Arbitrary Unions and Intersections

In topology, a topology on a set $X$ is by definition a collection $\tau$ of subsets closed under arbitrary unions and finite intersections — the asymmetry between "arbitrary" and "finite" here is exactly the content of these generalized laws, since infinite intersections of open sets can fail to be open (e.g. $\bigcap_{n} (-1/n, 1/n) = \{0\}$, not open in $\mathbb{R}$), which is precisely why the topology axioms restrict intersection to the finite case while allowing union to range over an arbitrary index set. In measure theory, $\sigma$-algebras are closed under countable unions and intersections (a restriction of "arbitrary" to countably-indexed $I$), and the generalized De Morgan laws are the tool used to pass between countable unions and countable intersections when verifying that a collection of sets is a $\sigma$-algebra. In computer science and logic, arbitrary unions and intersections model quantification over infinite state spaces, e.g. the union of all reachable states across an infinite family of runs of a program, or in database theory, the semantics of unbounded existential and universal queries over infinite domains, translating directly through the same $\exists/\forall$ correspondence proved above. In algebra, the intersection of an arbitrary (possibly infinite) family of subgroups of a group, or ideals of a ring, is again a subgroup or ideal — a fact whose proof is a direct instance of the generalized intersection definition applied to the subgroup/ideal axioms, without any finiteness assumption on the index set. Finally, these operations set the stage for two deeper foundational topics: the ZFC Axioms Overview, where the Axiom of Union is precisely the axiom asserting $\bigcup \mathcal{F}$ exists as a set for any set $\mathcal{F}$ of sets, and the Axiom of Choice, which is needed once one wants to choose an element from each set in an arbitrary indexed family $(A_i)_{i\in I}$ — a task visibly different from merely forming $\bigcup_{i\in I}A_i$ or $\bigcap_{i\in I}A_i$, since forming the union/intersection requires no choices at all, only the logical translation of $\exists$/$\forall$ given here.

Ask the tutor: How does the Axiom of Choice let us select one element from each set in an arbitrary indexed family, beyond just forming its union or intersection?

Picture the index set $I$ laid out as a set of "addresses," and at each address $i$ sits a set $A_i$ drawn as a region. The union $\bigcup_{i \in I} A_i$ is the shaded silhouette you'd see if you overlaid every region and painted anything touched by at least one of them. The intersection $\bigcap_{i \in I} A_i$ is the much smaller (often empty) core that survives only where every single region overlaps. The key conceptual jump from the finite case is that $I$ need not be listable. When $I = \{1, \dots, n\}$, $\bigcup_{i\in I} A_i$ is just $A_1 \cup \cdots \cup A_n$, computed by scanning a finite list. When $I = \mathbb{R}$ or $I$ is a proper class-sized index in disguise, there is no scanning: membership in the union is an existential fact ("some region catches $x$"), not a finite algorithmic check. This is why the rigorous definition routes through the logical quantifiers $\exists$ and $\forall$ rather than through repeated binary operations — the binary operation $\cup$ literally cannot be iterated infinitely many times without first defining what the infinite union means. The picture below shows a family indexed by $I = (0,1]$, with $A_t = [0, t]$ for each $t \in I$ — a nested "fan" of intervals shrinking toward $0$. The union recovers $(0,1]$'s worth of coverage stacked onto $[0,1]$, in fact $\bigcup_{t \in (0,1]} A_t = [0,1]$, while the intersection collapses to $\{0\}$, since $0$ is the only point in every $A_t$.

:::svg[The nested family $A_t = [0,t]$ for $t \in (0,1]$: union fills $[0,1]$, intersection is $\{0\}$]

Example 1 (an uncountable family indexed by $\mathbb{R}$). For each $r \in \mathbb{R}$, let

the closed disk of radius $1$ centered at $(r, 0)$. Take $I = \mathbb{R}$, so this is an uncountable family of sets in the plane, all subsets of $U = \mathbb{R}^2$. Computing $\bigcup_{r\in\mathbb{R}} A_r$. A point $(x,y)$ lies in the union iff there exists $r \in \mathbb{R}$ with $(x-r)^2 + y^2 \le 1$, i.e. iff $(x-r)^2 \le 1 - y^2$ for some real $r$. This is solvable for some real $r$ exactly when $1 - y^2 \ge 0$, i.e. $|y| \le 1$ (choose $r = x$ when $|y|\le 1$, giving $(x-r)^2 = 0 \le 1-y^2$; conversely if $|y|>1$ then $1-y^2<0$ so no real $r$ can make $(x-r)^2 \le 1-y^2$ since the left side is $\ge 0$). Hence

an infinite horizontal strip — every disk sweeps across the $x$-axis, and their union is exactly the strip of height $2$ they trace out. Computing $\bigcap_{r \in \mathbb{R}} A_r$. A point $(x,y)$ must lie in every disk $A_r$, for all $r \in \mathbb{R}$ simultaneously — including $r$ arbitrarily far from $x$. Fix $(x,y)$ and take $r = x + 2$ (a legitimate, if distant, index). Then $(x - r)^2 + y^2 = 4 + y^2 \ge 4 > 1$, so $(x,y) \notin A_{x+2}$. Since this holds for every choice of $(x,y) \in \mathbb{R}^2$ (the same argument works verbatim for any $x$), no point lies in all the disks, so

