Taking a Subspace in One Step or Two
A subset of a topological space can be given the topology it inherits from the whole space. If that subset lies inside another subspace, it can also inherit a topology in two stages: first from the intermediate space, then from that topology. The two descriptions should agree, but the point is to verify this by working with the sets that make up each topology. This transitivity lets us choose whichever ambient space makes a relative-topology calculation more convenient.
Throughout, when \(A\subseteq B\subseteq X\), assume that \(X\) has topology \(\mathcal{T}\). The topology on \(B\) inherited from \(X\) is denoted by \(\mathcal{T}_B\). The topology on \(A\) inherited directly from \(X\) is denoted by \(\mathcal{T}_A^{X}\), and the topology on \(A\) inherited from \(B\) is denoted by \(\mathcal{T}_A^{B}\).
In particular, the three topologies in the notation above are given by \(\mathcal{T}_B=\{B\cap U:U\in\mathcal{T}\}\), \(\mathcal{T}_A^{X}=\{A\cap U:U\in\mathcal{T}\}\), and \(\mathcal{T}_A^{B}=\{A\cap V:V\in\mathcal{T}_B\}\). The last formula takes intersections with \(A\) of sets that are already open in \(B\). The central question is whether it produces exactly the same collection as intersecting \(A\) with open sets of \(X\) in one step.
The Transitivity Theorem
Proof. We prove equality by showing that every set in either collection belongs to the other. If \(C\in\mathcal{T}_A^{B}\), then by definition \(C=A\cap V\) for some \(V\in\mathcal{T}_B\). Since \(V\) is open in \(B\), there is \(U\in\mathcal{T}\) such that \(V=B\cap U\). Therefore \(C=A\cap(B\cap U)=(A\cap B)\cap U=A\cap U\), because \(A\subseteq B\). This shows \(C\in\mathcal{T}_A^{X}\).
Conversely, if \(C\in\mathcal{T}_A^{X}\), then \(C=A\cap U\) for some \(U\in\mathcal{T}\). The set \(V=B\cap U\) belongs to \(\mathcal{T}_B\) by the definition of the subspace topology on \(B\). Also \(A\cap V=A\cap(B\cap U)=A\cap U=C\), again because \(A\subseteq B\). Hence \(C\in\mathcal{T}_A^{B}\), proving equality. If \(A=\varnothing\), both collections consist of \(\varnothing\), and the same formulas still apply; if \(B=\varnothing\), then \(A=\varnothing\) as well. Thus the equality includes these degenerate cases. \(\square\)
The proof rests on the set identity \(A\cap(B\cap U)=A\cap U\), which holds because every point of \(A\) is already in \(B\). No special properties of \(X\), such as a metric or a separation condition, are needed. The argument also explains how to translate a relative open set: an open set in \(B\) has the form \(B\cap U\), and intersecting it with \(A\) removes the intermediate restriction without changing the result.
Worked Example: A Finite Space with a Nontrivial Topology
Let \(X=\{a,b,c,d\}\) have topology \(\mathcal{T}=\{\varnothing,\{a,b\},\{a,b,c\},X\}\). Take \(B=\{a,b,c\}\) and \(A=\{b,c\}\). Compute the topology on \(A\) directly from \(X\), and then through \(B\).
Directly from \(X\). Intersect each open set of \(X\) with \(A\): \(\varnothing\cap A=\varnothing\), \(\{a,b\}\cap A=\{b\}\), \(\{a,b,c\}\cap A=\{b,c\}=A\), and \(X\cap A=A\). Thus \(\mathcal{T}_A^{X}=\{\varnothing,\{b\},\{b,c\}\}\).
First from \(X\) to \(B\). Intersect the same open sets with \(B\): \(\varnothing\cap B=\varnothing\), \(\{a,b\}\cap B=\{a,b\}\), \(\{a,b,c\}\cap B=B\), and \(X\cap B=B\). Hence \(\mathcal{T}_B=\{\varnothing,\{a,b\},B\}\).
Then from \(B\) to \(A\). Intersect each of these open sets with \(A\): \(\varnothing\cap A=\varnothing\), \(\{a,b\}\cap A=\{b\}\), and \(B\cap A=A=\{b,c\}\). So \(\mathcal{T}_A^{B}=\{\varnothing,\{b\},\{b,c\}\}\), which matches the direct calculation. This example shows that the induced topology on \(A\) need not contain every subset of \(A\): here \(\{c\}\) is not open in either description.
Check: Each listed set in the topology on \(A\) is an intersection with an open set at the relevant stage, and the complete lists from both routes are identical.
Consequences for Relative Open and Closed Sets
The theorem does more than equate two notations for a topology. It ensures that openness in \(A\) can be tested using either ambient space, as long as the set being tested is a subset of \(A\). For example, if \(C\subseteq A\), then \(C\) is open in \(A\) precisely when \(C=A\cap U\) for some open \(U\) in \(X\); it is also open in \(A\) precisely when \(C=A\cap V\) for some open \(V\) in \(B\). The choice between these descriptions is often a matter of convenience.
