The General Inclusion
For a finite family of sets, taking the closure commutes with union. An arbitrary family can behave differently: the closure of its union may contain points that are in none of the individual closures. The general rule is an inclusion in one direction, and the key question is when that inclusion becomes equality.
Let \((X,\mathcal{T})\) be a topological space. Recall that a set is closed if its complement is open, and that \(\overline{A}\) is the smallest closed subset of \(X\) containing \(A\). Let \((A_i)_{i\in I}\) be any family of subsets of \(X\), indexed by a set \(I\). The index set may be empty.
Proof. Write \(U=\bigcup_{i\in I}A_i\). For each \(i\in I\), \(A_i\subseteq U\), and \(\overline{U}\) is closed and contains \(U\). Thus \(\overline{U}\) is a closed set containing \(A_i\). Since \(\overline{A_i}\) is the smallest closed set containing \(A_i\), \(\overline{A_i}\subseteq\overline{U}\). This holds for every \(i\), so their union is contained in \(\overline{U}\). If \(I\) is empty, both unions in the asserted inclusion are empty, and \(\overline{\varnothing}=\varnothing\) because the empty set is closed. The inclusion holds in this case as well. \(\square\)
The proof uses only the smallest-closed-superset property: the closure of the whole union is a closed set containing each member of the family. It does not assert that the union of the individual closures is itself closed. That distinction matters, because an arbitrary union of closed sets need not be closed.
When Does Equality Hold?
There is a precise test for equality. It does not require a special property of the space: equality holds exactly when the union of the individual closures is closed.
Proof. Each \(A_i\subseteq\overline{A_i}\), so \(U\subseteq S\). The first theorem gives \(S\subseteq\overline{U}\). Since \(\overline{U}\) is closed and contains \(S\), the smallest-closed-superset property gives \(\overline{S}\subseteq\overline{U}\). On the other hand, \(U\subseteq S\subseteq\overline{S}\), and \(\overline{S}\) is closed. It is therefore a closed set containing \(U\), so \(\overline{U}\subseteq\overline{S}\). Hence \(\overline{U}=\overline{S}\). If \(S\) is closed, then \(\overline{S}=S\), giving \(\overline{U}=S\). Conversely, if \(\overline{U}=S\), then \(S\) is closed because \(\overline{U}\) is closed. \(\square\)
For a finite family, \(S\) is a finite union of closed sets and is therefore closed. Indeed, the complement of a finite union of closed sets is a finite intersection of open sets, which is open. Thus the general theorem gives equality for finite families. For an arbitrary family, that argument no longer applies: an infinite union of closed sets can fail to be closed.
Worked Example: Singletons Converging to Zero
In \(\mathbb{R}\) with its standard topology, let \(A_n=\{1/n\}\) for \(n\in\mathbb{N}\). Compute the union of the closures and the closure of the union.
Each singleton is closed in \(\mathbb{R}\). To see this, for any \(y\ne 1/n\), the open interval centered at \(y\) with radius \(|y-1/n|/2\) misses \(\{1/n\}\). Thus every point of its complement has an open neighborhood in the complement. It follows that \(\overline{A_n}=A_n\), and so \(S=\bigcup_{n\in\mathbb{N}}\overline{A_n}=\{1/n:n\in\mathbb{N}\}\).
Now set \(F=\{0\}\cup\{1/n:n\in\mathbb{N}\}\). Every \(1/n\) belongs to \(\overline{\bigcup_n A_n}\) because it belongs to the union itself. Also, every open neighborhood of \(0\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\). Choose \(n\in\mathbb{N}\) with \(n>1/\varepsilon\). Then \(0<1/n<\varepsilon\), so this neighborhood meets the union. The open-neighborhood criterion for closure gives \(0\in\overline{\bigcup_n A_n}\).
To check that there are no other closure points, we show that \(F\) is closed. If \(x<0\), an interval around \(x\) of radius \(-x/2\) misses \(F\). If \(x>0\) and \(x\notin F\), choose \(N\ge 2\) with \(1/N<x/2\). The terms \(1/n\) for \(n\ge N\) are all below \(x/2\). The finitely many terms \(1/n\) for \(1\le n<N\) differ from \(x\). Choose a positive radius smaller than or equal to \(x/2\) and smaller than each of the finitely many numbers \(|x-1/n|/2\) for \(1\le n<N\). The interval with this radius around \(x\) misses both the tail and the finite initial terms, as well as \(0\). Thus every point outside \(F\) has an open neighborhood disjoint from \(F\), so \(F\) is closed. Since \(F\) contains the union, its closure is contained in \(F\).
Combining the two directions gives \(\overline{\bigcup_n A_n}=F\), while \(\bigcup_n\overline{A_n}=\{1/n:n\in\mathbb{N}\}\). The containment is strict: \(0\) belongs to the closure of the union but not to the union of the closures.
