From Neighborhoods to Pointwise Data
In a topological space, neighborhoods record which sets surround each point. Instead of starting with open sets, we can ask whether a collection of sets assigned separately to each point contains enough information to recover a topology. The answer is yes, provided the collections satisfy a local compatibility condition as well as the expected closure properties.
Throughout, a neighborhood of \(x\) means a set containing an open set that contains \(x\); the neighborhood itself need not be open. For a topological space \((X,\mathcal{T})\), write \(\mathcal{N}(x)\) for the collection of all neighborhoods of \(x\).
- Containment: every \(V\in\mathcal{N}(x)\) contains \(x\), and \(X\in\mathcal{N}(x)\).
- Upward closure: if \(V\in\mathcal{N}(x)\) and \(V\subseteq W\subseteq X\), then \(W\in\mathcal{N}(x)\).
- Finite-intersection closure: if \(V,W\in\mathcal{N}(x)\), then \(V\cap W\in\mathcal{N}(x)\).
- Local coherence: if \(V\in\mathcal{N}(x)\), there is \(W\in\mathcal{N}(x)\) such that \(V\in\mathcal{N}(y)\) for every \(y\in W\).
The first three conditions say that at each point the neighborhoods behave like a filter of sets containing that point. Local coherence links the systems at different points: whenever \(V\) is a neighborhood of \(x\), some neighborhood \(W\) of \(x\) consists entirely of points for which \(V\) is also a neighborhood. This is the extra condition that allows the pointwise data to come from open sets.
Every Topology Gives These Axioms
Proof. By definition, a neighborhood of \(x\) contains an open set containing \(x\), so it contains \(x\); also \(X\) is open and contains \(x\), hence \(X\in\mathcal{N}(x)\). If \(V\in\mathcal{N}(x)\), choose an open \(U\) with \(x\in U\subseteq V\). For \(V\subseteq W\subseteq X\), the same \(U\) shows \(W\in\mathcal{N}(x)\), proving upward closure. If \(V,W\in\mathcal{N}(x)\), choose open sets \(U_1,U_2\) with \(x\in U_1\subseteq V\) and \(x\in U_2\subseteq W\). The open set \(U_1\cap U_2\) contains \(x\) and is contained in \(V\cap W\), so \(V\cap W\in\mathcal{N}(x)\). Finally, for \(V\in\mathcal{N}(x)\), choose open \(U\) with \(x\in U\subseteq V\). Set \(W=U\), which is in \(\mathcal{N}(x)\). For every \(y\in W=U\), the same open set \(U\) contains \(y\) and lies in \(V\), so \(V\in\mathcal{N}(y)\). This proves local coherence. \(\square\)
Worked Example: Neighborhood Systems on the Real Line
Let \(\mathbb{R}\) have its standard topology. For each \(x\in\mathbb{R}\), define \(\mathcal{N}(x)\) to consist of all \(A\subseteq\mathbb{R}\) for which there is an \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq A\). Verify the axioms.
Every such \(A\) contains \(x\), because \(x\in(x-\varepsilon,x+\varepsilon)\), and \(\mathbb{R}\in\mathcal{N}(x)\), for example using \(\varepsilon=1\). If \(A\in\mathcal{N}(x)\) is witnessed by \(\varepsilon>0\) and \(A\subseteq B\subseteq\mathbb{R}\), then \((x-\varepsilon,x+\varepsilon)\subseteq B\), so \(B\in\mathcal{N}(x)\).
If \(A,B\in\mathcal{N}(x)\), choose \(\varepsilon_1,\varepsilon_2>0\) with \((x-\varepsilon_1,x+\varepsilon_1)\subseteq A\) and \((x-\varepsilon_2,x+\varepsilon_2)\subseteq B\). Set \(\delta=\min(\varepsilon_1,\varepsilon_2)>0\). Then \((x-\delta,x+\delta)\subseteq A\cap B\), so \(A\cap B\in\mathcal{N}(x)\).
