and, if $I \neq \emptyset$,

Let $x$ be an element of $\bigcap\limits_{\iota \in I} X_\iota$. There exists an index $\iota \in I$ such that $x \in X_\iota$. Since $f\langle K \rangle = I$, there exists an index $\varkappa \in K$ such that $\iota = f(\varkappa)$, whence $x \in X_{f(\varkappa)}$ and consequently

Conversely, if $x \in \bigcup\limits_{\varkappa \in K} X_{f(\varkappa)}$, there exists $\varkappa \in K$ such that $x \in X_{f(\varkappa)}$, and therefore, since $f(\varkappa) \in I$, $x \in \bigcup\limits_{\iota \in I} X_\iota$. Hence

¶ Now suppose that $I \neq \emptyset$, and let $x$ be an element of $\bigcap\limits_{\iota \in I} X_\iota$. For each element $\varkappa$ of K we have $f(\varkappa) \in I$, hence $x \in X_{f(\varkappa)}$, and therefore

Conversely, let $x$ be an element of $\bigcap\limits_{\varkappa \in K} X_{f(\varkappa)}$. If $\iota$ is any element of I, there exists an element $\varkappa$ of K such that $\iota = f(\varkappa)$, whence $x \in X_\iota$ and consequently $x \in \bigcap_{\iota \in I} X_\iota$. Hence

For families of subsets of a given set, it is clear that the second part of Proposition 1 remains valid without the restriction $I \neq \emptyset$.

Apply Proposition 1 to the constant mapping $\iota \to \alpha$ of I onto $\{\alpha\}$.

and (if $L \neq \emptyset$ and $J_\lambda \neq \emptyset$ for each $\lambda \in L$)

("associativity" of union and intersection).

Let $x$ be an element of $\bigcup_{\iota \in I} X_\iota$. There exists an index $\iota \in I$ such that $x \in X_\iota$. Since I is the union of the family $(J_\lambda)_{\lambda \in L}$, there exists an index $\lambda \in L$ such that $\iota \in J_\lambda$, whence $x \in \bigcup_{\iota \in J_\lambda} X_\iota$, and consequently

Conversely, let $x$ be an element of this set. There exists an index $\lambda \in L$ such that $x \in \bigcup_{\iota \in J_\lambda} X_\iota$, hence there exists an index $\iota \in J_\lambda$ (and therefore $\iota \in I$) such that $x \in X_\iota$; it follows that

Now suppose that $L \neq \emptyset$ and $J_\lambda \neq \emptyset$ for each $\lambda \in L$. Then $I \neq \emptyset$. Let $x$ be an element of $\bigcap_{\iota \in I} X_\iota$. If $\lambda \in L$, we have $x \in X_\iota$ for each $\iota \in J_\lambda$ (since $J_\lambda \subset I$), whence $x \in \bigcap_{\iota \in J_\lambda} X_\iota$. Since the last inclusion is true for all $\lambda \in L$, $x$ belongs to $\bigcap_{\lambda \in L} \left( \bigcap_{\iota \in J_\lambda} X_\iota \right)$. Conversely, let $x$ be an element of this latter set, and let $\iota$ be any element of I. There exists $\lambda \in L$ such that $\iota \in J_\lambda$; since $x \in \bigcap_{\iota \in J_\lambda} X_\iota$, we have $x \in X_\iota$, which is true for all $\iota \in I$. Hence $x \in \bigcap_{\iota \in I} X_\iota$. This completes the proof.

For families of subsets of a given set the second part of Proposition 2 remains valid without restriction on L and $J_\lambda$.

The relation $(\exists x)\left(x \in \bigcup_{\iota \in I} X_\iota \text{ and } y \in \Gamma(x)\right)$ is equivalent to

hence to $(\exists \iota)(\iota \in I$ and $y \in \Gamma\langle X_\iota \rangle)$, hence is equivalent to $y \in \bigcup_{\iota \in I} \Gamma\langle X_\iota \rangle$; this proves the first formula. As to the second formula, we have $\bigcap_{\iota \in I} X_\iota \subset X_\iota$ for all $\iota \in I$, whence (§3, Proposition 2)

and consequently

¶ If $\Gamma$ is an arbitrary correspondence (and in particular an arbitrary function), the formula

is usually false.

\* For example, in the plane $\mathbf{R}^2$ the first projections of the lines $y = x$ and $y = x + 1$ are equal to $\mathbf{R}$, but the intersection of these lines is empty, and therefore so is the first projection of this intersection [^2]. \*

However, we have the following important result :

To prove this, let $x$ be an element of $\bigcap_{\iota \in I} \overset{-1}{f}\langle Y_\iota \rangle$. We have $f(x) \in Y_\iota$ for all $\iota \in I$, whence $f(x) \in \bigcap_{\iota \in I} Y_\iota$, and consequently

Therefore $\bigcap_{\iota \in I} \overset{-1}{f}(Y_\iota) \subset \overset{-1}{f}\left\langle \bigcap_{\iota \in I} Y_\iota \right\rangle$; this relation, together with Proposition 3, gives the result.

For we may write $f = i \circ g$, where $i$ is the canonical injection of $f\langle A \rangle$ into B and $g$ is a bijection of A onto $f\langle A \rangle$. If $h$ denotes the inverse mapping of $g$, we have $f\langle X \rangle = \overset{-1}{h}\langle X \rangle$ for every subset X of A, and we are therefore brought back to Proposition 4.

