Separating Points After Restricting a Space
In a Hausdorff space, distinct points have disjoint open neighborhoods. A subset inherits its open sets by intersecting open sets of the ambient space with that subset. The key question is whether this restriction preserves the neighborhoods needed to separate two points. It does: each point remains in its restricted neighborhood, and disjoint ambient neighborhoods still have disjoint intersections.
Let \((X,\mathcal{T})\) be a topological space, and let \(A\subseteq X\). The subspace topology on \(A\) is the collection of sets \(A\cap U\), where \(U\in\mathcal{T}\). A set open in this topology is called open in \(A\); it need not be open in \(X\).
The definition of the subspace topology gives a useful neighborhood test: if \(x\in A\), then every open neighborhood of \(x\) in \(A\) has the form \(A\cap U\), where \(U\) is open in \(X\) and contains \(x\). Conversely, whenever \(U\) is open in \(X\) and contains \(x\in A\), the set \(A\cap U\) is an open neighborhood of \(x\) in \(A\). In particular, a subspace neighborhood is a trace on \(A\), not necessarily an ambient open set.
The Inheritance Theorem
Proof. We must show that any two distinct points of \(A\) have disjoint open neighborhoods in \(A\). Let \(x,y\in A\) with \(x\ne y\). Since \(X\) is Hausdorff, there are open sets \(U,V\in\mathcal{T}\) such that \(x\in U\), \(y\in V\), and \(U\cap V=\varnothing\). Set \(U_A=A\cap U\) and \(V_A=A\cap V\). By the definition of the subspace topology, \(U_A\) and \(V_A\) are open in \(A\). Since \(x,y\in A\), we have \(x\in U_A\) and \(y\in V_A\). Also, \[ U_A\cap V_A=(A\cap U)\cap(A\cap V)=A\cap(U\cap V)=A\cap\varnothing=\varnothing. \] Thus the two points have disjoint open neighborhoods in \(A\). If \(A\) has fewer than two points, the condition is vacuous, so the conclusion holds in those cases as well. \(\square\)
The proof uses exactly the Hausdorff condition on the ambient space and the definition of the subspace topology. It does not require \(A\) to be open, closed, or nonempty. The ambient separating neighborhoods may contain many points outside \(A\); only their traces on \(A\) are needed.
Worked Applications
Worked Example: An Interval in the Real Line
Give \(A=(0,1]\) the subspace topology inherited from \(\mathbb{R}\) with its standard topology. Separate \(x=1/4\) and \(y=3/4\) by open neighborhoods in \(A\).
The distance between the points is \[ \left|\frac34-\frac14\right|=\frac12. \] Choose radius \(1/4\) around each point. The ambient open balls are the intervals \[ B\left(\frac14,\frac14\right)=\left(0,\frac12\right), \qquad B\left(\frac34,\frac14\right)=\left(\frac12,1\right). \] Their traces on \(A\) are the same intervals, since both already lie in \(A\). Each is open in \(A\), and they are disjoint: the first omits \(1/2\), while the second also omits \(1/2\). The first contains \(1/4\), and the second contains \(3/4\), so these are the required neighborhoods.
Check: the endpoints are consistent with open balls: the first interval has endpoints \(0\) and \(1/2\) excluded, and the second has endpoints \(1/2\) and \(1\) excluded. Thus no point is shared. This calculation illustrates the local separation; the theorem guarantees separation for every distinct pair in \(A\).
Worked Example: Two Points on the Circle
Let \(S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\) with the subspace topology inherited from the Euclidean plane. Separate \(p=(1,0)\) and \(q=(0,1)\) by open neighborhoods in \(S^1\).
The Euclidean distance between them is \[ \|p-q\|=\sqrt{(1-0)^2+(0-1)^2}=\sqrt{2}. \] Use ambient open balls of radius \(1/2\), and intersect each with \(S^1\): \[ N_p=S^1\cap B_{\|\cdot\|}\left(p,\frac12\right), \qquad N_q=S^1\cap B_{\|\cdot\|}\left(q,\frac12\right). \] These sets are open in \(S^1\) by the definition of its subspace topology. The point \(p\) belongs to \(N_p\) because \(\|p-p\|=0<1/2\); the point \(q\) belongs to \(N_q\) because \(\|q-q\|=0<1/2\).
To check disjointness, suppose \(z\in N_p\cap N_q\). Then \(\|p-z\|<1/2\) and \(\|z-q\|<1/2\). The Euclidean triangle inequality would give \[ \sqrt{2}=\|p-q\|\leq \|p-z\|+\|z-q\|<\frac12+\frac12=1, \] which is impossible since \(\sqrt{2}>1\). Hence \(N_p\cap N_q=\varnothing\).
Check: both neighborhood conditions use strict inequalities, as required for open balls. The contradiction proves that there is no point in both traces, not merely that their centers differ.
Worked Example: The Rationals Are Hausdorff
Give \(\mathbb{Q}\) the subspace topology inherited from the standard topology on \(\mathbb{R}\). The theorem implies that \(\mathbb{Q}\) is Hausdorff, even though it is not a closed subset of \(\mathbb{R}\). Here is an explicit separation for \(r=1/3\) and \(s=2/3\).
