Putting Spaces Side by Side
A disjoint union collects several topological spaces into one space while keeping their points in separate, labelled copies. Its defining feature is that openness is tested separately in each copy. This makes it possible, for example, for an infinite union of relatively open sets—one in each summand—to be open without requiring any uniform condition across the summands.
Let \(J\) be an index set, and for each \(j\in J\) let \((X_j,\mathcal{T}_j)\) be a topological space. We use tagged points so that copies remain distinct even when the original spaces have points in common. For \(U\subseteq X=\bigsqcup_{j\in J}X_j\), its slice in the \(j\)-th summand is the set \(U_j=\{x\in X_j:(j,x)\in U\}\). Each \(U_j\) is viewed as a subset of \(X_j\).
When \(J\) is empty, the disjoint union is the empty set, and its topology is \(\{\varnothing\}\). For a nonempty index set, \(X\) consists of all tagged points \((j,x)\) with \(j\in J\) and \(x\in X_j\). The tags matter: points \((j,x)\) and \((k,x)\) are different whenever \(j\ne k\), even if the untagged point \(x\) occurs in both spaces.
Why This Really Is a Topology
Proof. For every \(j\), the slice of \(\varnothing\) is \(\varnothing\), which belongs to \(\mathcal{T}_j\). The slice of \(X\) is \(X_j\), which also belongs to \(\mathcal{T}_j\). Thus \(\varnothing\) and \(X\) are open in the proposed collection.
Now let \(\{U^\alpha:\alpha\in K\}\) be any family of sets in the collection. For each \(j\), the slice of their union is \((\bigcup_{\alpha\in K}U^\alpha)_j=\bigcup_{\alpha\in K}U^\alpha_j\). Each \(U^\alpha_j\) is open in \(X_j\), so this union is open in \(X_j\), including when \(K\) is empty. Therefore \(\bigcup_{\alpha\in K}U^\alpha\) belongs to the collection.
Finally, for two sets \(U,V\) in the collection, \((U\cap V)_j=U_j\cap V_j\). Finite intersections of open sets are open in \(X_j\), so \(U\cap V\) belongs to the collection. These are the topology axioms, and they also cover the empty disjoint union. \(\square\)
The same slice test describes closed sets, by taking complements.
Proof. By definition, \(F\) is closed if and only if \(X\setminus F\) is open. The \(j\)-th slice of \(X\setminus F\) is \(X_j\setminus F_j\). The open-set definition says that \(X\setminus F\) is open if and only if \(X_j\setminus F_j\) is open in \(X_j\) for every \(j\). This is equivalent to \(F_j\) being closed in \(X_j\) for every \(j\), as required. \(\square\)
These tests apply to any index set, finite or infinite. In particular, they do not require a separate condition that relates the slices to one another. The choice of an open slice in one summand places no restriction on the slice in a different summand.
Each Summand Is Open and Closed
For each \(j\), the \(j\)-th summand inside \(X\) is the tagged subset \(X_j^\ast=\{(j,x):x\in X_j\}\). Its slice in summand \(j\) is all of \(X_j\); its slice in every other summand is empty. Both are open, so \(X_j^\ast\) is open in \(X\). Its complement also has only full or empty slices, so the complement is open and \(X_j^\ast\) is closed as well.
More generally, the topology induced on \(X_j^\ast\) by \(X\) is exactly the original topology on \(X_j\). Indeed, if \(U\) is open in \(X\), its slice in \(X_j\) is open in \(X_j\). Conversely, given \(V\in\mathcal{T}_j\), the subset of \(X\) whose \(j\)-th slice is \(V\) and whose other slices are empty is open in \(X\). Thus no open sets inside a summand are lost or added by placing it in the disjoint union.
Worked Example: Two Copies of a Closed Interval
Let \(X_1=X_2=[0,2]\), each with its subspace topology from \(\mathbb{R}\), and form \(X=X_1\bigsqcup X_2\). Consider the set \(A\) whose first slice is \([0,1)\) and whose second slice is \((1,2]\). Determine whether \(A\) is open or closed.
In the first copy, \([0,1)=(-1,1)\cap[0,2]\), so \([0,1)\) is open in the subspace topology on \([0,2]\). In the second copy, \((1,2]=(1,3)\cap[0,2]\), so \((1,2]\) is also open in \([0,2]\). Both slices of \(A\) are open, and the open-set test shows that \(A\) is open in \(X\).
To test closedness, take complements within each copy. The first slice of \(X\setminus A\) is \([1,2]\), and the second is \([0,1]\). The first slice is not relatively open at \(1\): every relative open neighborhood of \(1\) in \([0,2]\) contains points less than \(1\), which are not in \([1,2]\). The second slice is not relatively open at \(1\), since every relative open neighborhood of \(1\) contains points greater than \(1\), which are not in \([0,1]\). Thus the complement is not open, and \(A\) is not closed.
Check: The slice calculations give \(X_1\setminus[0,1)=[1,2]\) and \(X_2\setminus(1,2]=[0,1]\), including the endpoint \(1\) in both complements. This confirms that the failure of closedness follows from the closed-set test.
