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My question is the following:

Let $A$ be a set with a distinguished point $0$. Given a set $X$, let $A^{(X)}$ denote the set of maps $X \to A$ which evaluate to $0$ for all but finitely-many values $x \in X$. Assume that $A$ contains at least two points and that $X,Y$ are infinite sets such that $|X| < |Y|$. Does this imply that $A^{(X)} < A^{(Y)}$?

For context, this set-theoretic question actually comes from commutative algebra. It is known that if $A$ is a nonzero (commutative, unital) ring and $m,n$ are positive integers such that there exists a monomorphism of $A$-modules $A^m \hookrightarrow A^n$, then necessarily $m \leq n$. I was wondering if the corresponding statement is true when we replace $A^m$ and $A^n$ with free $A$-modules over arbitrary bases. Then it occurred to me that this might be true for purely set-theoretic reasons, hence my above question.

I am very unsure about whether the original question should true or not, even in simple cases. On the one hand, it seems intuitively clear that $\left\{0,1\right\}^{(\mathbb{R})}$ should be larger than $\left\{0,1\right\}^{(\mathbb{N})}$. But on the other, the statement fails when $X$ and $Y$ are finite, which raises some doubt.

Many thanks.

Andrés E. Caicedo
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user1892304
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    ${0,1}^{(X)}$ is just the set of functions from $X$ to ${0,1}$ with finite support, which has the same cardinality as $[X]^{<\omega}$, the set of finite subsets of $X$. If $X$ is infinite, $[X]^{<\omega}$ and $X$ have the same cardinality. In particular, $\left|{0,1}^{(\Bbb N)}\right|=|\Bbb N|=\omega<2^\omega=\left|{0,1}^{(\Bbb R)}\right|$. – Brian M. Scott Apr 05 '20 at 21:02

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Let's estimate $|A^{(X)}|$ in terms of $|A|$ and $|X|$.

It is well known that $|\mathcal P_{\mathrm{fin}}(X)|=|X|$ (for infinite $X$), where the former denotes the set of finite subsets of $X$, and clearly for $x\in\mathcal P_{\mathrm{fin}}(X)$ there are $|x|\cdot|A|=|A|$ many elements of $A^{(X)}$ with support $x$. So in total we get $|A^{(X)}|=|A|\cdot|\mathcal P_{\mathrm{fin}}(X)|=|A|\cdot|X|=\max\{|A|,|X|\}$, by looking at all the possible functions with all the possible supports.

Now it should be clear that if $|A|$ dominates $|X|$ and $|Y|$ it is possible to have $|X|<|Y|$ but $|A^{(X)}|=|A^{(Y)}|$, for example this happens for $|X|=\aleph_0$, $|Y|=\aleph_1$ and $|A|=\aleph_2$.

Alessandro Codenotti
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  • Thanks for this answer, Alessandro. It’s a good things that I don’t usually work over rings of cardinality $\aleph_2$! I wonder if the algebraic statement also fails then... – user1892304 Apr 05 '20 at 21:15
  • The algebraic statement sounds true to me over rings with the IBN (such as commutative unital rings) since it should boil down to the proof that all basis have the same cardinality also in the infinite rank case – Alessandro Codenotti Apr 05 '20 at 21:22
  • The statement for monomorphisms is actually much harder to prove than the corresponding statements for epi- and isomorphisms. The proofs that I know (in the finite case) are all intimately founded on the fact that the ranks are finite, cf. this thread – user1892304 Apr 05 '20 at 21:33