For local coherence, let \(A\in\mathcal{N}(x)\), witnessed by \(\varepsilon>0\), and put \(W=(x-\varepsilon/2,x+\varepsilon/2)\). This interval belongs to \(\mathcal{N}(x)\). If \(y\in W\), then \(|y-x|<\varepsilon/2\). For every \(z\in(y-\varepsilon/2,y+\varepsilon/2)\), the triangle inequality gives \(|z-x|\leq |z-y|+|y-x|<\varepsilon\). Thus \((y-\varepsilon/2,y+\varepsilon/2)\subseteq(x-\varepsilon,x+\varepsilon)\subseteq A\), proving \(A\in\mathcal{N}(y)\).
Check: the radius \(\varepsilon/2\) works at every \(y\in W\), not just at \(x\). This verifies the required uniform local coherence.
Reconstructing the Topology
Proof. First, \(\varnothing\in\mathcal{T}_{\mathcal{N}}\), since its defining condition has no points to check. Also \(X\in\mathcal{T}_{\mathcal{N}}\), since \(X\in\mathcal{N}(x)\) for every \(x\in X\). If \(\{U_i\}_{i\in J}\) is any collection in \(\mathcal{T}_{\mathcal{N}}\), let \(x\in\bigcup_{i\in J}U_i\). Some \(U_j\) contains \(x\), and \(U_j\in\mathcal{N}(x)\). Since \(U_j\subseteq\bigcup_{i\in J}U_i\), upward closure gives \(\bigcup_{i\in J}U_i\in\mathcal{N}(x)\). Thus the union is in \(\mathcal{T}_{\mathcal{N}}\). If \(U,V\in\mathcal{T}_{\mathcal{N}}\) and \(x\in U\cap V\), then \(U,V\in\mathcal{N}(x)\), so \(U\cap V\in\mathcal{N}(x)\). Hence \(U\cap V\in\mathcal{T}_{\mathcal{N}}\). These are the topology axioms.
Now fix \(x\in X\). If \(A\in\mathcal{N}(x)\), define \(O_A=\{y\in X:A\in\mathcal{N}(y)\}\). Containment gives \(O_A\subseteq A\), and \(x\in O_A\). We show \(O_A\) is open in \(\mathcal{T}_{\mathcal{N}}\). For any \(y\in O_A\), we have \(A\in\mathcal{N}(y)\). Local coherence supplies \(W_y\in\mathcal{N}(y)\) such that \(A\in\mathcal{N}(z)\) for every \(z\in W_y\). Therefore \(W_y\subseteq O_A\); upward closure gives \(O_A\in\mathcal{N}(y)\). This holds for every \(y\in O_A\), so \(O_A\in\mathcal{T}_{\mathcal{N}}\). We have found an open set \(O_A\) with \(x\in O_A\subseteq A\), proving that \(A\) is a neighborhood of \(x\) in the reconstructed topology.
Conversely, if \(A\) is a neighborhood of \(x\) in the reconstructed topology, it contains an open set \(U\in\mathcal{T}_{\mathcal{N}}\) with \(x\in U\). The definition of \(\mathcal{T}_{\mathcal{N}}\) gives \(U\in\mathcal{N}(x)\), and upward closure then gives \(A\in\mathcal{N}(x)\). Thus the neighborhood systems agree exactly.
Finally, suppose another topology on \(X\) has these same neighborhood systems. Any open set \(U\) in that topology is a neighborhood of each \(x\in U\), so \(U\in\mathcal{N}(x)\) for every such \(x\); hence it belongs to \(\mathcal{T}_{\mathcal{N}}\). Conversely, if \(U\in\mathcal{T}_{\mathcal{N}}\), then for each \(x\in U\), \(U\in\mathcal{N}(x)\). In the other topology, this means there is an open set \(O_x\) with \(x\in O_x\subseteq U\). Therefore \(U=\bigcup_{x\in U}O_x\) is open; for \(U=\varnothing\), it is open by the topology axiom. The two topologies are equal. If \(X=\varnothing\), the same construction gives \(\{\varnothing\}\), the unique topology on \(X\), and the neighborhood claims are vacuous. \(\square\)
Worked Example: Recovering the Sierpinski Topology
Let \(X=\{a,b\}\). Set \(\mathcal{N}(a)=\{A\subseteq X:a\in A\}\) and \(\mathcal{N}(b)=\{X\}\). These are the neighborhood systems for the topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\); we verify both the axioms and the reconstruction.