Let $x \in \complement_E\left(\bigcup_{\iota \in I} X_\iota\right)$. Then $x \in E$ and, for every $\iota \in I$, $x \notin X_\iota$, so that $x \in \complement_E X_\iota$; consequently

Conversely, let $x$ be an element of $\bigcap_{\iota \in I} (\complement_E X_\iota)$; by the definition of intersection we have $x \in E$. Furthermore, if we had $x \in \bigcup_{\iota \in I} X_\iota$, there would exist an index $\varkappa \in I$ such that $x \in X_\varkappa$, which contradicts the hypothesi $x \in \bigcap_{\iota \in I} (\complement_E X_\iota)$; hence

This proves the first formula; the second one is an immediate consequence, in view of the relation $\complement_E(\complement_E X) = X$ for every subset X of E.

For $x$ belongs to $\overset{-1}{f}\langle \mathrm{B} - \mathrm{Y} \rangle$ if and only if $f(x)$ belongs to B but not to Y, i.e., if and only if $x$ belongs to $\overset{-1}{f}\ \langle \mathrm{B} \rangle$ but not to $\overset{-1}{f}\ \langle \mathrm{Y} \rangle$.

Writing $f = i \circ g$, where $i$ is the canonical injection of $f\langle \mathrm{A} \rangle$ into B, we reduce the Corollary to Proposition 6 applied to $\overset{-1}{g}$.

¶ The intersection $\mathrm{X} \cap \mathrm{A}$ is sometimes called the trace of X on A. If $\mathfrak{F}$ is a family of sets, the set of traces on A of the sets belonging to $\mathfrak{F}$ is called the trace of $\mathfrak{F}$ on A.

(2) Let $(X_\iota)_{\iota \in I}$ be a family of sets and let $(f_\iota)_{\iota \in I}$ be a family of mappings with the same target F such that for each $\iota \in I$ the domain of $f_\iota$ is $X_\iota$, and for each pair $(\iota, \varkappa) \in I \times I$, $f_\iota$ and $f_\varkappa$ agree on $X_\iota \cap X_\varkappa$. Then there is exactly one function $f$ with domain $A = \bigcup_{\iota \in I} X_\iota$ and target F which extends each of the functions $f_\iota$ $(i \in I)$.

(1) Let $x$ be any element of E. Then there exists $\iota \in I$ such that $x \in X_\iota$, whence $f(x) = g(x)$ by hypothesis. (2) Let $G_\iota$ be the graph of $f_\iota$ and let $G = \bigcup_{\iota \in I} G_\iota$; let us show that G is a functional graph. If $(x, y) \in G$ and $(x, y') \in G$, there exist two indices $\iota$, $\varkappa$ in I such that $(x, y) \in G_\iota$ and $(x, y') \in G_\varkappa$. This implies $x \in X_\iota$, $x \in X_\varkappa$, $y = f_\iota(x)$, $y' = f_\varkappa(x)$; but since $x \in X_\iota \cap X_\varkappa$, we have

that is to say, $y = y'$. The graph G has domain $\mathrm{pr}_1 G = \bigcup_{\iota \in I} \mathrm{pr}_1 G_\iota = A$; the function $f = (G, A, F)$ therefore satisfies the required conditions. Its uniqueness follows from the first part of the Proposition.

This is a corollary of Proposition 7, since $f_\iota$ and $f_\varkappa$ clearly agree on $\mathrm{X}_\iota \cap \mathrm{X}_\varkappa = \emptyset$ whenever $\iota \neq \varkappa$.

Let $\mathrm{A} = \bigcup_{\iota \in \mathrm{I}} \mathrm{X}_\iota$. If $\iota \in \mathrm{I}$, the mapping $x \rightarrow (x, \iota)$ $(x \in \mathrm{X}_\iota)$ is a one-to-one mapping of $\mathrm{X}_\iota$ onto a subset $\mathrm{X}'_\iota$ of $\mathrm{A} \times \mathrm{I}$. Moreover, the image of $\mathrm{X}'_\iota$ under the second coordinate function on $\mathrm{A} \times \mathrm{I}$ is contained in the set $\{\iota\}$; it follows that $\mathrm{X}'_\iota \cap \mathrm{X}'_\varkappa = \emptyset$ whenever $\iota \neq \varkappa$. We may then take $\mathrm{X} = \bigcup_{\iota \in \mathrm{I}} \mathrm{X}'_\iota$.

For each $\iota \in \mathrm{I}$, let $f_\iota$ be a bijection of $\mathrm{X}_\iota$ onto $\mathrm{X}_\iota \times \{\iota\}$. By virtue of Proposition 8, there exists a mapping $f$ of A into S which extends all the mappings $f_\iota$. It is immediately verified that $f$ is a bijective mapping of A onto S.

By abuse of language, a set E is said to be the sum of a family of sets $(\mathrm{X}_\iota)_{\iota \in \mathrm{I}}$ if there exists a bijection of E onto the sum of the family, as defined in Definition 8.

Note that if $(\mathrm{X}_\iota)_{\iota \in \mathrm{I}}$ is an arbitrary family of sets, the argument of Proposition 10 shows that there exists a mapping of the sum S onto the union A.