Their distance is \(|s-r|=1/3\). Choose radius \(1/9\), which is less than half this distance. The ambient intervals are \[ \left(\frac13-\frac19,\frac13+\frac19\right)=\left(\frac29,\frac49\right), \qquad \left(\frac23-\frac19,\frac23+\frac19\right)=\left(\frac59,\frac79\right). \] Their traces on \(\mathbb{Q}\) are open in \(\mathbb{Q}\), contain \(r\) and \(s\), respectively, and are disjoint because \(4/9<5/9\).
For any distinct \(r,s\in\mathbb{Q}\), the same construction works with radius \(|r-s|/3\): each ambient interval contains its center, and the distance between the centers is \(3\) times the radius, so the intervals are disjoint. Their traces separate the points in \(\mathbb{Q}\). This is also a direct application of the theorem. Closedness is not a hypothesis for Hausdorffness to pass to a subspace.
Subspaces of Subspaces
The inheritance result can be applied in stages. If \(B\subseteq A\subseteq X\), then giving \(A\) the subspace topology from \(X\), and then giving \(B\) the subspace topology from \(A\), produces the same topology on \(B\) as restricting the topology of \(X\) directly to \(B\).
Proof. An open set of \(A\) has the form \(A\cap U\) for some open \(U\subseteq X\). An open set of \(B\) obtained from this set is \(B\cap(A\cap U)\). Since \(B\subseteq A\), this equals \(B\cap U\), which is open in the subspace topology on \(B\) inherited directly from \(X\). Conversely, any directly inherited open set \(B\cap U\), with \(U\) open in \(X\), is also \(B\cap(A\cap U)\), and \(A\cap U\) is open in \(A\). Thus the two collections of open sets on \(B\) are equal. The Hausdorff conclusion follows by applying the inheritance theorem to \(B\subseteq X\), or successively to \(A\subseteq X\) and \(B\subseteq A\). \(\square\)
Why the Converse Fails
The theorem says that a Hausdorff ambient space forces all its subspaces to be Hausdorff. It does not say that a Hausdorff subspace forces the ambient space to be Hausdorff. The distinction matters: the topology on a subspace may have fewer points to separate, and its open sets are only traces of ambient open sets.
For a concrete example, let \(X\) be an infinite set with the finite-complement topology: its open sets are \(\varnothing\) and all subsets whose complements in \(X\) are finite. The space \(X\) is not Hausdorff. Indeed, any two nonempty open sets \(U,V\) have finite complements, so \(X\setminus(U\cap V)=(X\setminus U)\cup(X\setminus V)\) is finite. Since \(X\) is infinite, \(U\cap V\ne\varnothing\). Thus no two nonempty open neighborhoods can be disjoint.
Now choose distinct \(a,b\in X\) and let \(A=\{a,b\}\). The set \(X\setminus\{b\}\) is open in \(X\), and its trace on \(A\) is \(\{a\}\). Likewise \(X\setminus\{a\}\) is open and its trace is \(\{b\}\). These two singleton sets are disjoint open neighborhoods in \(A\). Therefore \(A\) is Hausdorff, although \(X\) is not. The inheritance theorem is one-way.
A Common Pitfall: Choosing Neighborhoods Too Large
When separating two points in \(\mathbb{R}\), choosing open intervals around them is not enough; the intervals must also be disjoint. For \(x=2/5\) and \(y=3/5\), the distance is \(1/5\). A mistaken choice is to give both points radius \(1/5\). This produces \[ \left(\frac15,\frac35\right) \quad\text{and}\quad \left(\frac25,\frac45\right), \] which overlap in \((2/5,3/5)\). They are not separating neighborhoods.
Instead take radius \(1/15\), which is one third of the distance. The intervals become \[ \left(\frac13,\frac7{15}\right) \quad\text{and}\quad \left(\frac8{15},\frac23\right). \] Since \(7/15<8/15\), they are disjoint, and each contains its center. For a subset \(A\subseteq\mathbb{R}\), intersect these intervals with \(A\); the resulting sets are open in \(A\), still contain their respective points when those points lie in \(A\), and remain disjoint. The restriction step preserves separation, but it cannot repair a choice of ambient neighborhoods that already overlap.
What the Result Lets You Conclude
Every subset of \(\mathbb{R}^n\), with its subspace topology from the standard Euclidean topology, is Hausdorff. The same applies to subsets of any Hausdorff space, including subsets that are not open or closed. To use this efficiently, identify the ambient Hausdorff space, choose disjoint ambient neighborhoods for the two points, and take their traces on the subset. The proof guarantees that those traces have all the required properties.
The method is specific to inheritance: it establishes separation inside a subspace from separation in the ambient space. It does not claim that other separation properties have the same inheritance behavior, nor that a continuous image of a Hausdorff space must be Hausdorff.
Check Your Understanding
Use the definition of the subspace topology and the Hausdorff condition to justify your answers.
- Why is \(A\cap U\) open in \(A\) when \(U\) is open in \(X\)?
- In the inheritance proof, which fact ensures that the restricted neighborhoods still contain the two points?
- Give disjoint open neighborhoods in \(\mathbb{R}\) for \(1/4\) and \(3/4\), then describe their traces on any subset containing both points.
- Why does the finite-complement topology on an infinite set fail to be Hausdorff?
- Can a Hausdorff subspace have a non-Hausdorff ambient space? Explain using the two-point subset of an infinite set with the finite-complement topology.