This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Covering
Definition 1 A family \((A_i)_{i\in I}\) is a covering of a set \(A\) if \(A=\bigcup_{i\in I} A_i\). For two coverings \((A_i)_{i\in I}\) and \((A'_j)_{j\in J}\) of \(A\), we say that \((A'_j)_{j\in J}\) is finer than \((A_i)_{i\in I}\) if for every \(j\in J\), there exists \(i\in I\) such that \(A'_j\subseteq A_i\).
Suppose a covering \((A_i)_{i\in I}\) of a set \(A\) is given. Then for any function \(f:B \rightarrow A\), the family \((f^{-1}(A_i))_{i\in I}\) of subsets of \(B\) is a covering of \(B\). We call this the preimage of \((A_i)\) under \(f\). For any function \(g:A\rightarrow C\), the family \((g(A_i))_{i\in I}\) of subsets of \(C\) need not be a covering of \(C\), but if \(g\) is surjective then they do cover \(C\). We call this the image of \((A_i)\) under the surjection \(g\).
Proposition 2 Consider a set \(A\) with its covering \((A_i)_{i\in I}\), and pick an arbitrary set \(B\).
- Suppose functions \(f,g:A\rightarrow B\) satisfy \(f\vert_{A_i}=g\vert_{A_i}\) for every \(i\in I\). Then \(f=g\).
-
If a family of functions \((f_i:A_i\rightarrow B)_{i\in I}\) satisfies the condition
\[f_i|_{A_i\cap A_j}=f_j|_{A_i\cap A_j}\]then there exists a function \(f:A\rightarrow B\) extending all the \(f_i\).
Proof
To prove the first claim, let an arbitrary \(x\in A\) be given. Since \((A_i)_{i\in I}\) covers \(A\), there exists some \(i\in I\) such that \(x\in A_i\). Now
\[f(x)=(f|_{A_i})(x)=(g|_{A_i})(x)=g(x)\]and thus the first claim holds.
For the second claim, use the given functions \(f_i=(F_i,A_i,B)\) to form \(F=\bigcup F_i\), and consider the new triple \(f=(F,A,B)\). Then \(\pr_1F=A\) is obvious, so to show that \(f\) is a function it suffices to prove that for any \(x\in A\), there is a unique \(y\) such that \((x,y)\in F\).
Suppose \(y,y'\in B\) satisfy \((x,y)\in F\) and \((x,y')\in F\). Then there exist \(i,j\) such that \((x,y)\in F_i\) and \((x,y')\in F_j\). Now
\[y=(f_i)(x)=(f_i|_{A_i\cap A_j})(x)=(f_j|_{A_i\cap A_j})(x)=(f_j)(x)=y'\]and thus the second claim also holds.
It is obvious from the first claim that a function \(f\) satisfying condition 2 of the above proposition is unique. Also, if in particular \(A_i\cap A_j\) holds for all \(i,j\), then the hypothesis of the second claim is always satisfied. We define this as follows.
Definition 3 Sets \(A\) and \(B\) are disjoint if \(A\cap B=\emptyset\). More generally, a family \((A_i)_{i\in I}\) is pairwise disjoint if for any \(i, j\in I\) with \(i\neq j\), we have \(A_i\cap A_j=\emptyset\).
Definition 4 A partition of a set \(A\) is a pairwise disjoint covering of \(A\).
In general, since \(\emptyset\) plays no role among the members of this family, when we speak of a partition we assume that no member is empty.
Sum of Sets
Proposition 5 Let \((A_i)_{i\in I}\) be a family of sets. Then there exists a set \(S\) such that
- \(S\) is the union of a pairwise disjoint family \((S_i)_{i\in I}\), and
- for every \(i\in I\) there exists a bijection from \(A_i\) to \(S_i\).
Proof
Let \(S_i\) be the set of all pairs \((x, i)\) with \(x\in A_i\). Then \((S_i)_{i\in I}\) is pairwise disjoint. Also, for each \(i\), the map \(x\mapsto (x,i)\) is a bijection from \(A_i\) to \(S_i\). Therefore \(S=\bigcup_{i\in I} S_i\) satisfies the required conditions.
Definition 6 A set \(S\) satisfying the above conditions is called the sum of the family \((A_i)_{i\in I}\), and is denoted by \(\sum_{i\in I} A_i\).
This set is sometimes called the disjoint union, and is also written as \(\bigsqcup_{i\in I} A_i\). The following proposition shows that this name is quite reasonable.
Proposition 7 Consider a pairwise disjoint family \((A_i)_{i\in I}\). Let \(A\) be their union and \(S\) their sum. Then there exists a bijection between \(A\) and \(S\).
Proof
If \(f_i:A_i\rightarrow S_i\) are bijections satisfying the conditions of Proposition 5, then by Proposition 2 we can extend \((f_i)_{i\in I}\) to \(\bigcup_{i\in I} A_i=A\).