At \(a\), the sets in \(\mathcal{N}(a)\) are \(\{a\}\) and \(X\). They contain \(a\), include \(X\), are upward closed, and are closed under intersections: the possible intersections are \(\{a\}\cap\{a\}=\{a\}\), \(\{a\}\cap X=\{a\}\), and \(X\cap X=X\). At \(b\), the only neighborhood is \(X\), which satisfies containment, upward closure, and intersection closure.
For local coherence at \(a\), take any \(V\in\mathcal{N}(a)\) and choose \(W=\{a\}\). Then \(W\in\mathcal{N}(a)\), and the only \(y\in W\) is \(a\), for which \(V\in\mathcal{N}(a)\). At \(b\), the only choice of \(V\) is \(X\); choose \(W=X\). Both \(a\) and \(b\) have \(X\) as a neighborhood, as required.
To compute the reconstructed topology, \(\varnothing\) and \(X\) satisfy the defining condition. The set \(\{a\}\) also satisfies it because \(\{a\}\in\mathcal{N}(a)\). The set \(\{b\}\) does not: it contains \(b\), but \(\{b\}\notin\mathcal{N}(b)=\{X\}\). Hence the reconstructed topology is exactly \(\{\varnothing,\{a\},X\}\).
Check: in this topology the only open set containing \(b\) is \(X\), so the only set containing an open neighborhood of \(b\) is \(X\). This agrees with \(\mathcal{N}(b)\).
Why Local Coherence Cannot Be Omitted
Worked Example: Filter Conditions Without Local Coherence
Let \(X=\{a,b,c\}\), and define \(\mathcal{N}(a)\) to be all subsets containing \(\{a,b\}\), \(\mathcal{N}(b)\) all subsets containing \(\{b,c\}\), and \(\mathcal{N}(c)\) all subsets containing \(\{c\}\). Check which axioms hold.
At each point, every member contains that point, and \(X\) belongs to each collection. Upward closure holds because any superset still contains the required set. For intersections, if two members of \(\mathcal{N}(a)\) both contain \(\{a,b\}\), their intersection contains \(\{a,b\}\); the same argument applies to the required sets \(\{b,c\}\) at \(b\) and \(\{c\}\) at \(c\). Thus containment, upward closure, and finite-intersection closure all hold.
But \(V=\{a,b\}\) belongs to \(\mathcal{N}(a)\). Every \(W\in\mathcal{N}(a)\) contains \(b\). Local coherence would require \(V\in\mathcal{N}(y)\) for every \(y\in W\), in particular \(V\in\mathcal{N}(b)\). This is false: a member of \(\mathcal{N}(b)\) must contain \(c\), and \(V\) does not. So local coherence fails, and the characterization theorem shows that these collections cannot be the neighborhood systems of any topology.
Check: the obstruction is not merely that a particular candidate open set is missing. The neighborhood systems themselves violate a necessary condition for every topology, proved in the proposition above.
Common pitfall. Checking only upward closure and finite intersections is not enough. Those properties make each point's collection behave like a filter, but they do not guarantee that the collections at nearby points fit together. In the three-point example, \(\{a,b\}\) is declared a neighborhood of \(a\), yet it is not a neighborhood of \(b\), even though every proposed neighborhood of \(a\) must contain \(b\). Local coherence rules out precisely this mismatch.
The reconstruction formula also explains the role of the axioms. An open set must be a neighborhood of each of its points, while local coherence ensures that a declared neighborhood contains an open set around its point. Together these facts make the pointwise data and the topology equivalent descriptions.
Check Your Understanding
Use the axioms and reconstruction theorem to answer the following questions.
- Why does upward closure follow for neighborhoods induced by a topology, even when the neighborhood itself is not open?
- In the real-line example, explain why the radius \(\varepsilon/2\) establishes local coherence for every point in the chosen interval.
- For \(X=\{a,b\}\), define \(\mathcal{N}(a)=\mathcal{N}(b)=\{A\subseteq X:A\ne\varnothing\}\). Which neighborhood-system axiom fails?
- Show directly from the definition of \(\mathcal{T}_{\mathcal{N}}\) why the union of any collection of its open sets is open.
- In the three-point example, identify the precise quantifier in local coherence that fails for \(V=\{a,b\}\in\mathcal{N}(a)\).