Why Restricting the Domain Preserves Continuity
A continuous map controls how open sets in its target pull back to its domain. When we restrict the domain to a subset, each such preimage is simply cut down to that subset. The resulting set is open in the subset's relative topology, even when it is not open in the original space. This observation proves a general rule: continuity is preserved under restriction of the domain.
Let \((X,\mathcal{T})\) and \((Y,\mathcal{U})\) be topological spaces, and let \(A\subseteq X\). The subspace topology on \(A\) is \(\mathcal{T}_A=\{A\cap U:U\in\mathcal{T}\}\). Thus a set is open in \(A\) precisely when it is the intersection of \(A\) with an open set of \(X\). For a map \(f:X\to Y\), its restriction is the map \(f|_A:A\to Y\) defined by \(f|_A(a)=f(a)\) for each \(a\in A\). Continuity here means that the preimage of every open set in the target is open in the domain.
Proof. Let \(V\) be any open set in \(Y\). By the definition of restriction, a point \(a\in A\) belongs to \((f|_A)^{-1}(V)\) exactly when \(f(a)\in V\). Therefore \((f|_A)^{-1}(V)=A\cap f^{-1}(V)\). Since \(f\) is continuous, \(f^{-1}(V)\) is open in \(X\). By the definition of the subspace topology, its intersection with \(A\) is open in \(A\). Thus the preimage of every open \(V\subseteq Y\) is open in \(A\), which proves that \(f|_A\) is continuous. If \(A=\varnothing\), the preimage is empty and is open in the empty subspace as well. \(\square\)
The proof depends exactly on the form of the subspace topology: open sets in \(A\) are ambient open sets intersected with \(A\). It does not require \(A\) itself to be open or closed in \(X\). Nor does it assert that a preimage under the restriction must be open in \(X\); its required openness is relative to \(A\).
Computing a Restriction on an Interval
Worked Example: Squaring on a Half-Open Interval
Give \(X=Y=\mathbb{R}\) their standard topologies, let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\), and take \(A=[-1,2)\) with the subspace topology inherited from \(\mathbb{R}\). The restriction is the map \(f|_A:A\to\mathbb{R}\) given by \(f|_A(x)=x^2\) for \(-1\le x<2\). The function \(f\) is continuous on \(\mathbb{R}\), so the theorem guarantees that this restriction is continuous.
To see the relative-open-set calculation explicitly, take the open target set \(V=(0,1)\). The inequality \(x^2\in(0,1)\) means \(0<x^2<1\). The first inequality excludes \(x=0\), and the second is equivalent to \(-1<x<1\). Hence \(f^{-1}(V)=(-1,0)\cup(0,1)\), and
This preimage is open in \(A\), since it is the intersection of \(A\) with the open set \((-1,0)\cup(0,1)\) in \(\mathbb{R}\). Check: every \(x\) in the displayed preimage satisfies \(0<x^2<1\), while \(x=0\) and \(x=-1\) do not. Thus the strict inequalities and both excluded endpoints agree with the computed preimage.
Worked Example: An Affine Map on a Closed Interval
Let \(X=Y=\mathbb{R}\) with their standard topologies. Define \(g:\mathbb{R}\to\mathbb{R}\) by \(g(x)=3x-2\), and give \(A=[0,2]\) its subspace topology. The restriction \(g|_A:A\to\mathbb{R}\) is \(g|_A(x)=3x-2\) for \(0\le x\le2\). The affine function \(g\) is continuous, so \(g|_A\) is continuous by the theorem.
For a direct calculation, use the open target interval \(V=(-1,2)\). Solving the membership condition gives \(-1<3x-2<2\), hence \(1<3x<4\), and therefore \(1/3<x<4/3\). This interval lies inside \(A\), so
The result is open in \(A\), since it is the intersection of \(A\) with the ambient open interval \((1/3,4/3)\). Check: \(g(1/3)=3(1/3)-2=1-2=-1\), and \(g(4/3)=3(4/3)-2=4-2=2\). Both endpoint values are excluded from \(V\), so neither endpoint belongs in the preimage.
Restrictions on Euclidean Subsets
The same argument applies when the domain is a subset of a Euclidean space. The subset need not be an interval or an open region: it carries the subspace topology inherited from the ambient Euclidean space. A continuous map on the whole space can therefore be restricted to a line, a curve, or any other subset without losing continuity.
