$\DeclareMathOperator\Card{Card}$Let $A$ be an infinite set, and let $B=\{X\subset A \mid \Card(X)<\Card(A)\}$. Can it be proven that $\Card(A)=\Card(B)$?
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2$\begingroup$ Better say $A$ is not a finite set. $\endgroup$– Zach TeitlerCommented Dec 7, 2023 at 1:46
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7$\begingroup$ Not from $\mathsf{ZFC}$ alone: if $\mathsf{CH}$ fails, then taking $\vert A\vert=\aleph_1$ we get $\vert B\vert=2^{\aleph_0}>\vert A\vert$. But this would be more appropriate at math.stackexchange. $\endgroup$– Noah SchweberCommented Dec 7, 2023 at 2:04
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2$\begingroup$ This question has the potential for an interesting question if it is edited along the lines of Joel David's answer (namely, as a function of the cardinality). It should be edited rather than closed. $\endgroup$– Mikhail KatzCommented Dec 7, 2023 at 11:37
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$\begingroup$ How should it be rephrased? $\endgroup$– Vuk JovovicCommented Dec 7, 2023 at 23:07
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$\begingroup$ It is fine with me to close the question or migrate it to math.stackexchange. I posted an answer just to point out that the hypothesis does not have GCH or CH dependence, as it is refutable in ZFC for singular cardinals. But also, I find it interesting that it is consistent with ZFC that it fails for all infinite cardinals. $\endgroup$– Joel David HamkinsCommented Dec 8, 2023 at 0:17
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1 Answer
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This hypothesis is refutable in ZFC for the case where $A$ has singular cardinality $\kappa$, since by König's theorem $\kappa<\kappa^{\text{cof}(\kappa)}$.
Thus, we have provable counterexamples to the hypothesis, such as the case $A=\aleph_\omega$, for which the set $B$ will be strictly larger than $A$.
For other sets $A$, however, it can be true. Set theorists would usually write $\kappa^{<\kappa}=\kappa$ for the cardinals $\kappa$ for which it holds, and this is commonly seen as a hypothesis in theorems. If GCH holds, then this is true for every regular cardinal (but still false for every singular cardinal).
Let me also mention that set theorists have produced models with a global violation of the GCH, meaning that $2^\kappa>\kappa^+$ for every infinite cardinal. [Thanks for comment of Andrés.] We can have, for example, $2^{\aleph_\alpha}=\aleph_{\alpha+2}$ for every $\alpha$. This situation requires large cardinal strength to produce. We may also assume without loss that there are no inaccessible cardinals, by chopping the universe off at the least inaccessible.
In such a model, we will have $\kappa<\kappa^{<\kappa}$ for every uncountable cardinal $\kappa$. This will be immediate for successor cardinals by the global failure of the GCH. It will hold for singular limits by the König's theorem observation above. And there will be no uncountable regular limit cardinals by our ommission of the inaccessibles.
In such a model, there is not a single instance of $\text{card}(A)=\text{card}(B)$, to use your notation, except the countably infinite case, and the finite cases of size 0 and 1.
answered Dec 7, 2023 at 2:10
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$\begingroup$ Is $Card(A)=Card(B)$ equivalent to any axiom? $\endgroup$ Commented Dec 7, 2023 at 23:09
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1$\begingroup$ As I mentioned, this is the assertion $\kappa=\kappa^{<\kappa}$, where $\kappa$ is the cardinality of $A$. It is false for all singular $\kappa$, so if you mean the universal claim (i.e. "for all infinite $A$...), it wouldn't be an attractive axiom. Meanwhile, it can be true for other cardinals, as we've mentioned, and the hypothesis $\kappa=\kappa^{<\kappa}$ appears in many theorems, making it something like an axiom I suppose. $\endgroup$ Commented Dec 8, 2023 at 0:07
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2$\begingroup$ Your argument does not work for $\kappa$ the successor of a singular. To ensure that $\kappa^{<\kappa}>\kappa$ for all uncountable cardinals you need to violate $\mathsf{SCH}$ badly; the consistency strength should be that of the global violation of $\mathsf{GCH}$. $\endgroup$ Commented Dec 10, 2023 at 0:01
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2$\begingroup$ @AndrésE.Caicedo Ah, yes, you are right. I have edited. $\endgroup$ Commented Dec 10, 2023 at 0:49