2. $\lambda^{\lt\mu}\lt\kappa$ for every $\lambda\lt\kappa$.

Proof. You mentioned that 2 implies 1, and this is how one usually sees the $\Delta$ system lemma stated. For the converse, suppose that $\lambda^{\lt\mu}\geq\kappa$ for some $\lambda\lt\kappa$. Since $\kappa$ is regular and $\mu\lt\kappa$, this implies $\lambda^\eta\geq\kappa$ for some $\eta\lt\mu$. Let $T$ be the $\eta$-branching tree $\lambda^{\lt\eta}$, which has height $\eta$. Let $F$ be a family of $\kappa$ many branches through this tree, where we think of a branch as the set of nodes in the tree that lie on it, a maximal linearly ordered subset of the tree $T$. I claim that this family has no subfamily that is $\Delta$ system of size $\eta^+$. The reason is that because the tree is $\eta$-branching, if we have $\eta^+$ many branches with a common root, then at least two of them must extend that root to the next level in the same way, a contradiction to it being a root. Thus, the failure of 2 implies the failure of 1, as desired. QED

It is a very nice question! The answer is yes, natural instances of the $\Delta$ system property, which hold under GCH, are in fact equivalent to the GCH.

Theorem. $\Delta(\omega_2,\omega_1)$ is equivalent to CH.

Proof: You've pointed out that CH implies the principle, since the hypothesis you mention for this case amounts to $\omega_1^{\lt\omega_1}<\omega_2$, which amounts to CH. So let us consider what happens when CH fails. Let $T=2^{\lt\omega}$ be the tree of all finite binary sequences, and label the nodes of $T$ with distinct natural numbers. Let $F$ be the subsets of $\omega$ arising as the sets of labels occuring on any of $\omega_2$ many branches through $T$. Thus, $F$ has size $\omega_2$, and any two elements of $F$ have finite intersection. I claim that this family of sets can have no $\Delta$-system of size $\omega_2$, and indeed, it can have no $\Delta$-system even with three elements. If $r$ is the root of $a$, $b$ and $c$ in $F$, then $r=a\cap b=a\cap c$, and so $a$ and $b$ branch out at the same node that $a$ and $c$ do, in which case $b$ and $c$ must agree one step longer, so $b\cap c\neq r$. QED

The same idea works for higher cardinals as follows:

Theorem. For any infinite cardinal $\delta$, we have $\Delta(\delta^{++},\delta^+)$ is equivalent to $2^\delta=\delta^+$.

Proof. If $2^\delta=\delta^+$, then your criterion, which amounts to $(\delta^+)^{\lt\delta^+}<\delta^{++}$, is fulfilled, and so the $\Delta$ property holds. Conversely, consider the tree $T=2^{\lt\delta}$, the binary sequences of length less than $\delta$. Let $F$ be a family of $\delta^{++}$ many branches through $T$, regarding each branch $b$ as a subset of $T$, the set of its initial segments. Each such branch has size $\delta$, since the tree has height $\delta$. But for the same reason as before, there can be no $\Delta$ system with even three elements, since the tree is merely binary branching, and so three distinct branches cannot have a common root. This contradicts $\Delta(\delta^{++},\delta^+)$, as desired. QED

Corollary. The full GCH is equivalent to the assertion that $\Delta(\delta^{++},\delta^+)$ for every infinite cardinal $\delta$.

Update. The same idea shows that the hypothesis you mention is optimal: one can reverse the lemma from the conclusion to the hypothesis.

Theorem. The following are equivalent, for regular $\kappa$ and $\mu\lt\kappa$:

1. $\Delta(\kappa,\mu)$
2. $\lambda^{\lt\mu}\lt\kappa$ for every $\lambda\lt\kappa$.

Proof. You mentioned that 2 implies 1, and this is how one usually sees the $\Delta$ system lemma stated. For the converse, suppose that $\lambda^{\lt\mu}\geq\kappa$ for some $\lambda\lt\kappa$. Since $\kappa$ is regular and $\mu\lt\kappa$, this implies $\lambda^\eta\geq\kappa$ for some $\eta\lt\mu$. Let $T$ be the $\lambda$-branching tree $\lambda^{\lt\eta}$, which has height $\eta$. Let $F$ be a family of $\kappa$ many branches through this tree, where we think of a branch as the set of nodes in the tree that lie on it, a maximal linearly ordered subset of the tree $T$. Each such branch is a set of size $\eta$. I claim that this family has no subfamily that is $\Delta$ system of size $\lambda^+$. The reason is that because the tree is $\lambda$-branching, if we have $\lambda^+$ many branches with a common root, then at least two of them must extend that root to the next level in the same way, a contradiction to it being a root. Thus, the failure of 2 implies the failure of 1, as desired. QED

It is a very nice question! The answer is yes, natural instances of the $\Delta$ system property, which hold under GCH, are in fact equivalent to the GCH.

Theorem. $\Delta(\omega_2,\omega_1)$ is equivalent to CH.