This example shows the union/intersection asymmetry sharply: an uncountable union of bounded sets can still be an unbounded, "thick" region, while the corresponding intersection can collapse to nothing — because a single distant index is enough to exclude every point from the intersection. Checking generalized De Morgan on this example, restricted to $U = \mathbb{R}\times[-1,1]$ for concreteness. Inside $U$, $\left(\bigcup_r A_r\right)^c = U \setminus U = \emptyset$. On the other side, $\bigcap_r A_r^{\,c}$ (complements taken in $U$) requires, for a point $(x,y)\in U$, that $(x,y) \notin A_r$ for every $r$; but we just showed $(x,y) \in A_x$ always (taking $r=x$ gives distance $0$), so $(x,y) \in A_x$, hence $(x,y) \notin A_x^c$, hence $(x,y) \notin \bigcap_r A_r^c$. So $\bigcap_r A_r^c = \emptyset$ too, matching $\left(\bigcup_r A_r\right)^c = \emptyset$ exactly as the theorem guarantees.

Example 2 (a family indexed by $\mathbb{N}$, verifying a distributive law numerically). Let $I = \mathbb{N} = \{1,2,3,\dots\}$ and define, for $n \in I$,

an open (in $\mathbb{Q}$) rational interval shrinking down toward $[0,3]$ from outside. Let $B = \{q \in \mathbb{Q} : 1 \le q \le 5\}$, a fixed closed rational interval, and take $U = \mathbb{Q}$. Computing $\bigcap_{n\in\mathbb{N}} A_n$. A rational $q$ lies in every $A_n$ iff for all $n$, $-1/n < q < 3+1/n$. As $n \to \infty$, the bounds $-1/n \to 0^-$ and $3 + 1/n \to 3^+$, so the constraint "for all $n$" pins $q$ into the closed interval $[0,3]$: indeed if $0 \le q \le 3$ then $-1/n < 0 \le q \le 3 < 3+1/n$ for every $n\ge 1$, so $q \in A_n$ for all $n$; conversely if $q < 0$, pick $n$ large enough that $-1/n > q$ (possible since $-1/n \to 0$ and $q<0$), giving $q \notin A_n$, and symmetrically if $q>3$. Hence

Computing $B \cup \bigcap_n A_n$ directly. This is $\{q : 0 \le q \le 3\} \cup \{q : 1 \le q \le 5\} = \{q \in \mathbb{Q} : 0 \le q \le 5\}$, since $[0,3]$ and $[1,5]$ overlap on $[1,3]$ and together cover $[0,5]$ with no gap. Computing $\bigcap_n (B \cup A_n)$ and confirming it matches. For each fixed $n$, $B \cup A_n = [1,5] \cup (-\tfrac1n, 3+\tfrac1n) = (-\tfrac1n, 5]$ (since $-1/n < 1$ trivially for $n\ge1$ and the two rational intervals overlap on $[1, 3+1/n)$, merging into one interval from $-1/n$ up to $5$). So

by the same shrinking-bound argument as above (the lower bound $-1/n \to 0^-$ forces the intersection's lower edge to be the closed value $0$, while the upper bound $5$ is already common to every term). This matches $B \cup \bigcap_n A_n = [0,5]$ exactly, confirming the generalized distributive law (2) — $B \cup \bigcap_{i} A_i = \bigcap_i (B\cup A_i)$ — on a genuinely infinite, non-symmetric family where the bound on each side is a limit, not a finite computation.

Cantor's insight was that "infinite" is not a single undifferentiated notion, and that sets indexed by infinite — even uncountably infinite — collections behave lawfully, obeying the same logical translation of $\exists$/$\forall$ into $\cup$/$\cap$ that governs the finite case. This is the conceptual foundation on which the generalized distributive and De Morgan laws rest: they hold with literally the same proof regardless of whether $|I|$ is $2$, $17$, $\aleph_0$, or $2^{\aleph_0}$, because the proof only ever uses the logical form of $\exists i \in I$ and $\forall i \in I$, never a listing or count of the elements of $I$.