Check: \(0\) is the extra point, and every neighborhood of \(0\) contains some \(1/n\). The proposed closure is closed by the neighborhood argument above and contains every \(A_n\), so no points outside it can belong to the closure.
A Second Real-Line Calculation
Worked Example: A Family of Open Intervals
For \(n\ge 2\), let \(A_n=(1/n,1)\) in \(\mathbb{R}\) with the standard topology. Find both sides of the general inclusion.
The union of the intervals is \((0,1)\). Every point in any \(A_n\) lies in \((0,1)\). Conversely, if \(0<x<1\), choose \(n\ge2\) so large that \(1/n<x\); then \(x\in A_n\). Hence \(\bigcup_{n\ge2}A_n=(0,1)\), whose closure is \([0,1]\).
For each \(n\ge2\), the closure of \((1/n,1)\) is \([1/n,1]\): points strictly between the endpoints are in the interval, and every neighborhood of either endpoint meets the interval; points outside the closed interval have a neighborhood missing it. Therefore \(\bigcup_{n\ge2}\overline{A_n}=\bigcup_{n\ge2}[1/n,1]=(0,1]\). Indeed, each such interval lies in \((0,1]\); and for any \(x\in(0,1]\), choose \(n\ge2\) with \(1/n\le x\), which places \(x\) in \([1/n,1]\).
Thus \((0,1]\subsetneq[0,1]\). The endpoint \(0\) is in the closure of the union because intervals in the family contain points arbitrarily close to \(0\), but it lies in none of the individual closures.
Check: The union of the individual closures is missing \(0\), while every interval around \(0\) meets \((0,1)=\bigcup_{n\ge2}A_n\). This confirms both the strict containment and the location of the extra point.
Equality in a Finite Space
Worked Example: Alternating Sets in a Three-Point Space
Let \(X=\{p,q,r\}\) have topology \(\mathcal{T}=\{\varnothing,\{p\},X\}\). This is a topology: arbitrary unions of its open sets and finite intersections of its open sets again belong to the displayed collection. Its closed sets, obtained by taking complements, are \(\varnothing,\{q,r\},X\).
For \(n\in\mathbb{N}\), define \(A_n=\{p\}\) when \(n\) is even and \(A_n=\{q\}\) when \(n\) is odd. The only closed set containing \(p\) is \(X\), so \(\overline{\{p\}}=X\). The smallest closed set containing \(q\) is \(\{q,r\}\), so \(\overline{\{q\}}=\{q,r\}\). Both types of index occur, and consequently \(\bigcup_n\overline{A_n}=X\).
The union of the original sets is \(\{p,q\}\). The only closed set containing \(p\) is \(X\), so \(\overline{\bigcup_n A_n}=X\) as well. Equality holds in this example, despite there being infinitely many indices. What matters is that the union of the individual closures is closed; here it is \(X\).
Check: The family contains only two distinct sets, even though it has infinitely many indices. Their closures have union \(X\), which is closed, agreeing with the equality criterion.
Common Mistake and Practical Use
Common mistake. A tempting but incorrect calculation is to replace the closure of an arbitrary union by the union of the individual closures. For the singleton family above, that would give \(\overline{\bigcup_{n\in\mathbb{N}}\{1/n\}}=\bigcup_{n\in\mathbb{N}}\overline{\{1/n\}}=\{1/n:n\in\mathbb{N}\}\). The right-hand side omits \(0\), even though every neighborhood of \(0\) meets the union. The correction is to use the general inclusion and then check whether the union of the individual closures is closed. In this case it is not closed: every neighborhood of \(0\) meets \(\{1/n:n\in\mathbb{N}\}\), but \(0\) is not in that set.
This distinction is useful whenever a set is assembled from many pieces. Closing each piece captures points approached within that piece; closing the entire union can also capture points approached by points drawn from different pieces. The theorem guarantees that the union of the individual closures never includes a point outside the closure of the whole union, while the equality criterion identifies exactly when it captures every point.
Check Your Understanding
For each question, identify the family, compute the relevant union, and use the closure criterion to justify your conclusion.
- In \(\mathbb{R}\), let \(A_n=\{2+1/n\}\) for \(n\in\mathbb{N}\). Find \(\bigcup_n\overline{A_n}\) and \(\overline{\bigcup_n A_n}\).
- For a family with an empty index set, what are the union and its closure?
- Show directly from the smallest-closed-superset definition that \(\overline{\bigcup_i A_i}\) contains \(\overline{A_j}\) for each fixed index \(j\).
- Give a condition on \(\bigcup_i\overline{A_i}\) that is necessary and sufficient for equality in the general inclusion.
- In the three-point space above, replace every odd-indexed set by \(\varnothing\), leaving every even-indexed set equal to \(\{p\}\). Compute both sides of the inclusion.