From Ambient Open Sets to a Topology on a Subset
A subset of a topological space can be treated as a space in its own right. To do this, we need to specify which subsets of it count as open. The natural rule is to intersect the subset with open sets from the surrounding space. The essential point is that this rule satisfies the topology axioms, even when the subset is not itself open in the ambient space.
Let \((X,\mathcal{T})\) be a topological space, and let \(A\subseteq X\). Recall that a topology on a set \(A\) is a collection \(\mathcal{T}_A\subseteq\mathcal{P}(A)\) that contains \(\varnothing\) and \(A\), is closed under arbitrary unions, and is closed under finite intersections. Here, “arbitrary” includes unions indexed by an empty collection, and “finite” includes the intersection of no sets, which by convention is \(A\).
Each member \(A\cap U\) is a subset of \(A\), so \(\mathcal{T}_A\) is indeed a collection of subsets of the right set. The definition does not require \(A\) to be open or closed in \(X\). It also does not say that every member of \(\mathcal{T}_A\) must be open in \(X\): the members are formed by intersecting \(A\) with ambient open sets.
Why Intersections Preserve the Needed Operations
The topology axioms follow from simple identities for intersection with a fixed set. We establish those identities first so that each step in the axiom check is explicit.
Proof. For the union identity, take any \(x\in X\). The statement \(x\in A\cap\bigcup_{i\in J}U_i\) means that \(x\in A\) and that \(x\in U_i\) for at least one \(i\in J\). This is equivalent to saying that \(x\in A\cap U_i\) for at least one \(i\in J\), which means \(x\in\bigcup_{i\in J}(A\cap U_i)\). Thus the two sets have exactly the same elements. If \(J\) is empty, both sides are empty, so the identity still holds.
For the finite intersection identity, membership in the left side means \(x\in A\) and \(x\in U_k\) for every \(k=1,\dots,n\). Since \(x\in A\) in particular, this is equivalent to \(x\in A\cap U_k\) for every \(k=1,\dots,n\), which is membership in the right side. For \(n=0\), the left side is \(A\cap X=A\), and the right side is the empty intersection in \(\mathcal{P}(A)\), also \(A\). \(\square\)
Proof. We verify all three topology axioms. Since \(\varnothing\in\mathcal{T}\), we have \(A\cap\varnothing=\varnothing\in\mathcal{T}_A\). Since \(X\in\mathcal{T}\), we have \(A\cap X=A\in\mathcal{T}_A\).
Now let \((V_i)_{i\in J}\) be any family of members of \(\mathcal{T}_A\). For each \(i\in J\), the definition of \(\mathcal{T}_A\) gives an open set \(U_i\in\mathcal{T}\) with \(V_i=A\cap U_i\). Because \(\mathcal{T}\) is a topology, \(\bigcup_{i\in J}U_i\in\mathcal{T}\). By the lemma, $$ \bigcup_{i\in J}V_i =\bigcup_{i\in J}(A\cap U_i) =A\cap\bigcup_{i\in J}U_i. $$ The right side belongs to \(\mathcal{T}_A\), so \(\mathcal{T}_A\) is closed under arbitrary unions. If \(J\) is empty, this union is \(\varnothing\), which was already shown to belong to \(\mathcal{T}_A\).
Finally, take \(V_1,\dots,V_n\in\mathcal{T}_A\), and write \(V_k=A\cap U_k\) with \(U_k\in\mathcal{T}\). Since a topology is closed under finite intersections, \(\bigcap_{k=1}^n U_k\in\mathcal{T}\). The lemma gives $$ \bigcap_{k=1}^n V_k =A\cap\bigcap_{k=1}^n U_k, $$ which belongs to \(\mathcal{T}_A\). For \(n=0\), the intersection is \(A\), already shown to be in \(\mathcal{T}_A\). Thus all topology axioms hold. \(\square\)
Worked Examples
Worked Example: Intersections in the Real Line
Give \(\mathbb{R}\) its standard topology and let \(A=[0,1]\). For ambient open sets \(U=(-1,1/2)\) and \(V=(1/3,2)\), compute the corresponding members of the induced topology and their intersection.
First, \(A\cap U\) consists of the real numbers that satisfy both \(0\leq x\leq1\) and \(-1<x<1/2\). Combining these inequalities gives \(0\leq x<1/2\), so $$ A\cap U=[0,1/2). $$ Next, \(A\cap V\) consists of the real numbers satisfying both \(0\leq x\leq1\) and \(1/3<x<2\). Combining the inequalities gives \(1/3<x\leq1\), so $$ A\cap V=(1/3,1]. $$ Their intersection consists of the numbers satisfying both \(0\leq x<1/2\) and \(1/3<x\leq1\). The combined inequalities are \(1/3<x<1/2\), hence $$ (A\cap U)\cap(A\cap V)=(1/3,1/2). $$ As a check, \(U\cap V=(1/3,1/2)\), which is contained in \(A=[0,1]\); therefore \(A\cap(U\cap V)=(1/3,1/2)\), agreeing with the result.