Infinite Families of Summands
For an infinite disjoint union, the same definition and tests continue to hold. In particular, a set with an open slice in every summand is open, even if infinitely many slices are nonempty. A set with a closed slice in every summand is closed under the same unrestricted quantifier.
Worked Example: Infinitely Many Interval Copies
For each \(n\in\mathbb{N}\), let \(X_n=[0,1]\) with its usual subspace topology, and form \(X=\bigsqcup_{n\in\mathbb{N}}X_n\). Let \(A\) have slice \(A_n=[0,1/n)\) in the \(n\)-th summand. Decide whether \(A\) is open or closed.
For every \(n\ge 1\), \(A_n=(-1,1/n)\cap[0,1]\), so \(A_n\) is open in \(X_n\). This also holds for \(n=1\), when \(A_1=[0,1)\). Therefore \(A\) is open in the disjoint union, because every slice is open.
The complement has slice \(X_n\setminus A_n=[1/n,1]\). This slice is not open in \(X_n\): at \(1/n\), every relative open neighborhood contains points of \([0,1]\) smaller than \(1/n\), which are not in \([1/n,1]\). This reasoning applies for every \(n\ge1\), including \(n=1\), where the slice is \(\{1\}\). Hence \(X\setminus A\) is not open and \(A\) is not closed.
Now define \(B\) to have slice \(B_n=\{1/n\}\) in each summand. A singleton is closed in \([0,1]\), since its complement is relatively open: for each point different from \(1/n\), a sufficiently small interval around that point avoids \(1/n\). Thus every \(B_n\) is closed, and \(B\) is closed in \(X\). Each \(B_n\) is not open, because every relative neighborhood of \(1/n\) contains other points of \([0,1]\). Consequently \(B\) is not open.
Check: The \(n\)-th slice of \(X\setminus B\) is \([0,1]\setminus\{1/n\}\), which is open in \([0,1]\); this verifies that \(X\setminus B\) is open. The use of the index \(n\) is important: \(1/n\) is substituted separately in the slice for each \(n\), and each such singleton is closed in its own summand.
There is no conflict between this example and the fact that an infinite union of closed sets need not be closed in an arbitrary space. In the standard topology on \(\mathbb{R}\), each singleton \(\{1/n\}\) is closed, but \(\bigcup_{n\in\mathbb{N}}\{1/n\}\) is not closed: every open interval around \(0\) contains \(1/n\) for some sufficiently large \(n\), while \(0\) is not in the union. The set \(B\) above is different: its points lie in separate tagged summands, and its closedness follows from checking each slice in its own space.
Worked Example: A Real Line and a Singleton
Let \(X_1=\mathbb{R}\) with the standard topology and \(X_2=\{p\}\) with its unique topology \(\{\varnothing,\{p\}\}\). In their disjoint union, let \(C\) have first slice \((-\infty,0]\) and second slice \(\{p\}\). Determine whether \(C\) is open or closed.
The first slice is closed in \(\mathbb{R}\), because its complement \((0,\infty)\) is open. The second slice is the whole one-point space and is closed there. The closed-set test gives that \(C\) is closed in \(X\).
The first slice is not open in \(\mathbb{R}\): every open interval around \(0\) contains positive real numbers, which do not belong to \((-\infty,0]\). Therefore the open-set test shows that \(C\) is not open in \(X\). The singleton slice causes no obstruction to either conclusion: it is both open and closed in its one-point space.
Check: The complement of \(C\) has first slice \((0,\infty)\) and second slice \(\varnothing\). Both are open in their respective spaces, so the complement is open, confirming that \(C\) is closed. The failure of openness is specifically the failure of the first slice to be open.
A Common Mistake
Common pitfall: Do not decide whether a subset is open or closed by imagining that the summands have been merged into one ordinary space. For example, in the real line, the union of the singleton sets \(\{1/n\}\) is not closed because it omits the limit point \(0\). In a disjoint union of interval copies, the set with slice \(\{1/n\}\) in the \(n\)-th copy is closed, as the example above shows. These are different spaces and different sets: in the disjoint union, points carry tags, and closedness is tested separately in every summand.
The practical rule is to write down all the slices before deciding. For openness, verify that every slice is open in its own summand. For closedness, verify that every slice is closed there. A set may be open but not closed, closed but not open, both, or neither; the disjoint union definition does not force these two properties to coincide.
Check Your Understanding
Use the slice tests, and specify the topology on each summand in your reasoning.
- In \([0,1]\bigsqcup[0,1]\), take slices \([0,1/2]\) and \((1/2,1]\). Is the resulting set open, closed, both, or neither?
- Suppose every slice of \(D\subseteq\bigsqcup_{j\in J}X_j\) is closed. What does the closed-set test imply about \(D\), even if \(J\) is infinite?
- In a disjoint union of copies of \(\mathbb{R}\), take the slice \((-\infty,0)\) in one copy and \([0,\infty)\) in another. Is the resulting set open? Is it closed?
- Explain why each summand is both open and closed in the disjoint union, using its slices and the slices of its complement.
- Give an example of a closed subset of the disjoint union of infinitely many copies of \([0,1]\) that is not open, and verify both claims from the slice tests.