Worked Example: An Affine Map on a Line Segment
Give \(X=Y=\mathbb{R}^2\) their standard Euclidean topologies, and define \(F:\mathbb{R}^2\to\mathbb{R}^2\) by \(F(x,y)=(x+2y,3x-y)\). This affine map is continuous. Let \(A=\{(t,0):0\le t\le2\}\) have the subspace topology from \(\mathbb{R}^2\). Its restriction is obtained by substituting \(x=t\) and \(y=0\):
Take the open Euclidean ball \(V=B((0,0),\sqrt{10})\) in the target. A point \((t,0)\in A\) belongs to the preimage exactly when the Euclidean distance from \(F(t,0)\) to the origin is less than \(\sqrt{10}\). The squared distance is \(t^2+(3t)^2=10t^2\). Since \(t\ge0\), the inequality \(10t^2<10\) is equivalent to \(0\le t<1\). Consequently,
This is relatively open in \(A\): it equals \(A\cap\{(x,y)\in\mathbb{R}^2:x<1\}\), and the half-plane \(\{(x,y):x<1\}\) is open in \(\mathbb{R}^2\). Check: at \(t=0\), the image is \((0,0)\), whose distance from the origin is \(0<\sqrt{10}\). At \(t=1\), the image is \((1,3)\), whose distance is \(\sqrt{1^2+3^2}=\sqrt{10}\), so it is not in the open ball. This confirms the included and excluded endpoints of the parameter range.
Worked Example: A Constant Restriction on the Circle
Give \(X=\mathbb{R}^2\) its standard Euclidean topology and let \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) have the subspace topology. Define \(H:\mathbb{R}^2\to\mathbb{R}\) by \(H(x,y)=x^2+y^2\), where \(\mathbb{R}\) has its standard topology. This polynomial map is continuous. On \(S^1\), its restriction is the constant map \(H|_{S^1}(x,y)=1\), since every point of \(S^1\) satisfies \(x^2+y^2=1\).
For the open target set \(V=(0,2)\), every point of \(S^1\) maps to \(1\in V\), so \((H|_{S^1})^{-1}(V)=S^1\). This is open in \(S^1\), because it is \(S^1\cap\mathbb{R}^2\), with \(\mathbb{R}^2\) open in itself. For the open target set \(W=(2,3)\), no point of \(S^1\) maps into \(W\), so \((H|_{S^1})^{-1}(W)=\varnothing\), also open in \(S^1\). These two calculations illustrate the theorem; continuity for all open target sets follows from the restriction theorem, not just from checking these two sets.
A Pointwise Version
There is also a local form of the restriction principle. Say that \(f:X\to Y\) is continuous at \(a\in X\) if, for every open neighborhood \(V\) of \(f(a)\), there is an open neighborhood \(U\) of \(a\) in \(X\) such that \(f(U)\subseteq V\).
Proof. Let \(V\) be any open neighborhood of \(f(a)\). By continuity of \(f\) at \(a\), choose an open neighborhood \(U\) of \(a\) in \(X\) such that \(f(U)\subseteq V\). The set \(A\cap U\) is an open neighborhood of \(a\) in \(A\). For each \(z\in A\cap U\), we have \((f|_A)(z)=f(z)\in V\). Thus \((f|_A)(A\cap U)\subseteq V\), which is precisely continuity of the restriction at \(a\). \(\square\)
The global theorem is the pointwise statement applied at every point, but its open-preimage proof is often shorter. The local version is useful when continuity is established point by point rather than all at once.
Common Pitfall: Relative Openness Is Not Ambient Openness
A frequent error is to compute the correct preimage under a restriction and then require it to be open in the original domain. For example, let \(X=Y=\mathbb{R}\) have their standard topologies, let \(f(x)=x\), and restrict to \(A=[0,1]\). The open target set \(V=(-1,1)\) has preimage under the restriction
The set \([0,1)\) is not open in \(\mathbb{R}\): every ambient open interval containing \(0\) also contains negative real numbers, which are not in \([0,1)\). But \([0,1)\) is open in \(A\), because \([0,1)=A\cap(-1,2)\), and \((-1,2)\) is open in \(\mathbb{R}\). The restriction is continuous because its preimages are open in its domain \(A\), not because those preimages must be ambient open.
This distinction is useful whenever a continuous map is studied only on a chosen set of inputs. One may restrict a function to a curve, a boundary, or an interval and use the topology inherited by that subset. The restriction theorem justifies doing so without repeating a continuity proof from scratch.
Check Your Understanding
Use the definition of the subspace topology when deciding whether a preimage is open.
- State the restriction theorem and identify the topology on its domain.
- For \(f(x)=x^2\) on \(\mathbb{R}\), restrict to \(A=[0,3]\). Find the preimage of \((1,4)\) under \(f|_A\).
- Let \(A=\{(t,0):t\in\mathbb{R}\}\subseteq\mathbb{R}^2\), and let \(G(x,y)=x+y\). Compute \(G|_A\).
- Why does the proof of the restriction theorem not require \(A\) to be open in \(X\)?
- For \(A=[0,1]\subseteq\mathbb{R}\), explain why \([0,1)\) is open in \(A\) but not open in \(\mathbb{R}\).