Openness Depends on the Space
A set can be open when viewed inside a space and fail to be open in the larger space containing it. The distinction is not a change to the set itself; it is a change to which open sets are available as neighborhoods. The subspace topology makes this precise by defining open sets in a subset through intersections with open sets of the ambient space.
Let \((X,\mathcal{T})\) be a topological space and let \(A\subseteq X\). The subspace topology on \(A\) is the collection \(\mathcal{T}_A=\{A\cap G:G\in\mathcal{T}\}\). A set \(U\subseteq A\) is called open in \(A\), or relatively open in \(A\), when \(U\in\mathcal{T}_A\). Thus the defining description already suggests a criterion. The key point is that the criterion applies to every subset \(U\) of \(A\), whether or not \(U\) is open in \(X\).
Proof. Suppose first that \(U\) is open in \(A\). By the definition of the subspace topology, every member of \(\mathcal{T}_A\) has the form \(A\cap G\) for some \(G\in\mathcal{T}\). Since \(U\in\mathcal{T}_A\), there is such an open \(G\) with \(U=A\cap G\). Conversely, suppose \(G\in\mathcal{T}\) and \(U=A\cap G\). The definition of \(\mathcal{T}_A\) says that \(A\cap G\in\mathcal{T}_A\), so \(U\) is open in \(A\). These implications prove both directions. \(\square\)
The theorem is an existence criterion: to prove relative openness, it is enough to find one suitable ambient open set \(G\). That \(G\) need not be unique. Points outside \(A\) do not affect the intersection, so different ambient open sets may produce the same relatively open set. Conversely, to disprove relative openness, it is enough to show that no ambient open set can have the required intersection.
Working with the Criterion in the Real Line
Unless specified otherwise, \(\mathbb{R}\) has its standard topology, whose open neighborhoods contain open intervals around each of their points. When \(A\) is a subset of \(\mathbb{R}\), the set \(A\cap G\) keeps only the points of \(G\) that belong to \(A\). As a result, endpoints of \(A\) can belong to a relatively open set even though no ambient open interval can include such an endpoint while remaining inside \(A\).
Worked Example: An Endpoint Included Relatively
Let \(A=[-1,1]\subseteq\mathbb{R}\). Show that \(U=[-1,1)\) is open in \(A\), although it is not open in \(\mathbb{R}\).
Take the ambient set \(G=(-2,1)\). It is open in \(\mathbb{R}\). Intersecting with \(A\), the lower endpoint \(-1\) is retained because \(-2<-1\), and the upper endpoint \(1\) is excluded because \(1\notin(-2,1)\). Every point strictly between \(-1\) and \(1\) belongs to both sets. Therefore \(A\cap G=[-1,1)=U\), so the criterion proves that \(U\) is open in \(A\).
To check that this is not ambient openness, suppose \(U\) were open in \(\mathbb{R}\). Since \(-1\in U\), there would be some \(\varepsilon>0\) such that \((-1-\varepsilon,-1+\varepsilon)\subseteq U\). But \(-1-\varepsilon/2\) belongs to that interval and is less than \(-1\), so it does not belong to \(U=[-1,1)\). This contradiction proves that \(U\) is not open in \(\mathbb{R}\).
Check: the proposed representative gives exactly the target: \([-1,1]\cap(-2,1)=[-1,1)\). The ambient interval excludes \(1\) and includes \(-1\), as required.
A proposed set does not become relatively open just because it is contained in \(A\). In particular, the subspace cannot cut away points immediately beside an interior point of \(A\) if those points are forced into the intersection by ambient openness. The next example tests this obstruction directly.
Worked Example: A Set That Cannot Be Relatively Open
Let \(A=[-1,1]\subseteq\mathbb{R}\) and \(U=(-1,\tfrac12]\). Determine whether there is an ambient open set \(G\subseteq\mathbb{R}\) such that \(A\cap G=U\).
Assume, for contradiction, that such an open \(G\) exists. Since \(\tfrac12\in U=A\cap G\), the point \(\tfrac12\) belongs to \(G\). Openness of \(G\) gives an \(\varepsilon>0\) such that \((\tfrac12-\varepsilon,\tfrac12+\varepsilon)\subseteq G\). Set \(\delta=\min(\varepsilon/2,1/4)\) and \(y=\tfrac12+\delta\). Then \(\delta>0\), \(\delta<\varepsilon\), and \(\delta\leq1/4\), so \(\tfrac12<y\leq\tfrac34<1\). Hence \(y\in A\), and \(y\in G\) because \(y\in(\tfrac12-\varepsilon,\tfrac12+\varepsilon)\). Thus \(y\in A\cap G\). But \(y>\tfrac12\), so \(y\notin U\), contradicting \(A\cap G=U\).
Therefore no such \(G\) exists, and \(U\) is not open in \(A\). The obstruction is at the point \(\tfrac12\): every ambient open set containing it also contains points of \(A\) just to its right, but the proposed \(U\) excludes all of those points.
Check: the choice of \(y\) works for every \(\varepsilon>0\). It satisfies \(y-\tfrac12=\delta\leq\varepsilon/2<\varepsilon\), and \(y\leq3/4\), so \(y\) is both in the interval contained in \(G\) and in \(A\).
Worked Example: A Singleton as a Subspace
Let \(A=\{0\}\subseteq\mathbb{R}\) and \(U=\{0\}\). Verify that \(U\) is open in \(A\), but not open in \(\mathbb{R}\).