Proof. The theorem gives \(\mathcal{T}_A^{B}=\mathcal{T}_A^{X}\), so the two topologies on \(A\) have exactly the same open sets. A set \(C\subseteq A\) is closed in a topology on \(A\) when its complement \(A\setminus C\) is open in that topology. Since the open sets agree, \(A\setminus C\) is open under one topology exactly when it is open under the other. Therefore \(C\) is closed under one exactly when it is closed under the other. \(\square\)
Worked Example: A Relative Open Set Through an Intermediate Space
Work in \(\mathbb{R}\) with its standard topology, and let \(A=[1,2]\) and \(B=(0,3)\). These satisfy \(A\subseteq B\subseteq\mathbb{R}\). Consider the open set of \(\mathbb{R}\) \(U=(-2,3/2)\cup(7/4,4)\). Find the corresponding relatively open set in \(A\), both directly and through \(B\).
Direct calculation. Intersect \(U\) with \(A\). The first interval contributes the points of \([1,2]\) below \(3/2\), namely \([1,3/2)\). The second contributes the points above \(7/4\) through \(2\), namely \((7/4,2]\). Thus \(A\cap U=[1,3/2)\cup(7/4,2]\).
Calculation through \(B\). First intersect \(U\) with \(B=(0,3)\): \(B\cap U=(0,3/2)\cup(7/4,3)\). Now intersect with \(A=[1,2]\). The first part gives \([1,3/2)\), and the second gives \((7/4,2]\). Therefore \(A\cap(B\cap U)=[1,3/2)\cup(7/4,2]\), exactly as in the direct calculation.
Check: The endpoint \(1\) is included because it belongs to both \(A\) and \((-2,3/2)\); \(3/2\) is excluded because neither interval of \(U\) contains it. The endpoint \(7/4\) is excluded because the second interval is open there, while \(2\) is included because \(2\in A\cap(7/4,4)\). The two routes therefore give the same endpoint inclusions as well as the same intervals.
Worked Example: Nested Intervals with an Open Set Given in the Middle Space
Again use the standard topology on \(\mathbb{R}\). Let \(A=(2,4)\), \(B=(1,5)\), and \(X=\mathbb{R}\). This time start with a set open in \(B\), namely \(V=B\cap U\), where \(U=(0,3)\cup(7/2,6)\) is open in \(X\). Compute \(A\cap V\), and compare it with \(A\cap U\).
First, \(B\cap U=((1,5)\cap(0,3))\cup((1,5)\cap(7/2,6))=(1,3)\cup(7/2,5)\). Intersecting this set with \(A=(2,4)\) gives \(A\cap V=(2,3)\cup(7/2,4)\). For the direct route, \(A\cap U=((2,4)\cap(0,3))\cup((2,4)\cap(7/2,6))=(2,3)\cup(7/2,4)\). The results agree. The restrictions \(B\) adds to \(U\) have no effect after intersecting with \(A\), since \(A\) is already contained in \(B\).
Check: The first interval in \(U\) contributes points strictly above \(2\) and strictly below \(3\); the second contributes points strictly above \(7/2\) and strictly below \(4\). Neither endpoint is included in \(A\), so both computations give exactly the displayed open intervals.
A Common Pitfall
A common error is to drop the intersection with \(A\) when a set has already been restricted to \(B\). For example, with \(A=[1,2]\), \(B=(0,3)\), and \(U=(0,2)\), we have \(B\cap U=(0,2)\). It would be wrong to conclude that this is the corresponding open set in \(A\): it contains \(1/2\), which is not even a point of \(A\). The correct set is \(A\cap(B\cap U)=[1,2]\cap(0,2)=[1,2)\). The direct calculation gives \(A\cap U=[1,2)\) as well.
The theorem does not say that every open subset of \(B\) is itself a subset of \(A\), or that the topologies on \(B\) and \(A\) are equal. It says that after an open set of \(B\) is intersected with \(A\), the resulting collection of sets is exactly the topology on \(A\) obtained directly from \(X\). Keeping track of which space a set belongs to prevents the mistaken removal of a necessary intersection.
For a relatively open subset of \(A\), start with \(A\cap U\) for \(U\) open in \(X\), or \(A\cap V\) for \(V\) open in \(B\).
If \(V=B\cap U\), then \(A\cap V=A\cap(B\cap U)=A\cap U\), using \(A\subseteq B\).
In concrete examples, compute every intersection in order and verify that the final set contains only points of \(A\).
Check Your Understanding
Use the definition of the subspace topology and the transitivity theorem to justify your answers.
- Let \(A\subseteq B\subseteq X\). Express an arbitrary open set of \(A\), first using an open set of \(X\) and then using an open set of \(B\).
- In the finite-space example, why is \(\{c\}\) not open in \(A=\{b,c\}\)? Check it using both routes.
- For \(A=[0,1]\subseteq B=(-1,2)\subseteq\mathbb{R}\), compare the two routes for \(U=(-2,1/2)\).
- Explain why equality of the two topologies on \(A\) implies that the two notions of closed subset of \(A\) agree.
- What fails in the expression \(A\cap(B\cap U)=B\cap U\) when \(A\) is a proper subset of \(B\)? Give a concrete example.