One Subspace, Two Constructions
Suppose each space \(A_\alpha\) sits inside a topological space \(X_\alpha\). There are two natural ways to give \(\prod_{\alpha\in J} A_\alpha\) a topology: first give every \(A_\alpha\) its subspace topology and then form the product, or first form \(\prod_{\alpha\in J}X_\alpha\) and give \(\prod_{\alpha\in J} A_\alpha\) the subspace topology it inherits there. The key fact is that these constructions agree. This lets us describe a neighborhood by restricting finitely many coordinates, whether we work inside the factors or in the ambient product.
For each index \(\alpha\in J\), let \((X_\alpha,\mathcal{T}_\alpha)\) be a topological space, and let \(A_\alpha\subseteq X_\alpha\) have the subspace topology. Recall that a subset of \(A_\alpha\) is open in this subspace precisely when it has the form \(A_\alpha\cap U_\alpha\) for some \(U_\alpha\in\mathcal{T}_\alpha\).
We identify \(\prod_{\alpha\in J}A_\alpha\) with its natural subset of \(\prod_{\alpha\in J}X_\alpha\): a tuple belongs to the former exactly when its \(\alpha\)-coordinate belongs to \(A_\alpha\) for every \(\alpha\). We use the usual convention that a product over an empty index set is a one-point space.
The Product-Subspace Theorem
Proof. Write \(A=\prod_{\alpha\in J}A_\alpha\) and \(X=\prod_{\alpha\in J}X_\alpha\). A basic open set in the product topology on \(A\) has the form \(\bigcap_{\alpha\in F}\pi_\alpha^{-1}(V_\alpha)\), where \(F\) is finite and each \(V_\alpha\) is open in \(A_\alpha\). By the definition of the subspace topology, for each \(\alpha\in F\) there is an open \(U_\alpha\subseteq X_\alpha\) such that \(V_\alpha=A_\alpha\cap U_\alpha\). Thus the basic set in \(A\) is \(A\cap\bigcap_{\alpha\in F}\pi_\alpha^{-1}(U_\alpha)\), which is the intersection of \(A\) with a basic open set of \(X\). It is therefore open in the subspace topology. Conversely, a basic open set of \(X\) has the form \(\bigcap_{\alpha\in F}\pi_\alpha^{-1}(U_\alpha)\). Its intersection with \(A\) is \(\bigcap_{\alpha\in F}\pi_\alpha^{-1}(A_\alpha\cap U_\alpha)\), a basic open set in the product topology on \(A\), since each \(A_\alpha\cap U_\alpha\) is open in \(A_\alpha\). Every open set in either topology is a union of basic open sets, so the two topologies are equal. These identities also cover an empty factor, when \(A\) is empty, and an empty index set, when both constructions give the one-point space. \(\square\)
The proof is a comparison of neighborhood descriptions. In a product, a basic neighborhood restricts only finitely many coordinates. In a subspace, each restriction is intersected with the subset in that coordinate. Taking these intersections coordinate by coordinate is exactly the same as intersecting the ambient product neighborhood with the whole product of the subspaces.
Proof. The map \(j\) sends each tuple to the same tuple, now viewed in the ambient product. It is a bijection from \(\prod_{\alpha\in J}A_\alpha\) onto its image. The theorem says that the topology on its domain is exactly the topology inherited by that image. Consequently, a subset of the domain is open if and only if its image is open in the image subspace. This means \(j\) and its inverse are both continuous, so \(j\) is a homeomorphism onto its image. If the image is empty, the unique map between the empty spaces has this property as well. \(\square\)
Worked Neighborhoods
Worked Example: A Rectangle of Intervals
Give \(A=[0,2]\) and \(B=(-1,1]\) the subspace topologies inherited from \(\mathbb{R}\), and consider \(A\times B\subseteq\mathbb{R}\times\mathbb{R}\), where both real factors have their standard topology. Describe a product neighborhood of \((0,1)\) and verify it using the ambient product.
In \(A\), the set \((-1/4,1/4)\cap A=[0,1/4)\) is open in the subspace topology. In \(B\), the set \((3/4,5/4)\cap B=(3/4,1]\) is open in its subspace topology. Their product is therefore a basic open neighborhood of \((0,1)\) in \(A\times B\): it contains \((0,1)\), since \(0\in[0,1/4)\) and \(1\in(3/4,1]\).
Now intersect the ambient open rectangle \((-1/4,1/4)\times(3/4,5/4)\) with \(A\times B\). The result is \([0,1/4)\times(3/4,1]\), exactly the neighborhood just found in the product of the subspaces. The endpoint \(1\) is included because \(B\) includes \(1\), while the interval condition \(t>3/4\) is strict. This calculation exhibits the same neighborhood in both constructions.
Check: a point \((s,t)\in A\times B\) belongs to this neighborhood exactly when \(-1/4<s<1/4\) and \(3/4<t<5/4\). Using \(s\in[0,2]\) and \(t\in(-1,1]\), these conditions reduce to \(0\le s<1/4\) and \(3/4<t\le1\), as in the product calculation.