The intuition for why this is called the sum of sets will appear later. (§Operations of Cardinals, ⁋Definition 1)
Universal property
There is something we did not mention in Definition 6. The sum \(X\) of a family of sets \((A_i)\) is not unique. There are infinitely many sets satisfying the conditions of Proposition 5. For example, in the proof of that proposition we took \(S\) to be the set of pairs \((x,i)\), but one can see that taking the set of pairs \((i,x)\) also satisfies the definition of sum. Therefore, strictly speaking, writing the sum of the \(A_i\) as \(\sum A_i\) is not a well-defined expression.
First, let us examine the universal property of the sum as follows.
Theorem 8 (Universal property of sum) Let a family of sets \((A_i)\), a set \(S\) as defined in Proposition 5, and injections \(\iota_i:A_i\rightarrow S\) be given. Then, whenever another set \(B\) and maps \(f_i:A_i\rightarrow B\) are given, there exists a unique \(f:S\rightarrow B\) such that \(f_i=f\circ\iota_i\).
Proof
First, let us show that such a function \(f\) is unique (if it exists). For this, it suffices to show that for any \(x\in S\), its function value \(f(x)\) is always uniquely determined. Since \(S\) is the union of a pairwise disjoint family \((S_i)\), there exists a unique \(i\in I\) such that \(x\in S_i\). Then since \(\iota_i:A_i\rightarrow S_i\) is a bijection, there again exists a unique element \(x_i\) of \(A_i\) such that \(\iota_i(x_i)=x\). Now,
\[f(x)=f(\iota_i(x_i))=(f\circ\iota_i)(x_i)=f_i(x_i)\]so the function value \(f(x)\) at \(x\) must equal \(f_i(x_i)\), and therefore \(f\) is uniquely determined.
Now, taking a hint from the uniqueness proof, let us show the existence of the function \(f\). Define \(f(x)\) to be \(f_i(x_i)\) as in the above equation, and prove that \(f\) is actually a function. For instance, with this definition \(f\) will be defined for all elements of \(S\), and moreover a single \(x\) is assigned exactly one function value as discussed above.
In many cases the set \(S\) appearing in the proof of Proposition 5 is defined to be the sum of the \(A_i\), but in fact this is putting the cart before the horse. The reason we think of \(S\) as the sum of the \(A_i\) in many areas is not because the set \(S\) itself has any special meaning, but because of notational convenience. The properties of the sum do not come from the set \(S\), but from the universal property above.
Therefore, we could simply define it as follows from the start.
Definition 6\('\) The sum of a given family of sets \((A_i)\) is a set \(\sum A_i\) together with maps \(\iota_i:A_i\rightarrow \sum A_i\) satisfying the following condition:
Whenever a set \(B\) and maps \(f_i:A_i\rightarrow B\) are given, there exists a unique function \(f:\sum A_i\rightarrow B\) such that \(f_i=f\circ\iota_i\).
Of course, to use this as a definition we must show that at least one object satisfying the universal property exists. And Theorem 8 does exactly that.
We mentioned earlier that the set \(\sum A_i\) is not well-defined in the strict sense. But even if such a set itself is not well-defined, if several such sets are given then there exists a bijection between them. This situation is called unique up to bijection. From Definition 6\('\) one can show that the sum of sets is unique up to bijection.
Corollary 9 For a family of sets \((A_i)\), the sum \(\sum A_i\) is unique up to bijection.
Proof
Let two sums \(S\) and \(S'\) be given, and let \(\iota_i\) and \(\iota_i'\) be the injections from \(A_i\) into \(S\) and \(S'\) respectively. First, for the functions \(\iota_i':A_i\rightarrow S'\), applying the universal property of \(S\) yields a unique \(\phi':S\rightarrow S'\) such that \(\iota_i'=\phi'\circ\iota_i\). Similarly, applying the universal property of \(S'\) to the functions \(\iota_i\) yields a unique \(\phi:S'\rightarrow S\) such that \(\iota_i=\phi\circ\iota_i'\). Then
\[\iota_i'=\phi'\circ\iota_i=\phi'\circ(\phi\circ\iota_i')=(\phi'\circ\phi)\circ\iota_i'\]On the other hand, apply the universal property of \(S'\) to the functions \(\iota_i':A_i\rightarrow S'\) this time. Then there exists a unique function \(\psi:S'\rightarrow S'\) satisfying \(\iota_i'=\psi\circ\iota_i'\). This is obviously satisfied by \(\psi=\id_{S'}\), so by uniqueness every function \(\psi\) satisfying this equation equals \(\id_{S'}\). Therefore \(\phi'\circ\phi=\id_{S'}\), and since \(\id_{S'}\) is bijective, \(\phi'\) is surjective and \(\phi\) is injective. (§Retraction and Section, ⁋Proposition 3)
Likewise, one can show that \(\phi\circ\phi'=\id_S\), which implies that \(\phi\) is surjective and \(\phi'\) is injective. Thus they are both bijections, so there exists a bijection between \(S\) and \(S'\).
References
[Bou] N. Bourbaki, Theory of Sets. Elements of mathematics. Springer Berlin-Heidelberg, 2013.
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