Closure of Finite Unions

Why Finite Unions Are Especially Manageable

When a set is assembled from several pieces, it is often easier to find the closure of each piece and then combine the results. For finitely many pieces, this procedure always works: the closure of the union equals the union of the individual closures. The key reason is that a finite union of closed sets is closed. We will prove the identity and use it to calculate closures in $\mathbb{R}$ with its standard topology.

Recall that a set is closed if its complement is open. The closure $\overline{A}$ of a subset $A\subseteq X$ is the smallest closed subset of $X$ containing $A$. Equivalently, it is the intersection of all closed sets containing $A$. We will also use the open-neighborhood criterion: $x\in\overline{A}$ if and only if every open neighborhood of $x$ intersects $A$.

Definition: A finite union of sets $A_1,\ldots,A_n$ is $\bigcup_{i=1}^{n}A_i$. When $n=0$, the union is defined to be $\varnothing$. The corresponding union of closures is also $\varnothing$ when $n=0$.

The Finite-Union Identity

Theorem: If $A_1,\ldots,A_n$ are subsets of a topological space $X$, where $n$ is a nonnegative integer, then $$ \overline{\bigcup_{i=1}^{n}A_i}=\bigcup_{i=1}^{n}\overline{A_i}. $$ For $n=0$, both sides are $\varnothing$.

Proof. First suppose $n\geq 1$, and set $C=\bigcup_{i=1}^{n}\overline{A_i}$. Each $\overline{A_i}$ is closed. A finite union of closed sets is closed: for two closed sets $F,G$, the complement of $F\cup G$ is $(X\setminus F)\cap(X\setminus G)$, which is open because it is the intersection of two open sets. Repeating this argument proves that any finite union of closed sets is closed. Thus $C$ is closed. It contains every $A_i$, so it contains $\bigcup_{i=1}^{n}A_i$. By the smallest-closed-superset property of closure, $\overline{\bigcup_{i=1}^{n}A_i}\subseteq C$.

For the reverse inclusion, $\overline{\bigcup_{i=1}^{n}A_i}$ is closed and contains $\bigcup_{i=1}^{n}A_i$. Therefore it is a closed set containing each $A_i$. Since $\overline{A_i}$ is the smallest closed set containing $A_i$, we have $\overline{A_i}\subseteq\overline{\bigcup_{i=1}^{n}A_i}$ for every $i$. Taking the union over $i$ gives $C\subseteq\overline{\bigcup_{i=1}^{n}A_i}$. The two inclusions prove the identity when $n\geq1$. When $n=0$, the union is empty, and $\overline{\varnothing}=\varnothing$ because $\varnothing$ is itself closed. Hence the identity also holds in this case. $\square$

The proof gives a useful way to remember the two inclusions. The union of the individual closures is a closed set containing all the pieces, so it must contain the closure of their union. In the other direction, the closure of the whole union is a closed set containing each piece, so it must contain each individual closure.

For two sets, the result reads $\overline{A\cup B}=\overline{A}\cup\overline{B}$. Applying this two-set identity repeatedly gives the finite-family version. The finiteness is relevant to the proof: the step that a union of closed sets is closed applies to finite unions. The examples below use the identity only for finitely many pieces.

Worked Calculations in the Real Line

Worked Example: Two Intervals Missing a Shared Endpoint

In $\mathbb{R}$ with the standard topology, let $A=(0,1)$ and $B=(1,2)$. Calculate $\overline{A\cup B}$ using the finite-union identity.

First, $\overline{A}=[0,1]$. Every point $x\in(0,1)$ already belongs to $A$, so it belongs to $\overline{A}$. Every open neighborhood of $0$ contains some point of $(0,1)$, and every open neighborhood of $1$ contains some point of $(0,1)$, so both endpoints belong to $\overline{A}$. If $x<0$, the open neighborhood $(x-1,0)$ misses $A$; if $x>1$, the open neighborhood $(1,x+1)$ misses $A$. Thus no point outside $[0,1]$ belongs to $\overline{A}$.

Likewise, $\overline{B}=[1,2]$. Points in $(1,2)$ belong to $B$. Every open neighborhood of $1$ contains points greater than $1$ and less than $2$, and every open neighborhood of $2$ contains points less than $2$ and greater than $1$, so the endpoints belong to $\overline{B}$. Points below $1$ or above $2$ have open neighborhoods missing $B$.

Now substitute these closures into the theorem: $$ \overline{A\cup B}=\overline{A}\cup\overline{B}=[0,1]\cup[1,2]=[0,2]. $$ The last equality holds because the two closed intervals meet at $1$ and together contain every real number from $0$ through $2$, including both endpoints.

Check: $A\cup B$ contains every point of $(0,2)$ except $1$. Each neighborhood of $1$ contains points of $A\cup B$, and neighborhoods of $0$ and $2$ contain points of $A\cup B$ on the inside of $[0,2]$. Points outside $[0,2]$ have neighborhoods disjoint from $A\cup B$. This confirms $\overline{A\cup B}=[0,2]$.

Worked Example: An Interval Together with a Finite Set

Let $A=[-1,1)$ and $B=\{2,4\}$ in $\mathbb{R}$ with the standard topology. Find the closure of their union.