Worked Example: A Circle Factor and an Interval
Let \(S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\), with the subspace topology from standard \(\mathbb{R}^2\), and let \(I=[0,1]\) have the subspace topology from standard \(\mathbb{R}\). Regard \(S^1\times I\) as a subset of \(\mathbb{R}^2\times\mathbb{R}\). At the point \(((1,0),1)\), take the product neighborhood \(\bigl(S^1\cap B((1,0),1/4)\bigr)\times\bigl(I\cap(3/4,5/4)\bigr)\).
Here \(B((1,0),1/4)\) is the open Euclidean ball in \(\mathbb{R}^2\), and \((3/4,5/4)\) is open in \(\mathbb{R}\). Each factor of the displayed neighborhood is open in its respective subspace. The point belongs to it because the distance from \((1,0)\) to itself is \(0<1/4\), and \(3/4<1\le1\). Written as a subset of the ambient product, the same neighborhood is \((S^1\times I)\cap\bigl(B((1,0),1/4)\times(3/4,5/4)\bigr)\).
For an explicit membership check, a point \(((u,v),t)\in S^1\times I\) lies in this set precisely when \((u-1)^2+v^2<1/16\) and \(3/4<t<5/4\). Since \(t\in[0,1]\), the second condition is \(3/4<t\le1\). Thus this is an ambient open product neighborhood intersected with \(S^1\times I\), and also a basic neighborhood formed within the factors.
Worked Example: Two Circle Factors
Use the subspace topology on each copy of \(S^1\subseteq\mathbb{R}^2\). In \(S^1\times S^1\subseteq\mathbb{R}^2\times\mathbb{R}^2\), consider the point \(((1,0),(0,1))\). The product of the relative neighborhoods cut out by \(B((1,0),1/3)\) and \(B((0,1),1/2)\) is \(\bigl(S^1\cap B((1,0),1/3)\bigr)\times\bigl(S^1\cap B((0,1),1/2)\bigr)\).
This contains the chosen point: both center-to-center distances are \(0\), which is less than \(1/3\) and \(1/2\), respectively. In the ambient product it is \((S^1\times S^1)\cap\bigl(B((1,0),1/3)\times B((0,1),1/2)\bigr)\). For a point \(((u,v),(s,t))\in S^1\times S^1\), membership is equivalent to \(u^2+v^2=1\), \(s^2+t^2=1\), \((u-1)^2+v^2<1/9\), and \(s^2+(t-1)^2<1/4\). The strict inequalities come from the open balls. This shows directly how the product neighborhoods on the two circle factors become relative neighborhoods in the ambient product, often viewed as \(\mathbb{R}^4\).
Relative Openness Is Not Ambient Openness
A subset can be open in a subspace without being open in its ambient space. For instance, \(S^1\) is open in itself, but it is not open in \(\mathbb{R}^2\). To verify the latter claim, take any radius \(r>0\). The point \((1+r/2,0)\) lies within distance \(r\) of \((1,0)\), but \((1+r/2)^2\ne1\), so it does not lie in \(S^1\). Thus no open ball about \((1,0)\) is contained in \(S^1\).
Common pitfall. One might incorrectly conclude that \(\bigl(S^1\cap B((1,0),1/4)\bigr)\times\bigl(I\cap(3/4,5/4)\bigr)\) is not open in \(S^1\times I\), on the grounds that \(S^1\) is not open in \(\mathbb{R}^2\). That tests the wrong topology: the question is openness in the subspace \(S^1\times I\), not openness in \(\mathbb{R}^2\times\mathbb{R}\). The correct calculation is \((S^1\times I)\cap\bigl(B((1,0),1/4)\times(3/4,5/4)\bigr)\). The set in parentheses is open in the ambient product, so its intersection with the subspace is open in the subspace topology. The theorem also shows it is open in the product topology on the two subspaces.
This agreement is useful whenever a product of constrained objects is studied inside an unconstrained ambient product. A neighborhood can be described using open sets in the factors, then translated into an ambient relative neighborhood without changing which subsets are open. The coordinate restrictions remain finite in an arbitrary product, exactly as the product-topology basis requires.
Check Your Understanding
Use the basis descriptions to justify each answer, paying attention to which topology is being used.
- For \(A=[0,1]\subseteq\mathbb{R}\) and \(B=\{0\}\subseteq\mathbb{R}\), describe a basic neighborhood of \((1,0)\) in \(A\times B\) as an intersection with an ambient open rectangle.
- In \(S^1\times I\), write a relative basic neighborhood of \(((0,1),0)\) using an open ball in \(\mathbb{R}^2\) and an open interval in \(\mathbb{R}\).
- Explain why the theorem remains valid if one of the subspaces \(A_\alpha\) is empty.
- For an empty index set, identify the underlying set in each construction and explain why the topologies agree.
- Give an example of a set open in a subspace but not open in its ambient space, and express it as an intersection with an ambient open set.