If the functions $f$ and $g$ are functions that their output is the cardinality of some set associated to its input (the exact definition of these functions is irrelevant here), does $\forall x\in X:f(x)\leq g(x)$ and $\sum_{x\in X}f(x)=\sum_{x\in X}g(x)$ imply that $f(x)=g(x),\,\forall x\in X$? Where $X$ is a finite set. The only reason here to talk about cardinalities and not just numbers is that cardinalities may be infinite, and this is the case I'm looking for.
I start by making the assumption that $\forall x\in X:f(x)\leq g(x)$ and $\sum_{x\in X}f(x)=\sum_{x\in X}g(x)$ are true. If there exists an $x\in X$ such that $f(x) Is this line of reasoning correct? I do not feel confident about the step of defining an $x\in X$ with some property, and then getting a contradiction based on the assumption. I don't know if this really completes the proof, there is something missing, or it's merely a statement like "If X is true, then X is true".
Answer
Say $X=\Bbb N$, $f(n)=n$, $g(n)=2^n$. Now, obviously $2^n>n$ and both the sums are $\aleph_0$.
Or even $X=\{1,2\}$, $f(1)=0,\ g(1)=1,\ f(2)=\aleph_0,\ g(2)=\aleph_0$
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