# Are there situations outside of set theory where it would be useful if $\mathrm{ICF}$ were true?

Write $\mathrm{ICF}$ for the "injective continuum function hypothesis" i.e. the sentence of ZFC expressing that

$$2^X \cong 2^Y \rightarrow X \cong Y$$

for all sets $X$ and $Y$, where $\cong$ denotes equipotency.

If I understand correctly:

1. $\mathrm{ICF}$ is independent of $\mathrm{ZFC}$.
2. $\mathrm{GCH}$ implies $\mathrm{ICF}.$ (In the presence of the $\mathrm{ZFC}$ axioms.)
3. The converse of 2 is false.

Question. Are there situations outside set theory (e.g. group theory, measure theory, etc.) where it would be useful if $\mathrm{ICF}$ were true, irrespective of whether or not $\mathrm{GCH}$ is true?

• "Some reason" :-) Those are two different $\rm C$'s. – Asaf Karagila Mar 30 '14 at 9:31
• Also, I recall some talk about matroids and there was an example which required $\sf ICF$ to hold and $\sf CH$ to fail. I'll try to dig out the details. – Asaf Karagila Mar 30 '14 at 9:31
• @AsafKaragila, that would be great! I thought matroids were finite though, so I'm surprised ICF matters in that field. – goblin Mar 30 '14 at 9:35
• You probably thought that graphs were finite too, before you knew the harsh truth of reality! – Asaf Karagila Mar 30 '14 at 9:36
• Here's one from, uh, Lattice Theory. If two complete atomic Boolean algebras have the same cardinality, then they are isomorphic. – bof Mar 30 '14 at 9:55

In the theory of infinite matroids one can talk about a basis of a matroid. Higgs proved that $\sf GCH$ implies that if $\cal M$ is a matroid on a set $E$, then every two bases of $\cal M$ have the same cardinality.

However a close inspection of the proof shows that in fact we need two facts which follow from $\sf GCH$ (but do not imply it, even in conjunction):

1. $\sf ICF$, and
2. For every infinite set $X$ there exists a chain of subsets of size $2^{|X|}$ in $\mathcal P(X)$.

I'm not 100% sure whether or not the second is implied by the first, and I am inclined to believe that the answer to that is negative. But this is an example of somewhere that $\sf ICF$ explicitly shows up in the proof.

For some more information: Stefan Geschke - An invitation to infinite matroids (slides)

• Sweet. Does Axiom 2 have a name? I'd like to read more about it. – goblin Mar 30 '14 at 9:49
• Not that I know of. It's more of a model theoretic thing, methinks. – Asaf Karagila Mar 30 '14 at 9:57
• If its not very complicated, can you explain why GCH proves 2? I get that under GCH it reads "For every ordinal $\alpha,$ there exists a subchain of $\mathcal{P}(\aleph_\alpha)$ of cardinality $\aleph_{\alpha+1}$," but why should that be true? – goblin Mar 30 '14 at 13:54
• As I said, this is somewhat model theoretic, as far as I know. So I can't really give you a short explanation... Sorry – Asaf Karagila Mar 30 '14 at 14:06
• Fair enough, but is the case $\alpha=0$ any easier? Sorry if I'm hassling you. – goblin Mar 30 '14 at 14:07

a) The reason why GCH gives long chains in $\mathcal P(X)$ is the following: For an infinite cardinal $\kappa$ let $L=\{0,1\}^\kappa$ and let $B$ be the set of all sequences in $\{0,1\}^\kappa$ that are eventually 0 (or constant, it doesn't matter). The size of $B$ is $\sup\{2^\lambda:\lambda<\kappa\}$.
$L$ is linearly ordered by the lexicographic ordering and $B$ is dense in $L$. Assuming GCH we have $2^\lambda\leq\kappa$ for all $\lambda<\kappa$ (this is all of GCH that we need). Hence $|B|=\kappa$. Let $X=B$. Now the family of sets $\{x\in B:x<y\}$, $y\in L$, is a chain in $\mathcal P(X)$ of size $2^{|X|}$.
b) (2) is equivalent to (2'). One half of this is already hidden in my answer a): If there is an infinite linear order of size $|X|$ whose completion is of size $2^{|X|}$, then $\mathcal P(X)$ has a chain of length $2^{|X|}$.
On the other hand, assume $\mathcal P(X)$ has a chain $\mathcal C$ of length $2^{|X|}$. We may assume that $\mathcal C$ is closed under infinite intersections and that it contains $X$. Now for each $x\in X$ consider the first $A\in\mathcal C$ with $x\in A$ and call it $A_x$. By our assumptions on $\mathcal C$, $A_x$ exists for all $x\in X$. It is just the intersection of all $A\in\mathcal C$ with $x\in A$.
Now $\mathcal A=\{A_x:x\in X\}$ is a dense subset of the linearly ordered set $\mathcal C$ ($\mathcal C$ is ordered by $\subseteq$, of course). $\mathcal A$ is a linear order of size $|X|$ with a completion of size $2^{|X|}$.