Topology Axioms

From Open Sets to Local Data

A topology specifies open sets globally: it tells us which subsets are open and how they behave under unions and intersections. The same information can be recorded locally, point by point, by saying which sets count as neighborhoods of each point. This viewpoint is useful because many topological questions concern what happens near a particular point, rather than the entire open-set collection at once.

Recall that a topology $\mathcal{T}$ on $X$ contains $\varnothing$ and $X$, is closed under arbitrary unions, and is closed under finite intersections. The empty union is $\varnothing$, and the intersection of an empty finite family is $X$. These conventions make the axioms apply uniformly, including at the boundary cases. We now translate this global description into neighborhood axioms and prove that the translation loses no information.

Neighborhoods and Their Axioms

A neighborhood of a point need not itself be open. It is enough that it contain an open set around that point. For example, in the usual topology on $\mathbb{R}$, the closed interval $[x-1,x+1]$ is a neighborhood of $x$, because it contains the open interval $(x-\tfrac12,x+\tfrac12)$, which contains $x$. The interval $[x-1,x+1]$ is not open.

Definition: Let $(X,\mathcal{T})$ be a topological space and $x\in X$. A set $N\subseteq X$ is a neighborhood of $x$ if there is an open set $U\in\mathcal{T}$ such that $x\in U\subseteq N$. Write $\mathcal{N}(x)$ for the collection of all neighborhoods of $x$.

The definition immediately gives several rules. Every neighborhood contains its point, and any superset of a neighborhood is again a neighborhood. The intersection of two neighborhoods of the same point is a neighborhood, since the corresponding open sets have an open intersection. There is also a local coherence rule: if $N$ is a neighborhood of $x$, then some neighborhood of $x$ consists entirely of points for which $N$ is a neighborhood.

Definition: A neighborhood system on a set $X$ assigns to each $x\in X$ a collection $\mathcal{N}(x)$ of subsets of $X$ satisfying the following axioms:

$X\in\mathcal{N}(x)$, and every $N\in\mathcal{N}(x)$ contains $x$.
If $N\in\mathcal{N}(x)$ and $N\subseteq M\subseteq X$, then $M\in\mathcal{N}(x)$.
If $N_1,N_2\in\mathcal{N}(x)$, then $N_1\cap N_2\in\mathcal{N}(x)$.
If $N\in\mathcal{N}(x)$, there is an $M\in\mathcal{N}(x)$ such that, for every $y\in M$, $N\in\mathcal{N}(y)$.

The first axiom combines two requirements: there is at least one neighborhood at every point, and every designated neighborhood contains that point. The intersection axiom implies closure under all nonempty finite intersections by repeated application. The empty intersection is $X$, already included by the first axiom. The last axiom is the condition that ensures neighborhood information is compatible from point to point; the first three rules alone do not guarantee that the assigned collections come from a topology.

Examples of Neighborhood Systems

Worked Example: Neighborhoods in the Usual Topology on the Real Line

In the usual topology on $\mathbb{R}$, a set $N$ is a neighborhood of $x$ exactly when it contains an open interval $(x-r,x+r)$ for some $r>0$. The forward direction follows because every open set containing $x$ contains such an interval. Conversely, such an interval is open and contains $x$, so any set containing it is a neighborhood of $x$.

For example, $N=[x-1,x+1]$ is a neighborhood of $x$, since $$ x\in(x-\tfrac12,x+\tfrac12)\subseteq[x-1,x+1]. $$ It is not open: no open interval centered at either endpoint $x-1$ or $x+1$ is contained in $N$. This illustrates why the definition requires a neighborhood to contain an open set, rather than requiring the neighborhood itself to be open.

Worked Example: Discrete and Indiscrete Neighborhoods

Let $X$ have the discrete topology, so every subset of $X$ is open. A set $N$ is a neighborhood of $x$ if and only if $x\in N$. If $x\in N$, then $N$ itself is an open set containing $x$; if $N$ is a neighborhood, its definition requires $x\in N$.

For the indiscrete topology $\{\varnothing,X\}$, the only open set containing any given point $x$ is $X$. Therefore $N$ is a neighborhood of $x$ exactly when $N=X$. These two cases show how the topology affects the neighborhood system even when the underlying set is unchanged.

Recovering a Topology from Neighborhood Data

The neighborhood axioms are not merely consequences of a topology; they are an equivalent way to specify one. Given a neighborhood system, declare a set open when it is a neighborhood of each of its points. The local coherence axiom is essential in proving that the neighborhoods recovered this way are precisely the neighborhoods originally assigned.

Theorem (Neighborhood Systems Determine Topologies): Let $X$ be a set with a neighborhood system $\{\mathcal{N}(x):x\in X\}$. Define $$ \mathcal{T}=\{U\subseteq X: U\in\mathcal{N}(x)\text{ for every }x\in U\}. $$ Then $\mathcal{T}$ is a topology on $X$, and a set $N$ belongs to $\mathcal{N}(x)$ exactly when $N$ contains a member of $\mathcal{T}$ that contains $x$.

Proof. First, $\varnothing\in\mathcal{T}$, because its defining condition has no points to check. Also $X\in\mathcal{T}$, since $X\in\mathcal{N}(x)$ for every $x\in X$.