Worked Example: A Subset of a Finite Topological Space
Let \(X=\{a,b,c\}\) and \(\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}\). First, this is a topology: it contains \(\varnothing\) and \(X\); unions of its listed sets are again among the four listed sets; and intersections of any two listed sets are again among them. For example, \(\{a\}\cup\{a,b\}=\{a,b\}\) and \(\{a\}\cap\{a,b\}=\{a\}\). Since there are only four sets, these checks cover every possible union and intersection of listed sets.
Take \(A=\{b,c\}\). Intersect each ambient open set with \(A\): $$ A\cap\varnothing=\varnothing,\qquad A\cap\{a\}=\varnothing,\qquad A\cap\{a,b\}=\{b\},\qquad A\cap X=\{b,c\}. $$ Thus the induced topology is \(\mathcal{T}_A=\{\varnothing,\{b\},\{b,c\}\}\). To check its axioms directly, it contains \(\varnothing\) and \(A=\{b,c\}\). A union of members is either \(\varnothing\), \(\{b\}\), or \(\{b,c\}\), so all unions remain in the collection. An intersection of members is also one of those three sets, so finite intersections remain in the collection. The general theorem guarantees these properties, and the calculation confirms the result in this example.
Worked Example: An Indiscrete Ambient Topology
Let \(X=\{p,q,r\}\) with the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\), and take \(A=\{p,q\}\). The two possible ambient open sets give $$ A\cap\varnothing=\varnothing,\qquad A\cap X=A=\{p,q\}. $$ There are no other ambient open sets to intersect with \(A\), so the induced topology is exactly \(\mathcal{T}_A=\{\varnothing,\{p,q\}\}\). In particular, \(\{p\}\) is not in \(\mathcal{T}_A\): it is neither \(A\cap\varnothing=\varnothing\) nor \(A\cap X=A\). The collection \(\{\varnothing,A\}\) satisfies the axioms because any union or finite intersection of its members is again \(\varnothing\) or \(A\). This calculation also checks that the induced topology need not contain every subset of \(A\).
Common Pitfalls and the Role of the Axioms
Common mistake. A frequent error is to reject \(A\cap U\) because it is not open in the ambient space. In the first example, \(A=[0,1]\) and \(U=(-1,1/2)\) give \(A\cap U=[0,1/2)\). This set is not open in \(\mathbb{R}\): every open interval around \(0\) contains negative numbers, which are not in \([0,1/2)\). But the definition of \(\mathcal{T}_A\) asks whether the set is an intersection \(A\cap U\) for an ambient open \(U\), not whether the resulting set is itself ambient open. Here it is exactly such an intersection, so it belongs to the induced topology.
A second point to keep separate is arbitrary unions versus finite intersections. The axioms require closure under unions of any size, but only under intersections of finitely many sets. For instance, in the standard topology on \(\mathbb{R}\), each set \(U_n=(-1/n,1/n)\), for \(n\in\mathbb{N}\), is open. Their intersection is \(\{0\}\): \(0\) belongs to every \(U_n\), while if \(x\ne0\), choose \(n\in\mathbb{N}\) with \(n>1/|x|\); then \(1/n<|x|\), so \(x\notin U_n\). The singleton \(\{0\}\) is not open in \(\mathbb{R}\), because every open interval containing \(0\) also contains points other than \(0\). Thus arbitrary intersections are not required to remain open, even in the ambient topology itself.
The induced-topology construction is useful whenever a space is studied through a distinguished subset: the subset receives a topology using only the open sets already available in the surrounding space. The proof depends on no special features of \(\mathbb{R}\) or of finite spaces. It works for every topological space and every subset, including \(A=\varnothing\) and \(A=X\). If \(A=\varnothing\), the formula gives \(\mathcal{T}_A=\{\varnothing\}\), the topology on the empty set. If \(A=X\), it gives \(\mathcal{T}_A=\mathcal{T}\), since \(X\cap U=U\) for every \(U\in\mathcal{T}\).
Check Your Understanding
Use the definition by intersections and the topology axioms to justify your answers.
- If \(A\subseteq X\), which ambient open sets show that \(\varnothing\) and \(A\) belong to the induced topology?
- Let \(X=\{u,v,w\}\), \(\mathcal{T}=\{\varnothing,\{u\},X\}\), and \(A=\{u,w\}\). Compute every member of the induced topology.
- Explain why the proof of closure under arbitrary unions includes the case of a union indexed by the empty set.
- For \(A=[0,1]\subseteq\mathbb{R}\), compute \(A\cap(-1/2,3/4)\) and \(A\cap(1/4,2)\), then compute their intersection.
- Why does the topology axiom require finite intersections rather than arbitrary intersections? Give the real-line example used above in your explanation.