Choose \(G=(-1,1)\), which is open in \(\mathbb{R}\). Its intersection with \(A\) is \(\{0\}\), since \(0\in(-1,1)\) and \(A\) has no other points. Thus \(A\cap G=U\), and \(U\) is open in \(A\).
For ambient openness, any open set containing \(0\) must contain an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\). The point \(\varepsilon/2\) is nonzero and belongs to that interval, so no such open set can equal \(\{0\}\). Hence \(U\) is not open in \(\mathbb{R}\).
Check: in the subspace intersection, points outside \(A\) are discarded: \(\{0\}\cap(-1,1)=\{0\}\). In the ambient space, those nearby nonzero points are still present, which is why \(\{0\}\) is not open there.
The Criterion in a Finite Space
The same characterization works in spaces where openness is not defined by intervals. One lists the open sets of the ambient topology and checks their intersections with the chosen subset. This keeps the reasoning tied to the actual topology rather than to geometric intuition.
Worked Example: A Three-Point Topological Space
Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}\), and let \(A=\{b,c\}\). Determine whether \(U=\{b\}\) is open in \(A\), and whether it is open in \(X\).
The ambient open set \(G=\{a,b\}\) gives \(A\cap G=\{b,c\}\cap\{a,b\}=\{b\}\). Thus \(U\) is open in \(A\). It is not open in \(X\), because the list of members of \(\mathcal{T}\) is \(\varnothing,\{a\},\{a,b\},\{a,b,c\}\), and \(\{b\}\) is not on that list.
For a full check, intersect \(A\) with each ambient open set: \(A\cap\varnothing=\varnothing\), \(A\cap\{a\}=\varnothing\), \(A\cap\{a,b\}=\{b\}\), and \(A\cap X=\{b,c\}\). So the topology on \(A\) consists exactly of \(\varnothing,\{b\},\{b,c\}\), confirming that \(U\) is relatively open.
Check: \(G\) is in the stated ambient topology, and the computed intersection is exactly \(\{b\}\). The fact that \(\{b\}\notin\mathcal{T}\) rules out ambient openness without affecting relative openness.
How the Representation Behaves
The criterion also lets us verify directly how relative open sets combine. The following proposition is a useful check on the representation: arbitrary unions and finite intersections of sets represented in this way are still represented in the same way. The finiteness condition on intersections matters because a topology is required to be closed under finite intersections, not arbitrary intersections.
Proof. By the ambient-open-set criterion, for each \(i\in J\) there is an open \(G_i\subseteq X\) with \(U_i=A\cap G_i\). Since arbitrary unions of open sets in \(X\) are open, \(\bigcup_{i\in J}G_i\) is open in \(X\). Distributing intersection over union gives \(\bigcup_{i\in J}U_i=\bigcup_{i\in J}(A\cap G_i)=A\cap\bigl(\bigcup_{i\in J}G_i\bigr)\), which is relatively open by the criterion. For the finite-intersection claim, choose open \(G_1,\ldots,G_n\subseteq X\) with \(U_k=A\cap G_k\) for each \(k\). The finite intersection \(\bigcap_{k=1}^nG_k\) is open in \(X\), and \(\bigcap_{k=1}^nU_k=\bigcap_{k=1}^n(A\cap G_k)=A\cap\bigl(\bigcap_{k=1}^nG_k\bigr)\). The criterion again gives relative openness. \(\square\)
For the empty union, the result is \(\varnothing\), which is relatively open because \(\varnothing=A\cap\varnothing\). The proposition's finite-intersection statement is made for \(n\geq1\); the intersection of no sets is conventionally \(X\), and its counterpart within the subspace is \(A=A\cap X\), also relatively open.
A Common Mistake
Common pitfall. A frequent error is to take \(G=U\) automatically and conclude that \(U=A\cap G\), without checking that \(G\) is open in \(X\). For example, with \(A=[-1,1]\) and \(U=[-1,1)\), the equality \(A\cap U=U\) is true, but \(U\) is not open in \(\mathbb{R}\). That calculation alone does not establish relative openness by the criterion, because it used a set \(G=U\) that is not an ambient open set. The correction is to choose \(G=(-2,1)\), which is ambient open and satisfies \(A\cap G=[-1,1)=U\).
The distinction can also matter in the opposite direction: not every subset of \(A\) is relatively open. In the earlier interval example, \((-1,\tfrac12]\subseteq[-1,1]\), but the point \(\tfrac12\) forces any ambient open representative to include points of \(A\) greater than \(\tfrac12\). Checking the representative, rather than relying only on containment in \(A\), resolves both kinds of error.
Check Your Understanding
For each question, identify the ambient space and use the intersection criterion where appropriate.
- Let \(A=[0,2]\subseteq\mathbb{R}\). Find an ambient open set \(G\) such that \(A\cap G=[0,1)\), or explain why none exists.
- In \(A=[-1,1]\subseteq\mathbb{R}\), decide whether \((-1,0]\) is open in \(A\). Give a proof using the criterion or show why no representative exists.
- Let \(A=\{0,1\}\subseteq\mathbb{R}\). Is \(\{0\}\) open in \(A\)? Give an ambient open representative.
- In the finite space \(X=\{a,b,c\}\) with topology \(\{\varnothing,\{a\},\{a,b\},X\}\), list the relatively open subsets of \(A=\{b,c\}\).
- Explain why the equality \(U=A\cap U\), by itself, does not prove that \(U\) is open in \(A\).