Proof: You've pointed out that CH implies the principle, since the hypothesis you mention for this case amounts to $\omega_1^{\lt\omega_1}<\omega_2$, which amounts to CH. So let us consider what happens when CH fails. Let $T=2^{\lt\omega}$ be the tree of all finite binary sequences, and label the nodes of $T$ with distinct natural numbers. Let $F$ be the subsets of $\omega$ arising as the sets of labels occuring on any of $\omega_2$ many branches through $T$. Thus, $F$ has size $\omega_2$, and any two elements of $F$ have finite intersection. I claim that this family of sets can have no $\Delta$-system of size $\omega_2$, and indeed, it can have no $\Delta$-system even with three elements. If $r$ is the root of $a$, $b$ and $c$ in $F$, then $r=a\cap b=a\cap c$, and so $a$ and $b$ branch out at the same node that $a$ and $c$ do, in which case $b$ and $c$ must agree one step longer, so $b\cap c\neq r$. QED

The same idea works for higher cardinals as follows:

Theorem. For any infinite cardinal $\delta$, we have $\Delta(\delta^{++},\delta^+)$ is equivalent to $2^\delta=\delta^+$.

Proof. If $2^\delta=\delta^+$, then your criterion, which amounts to $(\delta^+)^{\lt\delta^+}<\delta^{++}$, is fulfilled, and so the $\Delta$ property holds. Conversely, consider the tree $T=2^{\lt\delta}$, the binary sequences of length less than $\delta$. Let $F$ be a family of $\delta^{++}$ many branches through $T$, regarding each branch $b$ as a subset of $T$, the set of its initial segments. Each such branch has size $\delta$, since the tree has height $\delta$. But for the same reason as before, there can be no $\Delta$ system with even three elements, since the tree is merely binary branching, and so three distinct branches cannot have a common root. This contradicts $\Delta(\delta^{++},\delta^+)$, as desired. QED

Corollary. The full GCH is equivalent to the assertion that $\Delta(\delta^{++},\delta^+)$ for every infinite cardinal $\delta$.

It is a very nice question! The answer is yes, natural instances of the $\Delta$ system property, which hold under GCH, are in fact equivalent to the GCH.

Theorem. $\Delta(\omega_2,\omega_1)$ is equivalent to CH.

Proof: You've pointed out that CH implies the principle, since the hypothesis you mention for this case amounts to $\omega_1^{\lt\omega_1}<\omega_2$, which amounts to CH. So let us consider what happens when CH fails. Let $T=2^{\lt\omega}$ be the tree of all finite binary sequences, and label the nodes of $T$ with distinct natural numbers. Let $F$ be the subsets of $\omega$ arising as the sets of labels occuring on any of $\omega_2$ many branches through $T$. Thus, $F$ has size $\omega_2$, and any two elements of $F$ have finite intersection. I claim that this family of sets can have no $\Delta$-system of size $\omega_2$, and indeed, it can have no $\Delta$-system even with three elements. If $r$ is the root of $a$, $b$ and $c$ in $F$, then $r=a\cap b=a\cap c$, and so $a$ and $b$ branch out at the same node that $a$ and $c$ do, in which case $b$ and $c$ must agree one step longer, so $b\cap c\neq r$. QED

The same idea works for higher cardinals as follows:

Theorem. For any infinite cardinal $\delta$, we have $\Delta(\delta^{++},\delta^+)$ is equivalent to $2^\delta=\delta^+$.

Proof. If $2^\delta=\delta^+$, then your criterion, which amounts to $(\delta^+)^{\lt\delta^+}<\delta^{++}$, is fulfilled, and so the $\Delta$ property holds. Conversely, consider the tree $T=2^{\lt\delta}$, the binary sequences of length less than $\delta$. Let $F$ be a family of $\delta^{++}$ many branches through $T$, regarding each branch $b$ as a subset of $T$, the set of its initial segments. Each such branch has size $\delta$, since the tree has height $\delta$. But for the same reason as before, there can be no $\Delta$ system with even three elements, since the tree is merely binary branching, and so three distinct branches cannot have a common root. This contradicts $\Delta(\delta^{++},\delta^+)$, as desired. QED

Corollary. The full GCH is equivalent to the assertion that $\Delta(\delta^{++},\delta^+)$ for every infinite cardinal $\delta$.

Update. It seems that the same idea shows that the hypothesis you mention is optimal, for one can reverse the lemma from the conclusion to this hypothesis.

Theorem. The following are equivalent, for regular $\kappa$ and $\mu\lt\kappa$:

1. $\Delta(\kappa,\mu)$
2. $\lambda^{\lt\mu}\lt\kappa$ for every $\lambda\lt\kappa$.