The closure of $A$ is $[-1,1]$. The points of $[-1,1)$ are in $A$, and every open neighborhood of $1$ meets $A$ in points less than $1$. No point $x<-1$ belongs to $\overline{A}$, since $(x-1,-1)$ is a neighborhood of $x$ disjoint from $A$; no point $x>1$ belongs to $\overline{A}$, since $(1,x+1)$ is a neighborhood of $x$ disjoint from $A$.

The set $B$ is closed. To check this directly, take any $x\in\mathbb{R}\setminus\{2,4\}$. Both distances $|x-2|$ and $|x-4|$ are positive. The interval centered at $x$ with radius $r=\frac12\min(|x-2|,|x-4|)$ contains neither $2$ nor $4$. Thus the complement of $B$ is open, so $\overline{B}=B=\{2,4\}$.

The finite-union identity now gives $$ \overline{A\cup B}=\overline{A}\cup\overline{B}=[-1,1]\cup\{2,4\}. $$ The points $2$ and $4$ are included because they are in $B$, not because they are limit points of $A$. For example, the open interval $(1.5,2.5)$ meets $A\cup B$ only at $2$, and $(3.5,4.5)$ meets it only at $4$.

Check: The displayed set is closed: its complement is $(-\infty,-1)\cup(1,2)\cup(2,4)\cup(4,\infty)$, a union of open intervals. It contains $A\cup B$, and the endpoint $1$ is forced into every closed superset of $A$ because every neighborhood of $1$ meets $A$. The isolated points $2$ and $4$ are already in the union. This agrees with the calculated closure.

Worked Example: Positive and Negative Reciprocal Sequences

In $\mathbb{R}$ with the standard topology, let $E=\{1/n:n\in\mathbb{N}\}$ and $F=\{-1/n:n\in\mathbb{N}\}$. Find the closure of $E\cup F$.

We first determine $\overline{E}$. Each point $1/n$ belongs to $E$, so it belongs to $\overline{E}$. Every open neighborhood $(-\varepsilon,\varepsilon)$ of $0$, where $\varepsilon>0$, contains some $1/n$: choose $n\in\mathbb{N}$ with $n>1/\varepsilon$, which gives $0<1/n<\varepsilon$. Thus $0\in\overline{E}$.

Now let $x\ne0$ and $x\notin E$. If $x<0$, the neighborhood $(x-|x|/2,x+|x|/2)$ lies below $0$, so it misses $E$. If $x>0$, choose $N\in\mathbb{N}$ such that $1/N<x/2$. For $n\geq N$, $1/n\leq1/N<x/2$, so these terms are at distance greater than $x/2$ from $x$. The finitely many terms $1/n$ with $1\leq n<N$ are all different from $x$. Choose a positive radius smaller than $x/2$ and smaller than every distance $|x-1/n|$ for $1\leq n<N$. (If there are no such indices, only the bound $x/2$ is needed.) The resulting neighborhood of $x$ misses all of $E$. Therefore $x\notin\overline{E}$, and $\overline{E}=E\cup\{0\}$.

Reflection across $0$, which sends each $1/n$ to $-1/n$, gives $\overline{F}=F\cup\{0\}$; the same neighborhood argument applies with signs reversed. Therefore $$ \overline{E\cup F}=\overline{E}\cup\overline{F} =(E\cup\{0\})\cup(F\cup\{0\})=E\cup F\cup\{0\}. $$ The only new point in the closure is $0$, approached by terms from both sets.

Check: Every neighborhood of $0$ contains $1/n\in E$ and $-1/n\in F$ for all sufficiently large $n$. Each point of $E\cup F$ is in the union itself. For any other $x$, the preceding neighborhood argument gives a neighborhood missing $E$ when $x>0$ or $x<0$, using the corresponding reciprocal sequence; if $x=0$, it is already included. Hence the computed set contains exactly the points of the closure.

A Common Endpoint Error

Common pitfall. For $A=(0,1)$ and $B=(1,2)$, one might incorrectly write $\overline{A\cup B}=(0,1)\cup(1,2)$, reasoning that the union itself has a gap at $1$. This leaves out points that every neighborhood of the gap point approaches from the two intervals.

The correction is to take the closures of the pieces before combining them: $\overline{A}=[0,1]$ and $\overline{B}=[1,2]$. Their union is $[0,2]$, so the theorem gives $\overline{A\cup B}=[0,2]$. In particular, omitting $1$ from $A\cup B$ does not exclude $1$ from its closure. The neighborhood test verifies this directly: every open neighborhood of $1$ contains points of $A\cup B$.

The finite-union identity is useful whenever a set is presented as a finite collection of pieces whose closures are easier to calculate separately. It preserves the exact endpoint information: an endpoint included in a piece is already in that piece, while an omitted endpoint can still enter its closure if every neighborhood meets the piece.

Check Your Understanding

Use the finite-union identity and, where appropriate, check the result with open neighborhoods.

In $\mathbb{R}$ with the standard topology, calculate the closure of $(-2,0]\cup(0,3)$.
Find the closure of $(1,2)\cup\{5\}$ in $\mathbb{R}$.
Explain why the union of finitely many closed sets is closed, and identify where this fact is used in the finite-union theorem.
For $A=\{1/n:n\in\mathbb{N}\}$ and $B=\{-1/n:n\in\mathbb{N}\}$, which point outside $A\cup B$ belongs to its closure?
For the empty family of subsets of $X$, verify both sides of the finite-union identity.

What You'll Learn