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Suppose we have two functions $f=\{(1,2),(3,4)\}$ and $g=\{(1,2),(3,4)\}$ with codomains $A$ and $B$ respectively so that $A\not=B.$ Then since $f$ and $g$ are sets, then the definition of equality of sets as defined here implies that the two sets-functions-are equal. But following the defination of equality of functions as defined here the two functions-sets-are not equal. Contradiction. Both definitions are contrdictiary! Have I done some mistake? Thanks.

Osmium
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  • @Joe So mathematicians define terms differently according to their needs and mood? This is not mentioned in my math book. – Osmium Mar 05 '22 at 14:42
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    In short, yes. Definitions are not universal truths etched into stone; they are simply convenient abbreviations. For example, saying that a set $S$ is "countable" is an abbreviation of saying that there is a bijection between $\mathbb N$ and $S$. But wait ... some authors use countable to mean there is a bijection between a subset of $\mathbb N$ and $S$. That might be a more useful convention in a certain context. – Joe Mar 05 '22 at 14:43
  • @Joe so what definition of equality of sets is used by the people who use the equality of function mentioned in the question? – Osmium Mar 05 '22 at 15:08
  • A set is not a function until a mapping is established. It can also be undefined for some (or even all) elements. Concerning Joe's comment : Finite sets are also countable , but usually the set is already infinite if we ask whether it is countable, and then the issue vanishes. – Peter Mar 05 '22 at 16:17
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    @Osmium: Everyone agrees that two sets $A$ and $B$ are equal if and only if $$(\forall x)(x\in A \iff x\in B) , .$$This is in fact an axiom of ZFC. However, some authors define a function as a set of ordered pairs blah blah blah... (like you did in your question), whereas others would call the aforementioned set the graph of a function—the actual function consists of a domain, codomain, and the graph. It is only in the second approach where the concept of a "codomain" really makes any sense. – Joe Mar 06 '22 at 23:20
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    @Osmium: For instance, according to the first definition, the following functions are identical, since they are the same set:$$ f:\mathbb R\to\mathbb R, f(x)=x^2 \quad g:\mathbb R\to[0,\infty),g(x)=x^2 $$Therefore, it makes no sense to call $g$ surjective and $f$ non-surjective. But according to the second defintion, $f$ and $g$ are different functions as they have different codomains. – Joe Mar 06 '22 at 23:23
  • @Osmium: Finally, it's worth noting that while it is true that functions are sets in the context of set theory, there's no point in getting too hung up on the precise set-theoretic formulation of a mathematical concept. For example, in set theory, numbers like $\sqrt{2}$ and $\pi$ are also sets, but a statement such as $\sqrt{2}\subset\pi$ would make little sense to the working mathematician (even if such a statement would technically be true in the set-theoretic abstraction of the theory). – Joe Mar 06 '22 at 23:27
  • @Peter: It is fairly common to define a countable set as a set with exactly the same cardinality as the natural numbers (so finite sets are not considered countable). See, for instance, Walter Rudin's Principles of Mathematical Analysis (chapter 2) or Terence Tao's Analysis I (chapter 8). In this usage, a set with a cardinality that is less than or equal to $\mathbb N$ is called at most countable. – Joe Mar 07 '22 at 15:01

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