Proof. You mentioned that 2 implies 1, and this is how one usually sees the $\Delta$ system lemma stated. For the converse, suppose that $\lambda^{\lt\mu}\geq\kappa$ for some $\lambda\lt\kappa$. Since $\kappa$ is regular and $\mu\lt\kappa$, this implies $\lambda^\eta\geq\kappa$ for some $\eta\lt\mu$. Let $T$ be the $\eta$-branching tree $\lambda^{\lt\eta}$, which has height $\eta$. Let $F$ be a family of $\kappa$ many branches through this tree, where we think of a branch as the set of nodes in the tree that lie on it, a maximal linearly ordered subset of the tree $T$. I claim that this family has no subfamily that is $\Delta$ system of size $\eta^+$. The reason is that because the tree is $\eta$-branching, if we have $\eta^+$ many branches with a common root, then at least two of them must extend that root to the next level in the same way, a contradiction to it being a root. Thus, the failure of 2 implies the failure of 1, as desired. QED

It is a very nice question!

Theorem. $\Delta(\omega_2,\omega_1)$ is equivalent to CH.

Proof: You've pointed out that CH implies the principle, since the hypothesis you mention for this case amounts to $\omega_1^{\lt\omega_1}<\omega_2$, which amounts to CH. So let us consider what happens when CH fails. Let $T=2^{\lt\omega}$ be the tree of all finite binary sequences, and label the nodes of $T$ with distinct natural numbers. Let $F$ be the subsets of $\omega$ arising as the sets of labels occuring on any of $\omega_2$ many branches through $T$. Thus, $F$ has size $\omega_2$, and any two elements of $F$ have finite intersection. I claim that this family of sets can have no $\Delta$-system of size $\omega_2$, and indeed, it can have no $\Delta$-system even with three elements. If $r$ is the root of $a$, $b$ and $c$ in $F$, then $r=a\cap b=a\cap c$, and so $a$ and $b$ branch out at the same node that $a$ and $c$ do, in which case $b$ and $c$ must agree one step longer, so $b\cap c\neq r$. QED

The same idea works for higher cardinals as follows:

Theorem. For any infinite cardinal $\delta$, we have $\Delta(\delta^{++},\delta^+)$ is equivalent to $2^\delta=\delta^+$.

Proof. If $2^\delta=\delta^+$, then your criterion of $(\delta^+)^{\lt\delta^+}<\delta^{++}$ is fulfilled, and so the $\Delta$ property holds. Conversely, consider the tree $T=2^{\lt\delta}$. Label the nodes of this tree with $2^{\lt\delta}$ many distinct labels (perhaps this is more than $\delta$, but it is no matter). Let $F$ be a family of $\delta^{++}$ many sets arising as the label sets of this many branches through $T$. Each of these sets has size $\delta$, since a branch has $\delta$ many labels. But again, for the same reason as above, there can be no $\Delta$ system with even three elements, since the tree is only binary branching, and this contradicts $\Delta(\delta^{++},\delta^+)$. QED

It is a very nice question! The answer is yes, natural instances of the $\Delta$ system property, which hold under GCH, are in fact equivalent to the GCH.

Theorem. $\Delta(\omega_2,\omega_1)$ is equivalent to CH.

Proof: You've pointed out that CH implies the principle, since the hypothesis you mention for this case amounts to $\omega_1^{\lt\omega_1}<\omega_2$, which amounts to CH. So let us consider what happens when CH fails. Let $T=2^{\lt\omega}$ be the tree of all finite binary sequences, and label the nodes of $T$ with distinct natural numbers. Let $F$ be the subsets of $\omega$ arising as the sets of labels occuring on any of $\omega_2$ many branches through $T$. Thus, $F$ has size $\omega_2$, and any two elements of $F$ have finite intersection. I claim that this family of sets can have no $\Delta$-system of size $\omega_2$, and indeed, it can have no $\Delta$-system even with three elements. If $r$ is the root of $a$, $b$ and $c$ in $F$, then $r=a\cap b=a\cap c$, and so $a$ and $b$ branch out at the same node that $a$ and $c$ do, in which case $b$ and $c$ must agree one step longer, so $b\cap c\neq r$. QED

The same idea works for higher cardinals as follows:

Theorem. For any infinite cardinal $\delta$, we have $\Delta(\delta^{++},\delta^+)$ is equivalent to $2^\delta=\delta^+$.

Proof. If $2^\delta=\delta^+$, then your criterion, which amounts to $(\delta^+)^{\lt\delta^+}<\delta^{++}$, is fulfilled, and so the $\Delta$ property holds. Conversely, consider the tree $T=2^{\lt\delta}$, the binary sequences of length less than $\delta$. Let $F$ be a family of $\delta^{++}$ many branches through $T$, regarding each branch $b$ as a subset of $T$, the set of its initial segments. Each such branch has size $\delta$, since the tree has height $\delta$. But for the same reason as before, there can be no $\Delta$ system with even three elements, since the tree is merely binary branching, and so three distinct branches cannot have a common root. This contradicts $\Delta(\delta^{++},\delta^+)$, as desired. QED

Corollary. The full GCH is equivalent to the assertion that $\Delta(\delta^{++},\delta^+)$ for every infinite cardinal $\delta$.