Let $\{U_i:i\in I\}$ be any family in $\mathcal{T}$, and put $U=\bigcup_{i\in I}U_i$. If $x\in U$, then $x\in U_j$ for some $j\in I$. Since $U_j\in\mathcal{T}$, we have $U_j\in\mathcal{N}(x)$. As $U_j\subseteq U$, the upward-closure axiom gives $U\in\mathcal{N}(x)$. Thus $U\in\mathcal{T}$. If $I$ is empty, this union is $\varnothing$, which has already been shown to be in $\mathcal{T}$.

Now let $U_1,U_2\in\mathcal{T}$. If $x\in U_1\cap U_2$, then $U_1,U_2\in\mathcal{N}(x)$, so the intersection axiom gives $U_1\cap U_2\in\mathcal{N}(x)$. Hence $U_1\cap U_2\in\mathcal{T}$. Repeating this argument proves closure under nonempty finite intersections; the empty intersection is $X$. Therefore $\mathcal{T}$ is a topology.

It remains to recover the assigned neighborhoods. Suppose first that $N\in\mathcal{N}(x)$. Define $$ W=\{y\in X:N\in\mathcal{N}(y)\}. $$ The local coherence axiom supplies $M\in\mathcal{N}(x)$ such that every $y\in M$ satisfies $N\in\mathcal{N}(y)$. Thus $M\subseteq W$, so $x\in W$, because $x\in M$. For any $y\in W$, apply local coherence to $N\in\mathcal{N}(y)$. It gives $M_y\in\mathcal{N}(y)$ such that every $z\in M_y$ has $N\in\mathcal{N}(z)$. Therefore $M_y\subseteq W$, and upward closure implies $W\in\mathcal{N}(y)$. This holds for every $y\in W$, so $W\in\mathcal{T}$. Also, every $y\in W$ belongs to $N$, since $N\in\mathcal{N}(y)$ and neighborhoods contain their points. We have proved $x\in W\subseteq N$.

Conversely, suppose $N$ contains some $W\in\mathcal{T}$ with $x\in W$. By the definition of $\mathcal{T}$, $W\in\mathcal{N}(x)$. Upward closure then gives $N\in\mathcal{N}(x)$. Together, these implications show that the original neighborhoods are exactly the sets containing an open set around the point. $\square$

Worked Example: Why Local Coherence Is Needed

Let $X=\{a,b,c\}$. For each point, choose a set $S_x$ containing it, as follows: $$ S_a=\{a,b\},\qquad S_b=\{b,c\},\qquad S_c=\{c\}. $$ Set $\mathcal{N}(x)=\{N\subseteq X:S_x\subseteq N\}$. Each collection contains $X$, all its members contain its assigned point, and it is upward closed. Intersections also remain in the collection: if $S_x\subseteq N_1$ and $S_x\subseteq N_2$, then $S_x\subseteq N_1\cap N_2$.

But local coherence fails. The set $S_a=\{a,b\}$ belongs to $\mathcal{N}(a)$. Any $M\in\mathcal{N}(a)$ must contain both $a$ and $b$. For the required conclusion to hold, $S_a$ would have to belong to $\mathcal{N}(b)$. That would require $S_b=\{b,c\}\subseteq S_a=\{a,b\}$, which is false because $c\notin S_a$. Thus no such $M$ can satisfy the local coherence axiom.

The first three axioms do imply that the pointwise-open sets form a topology, but they do not ensure that its neighborhoods match the originally assigned collections. Here the set $\{a,b\}$ was designated a neighborhood of $a$, yet no open set in the reconstructed topology can contain $a$ and remain inside $\{a,b\}$: any set open at $a$ must also be a neighborhood of $b$, which forces it to contain $c$. The final axiom prevents this mismatch.

Comparing Topologies Locally

On a fixed set, a topology with more open sets is called finer, and one with fewer open sets is called coarser. Neighborhood systems translate this comparison into a pointwise statement. A finer topology has more neighborhoods at each point: every neighborhood from the coarser topology remains a neighborhood in the finer one.

Theorem (Topology Inclusion and Neighborhood Inclusion): Let $\mathcal{T}_1$ and $\mathcal{T}_2$ be topologies on the same set $X$, with neighborhood systems $\mathcal{N}_1(x)$ and $\mathcal{N}_2(x)$. Then $$ \mathcal{T}_1\subseteq\mathcal{T}_2 \quad\Longleftrightarrow\quad \mathcal{N}_1(x)\subseteq\mathcal{N}_2(x)\text{ for every }x\in X. $$

Proof. Suppose $\mathcal{T}_1\subseteq\mathcal{T}_2$. If $N\in\mathcal{N}_1(x)$, there is $U\in\mathcal{T}_1$ such that $x\in U\subseteq N$. Since $\mathcal{T}_1\subseteq\mathcal{T}_2$, $U$ is open in $\mathcal{T}_2$ as well. Thus $N\in\mathcal{N}_2(x)$, proving the neighborhood inclusions.

For the reverse direction, suppose $\mathcal{N}_1(x)\subseteq\mathcal{N}_2(x)$ for every $x$. Take $U\in\mathcal{T}_1$. For each $x\in U$, the set $U$ is a neighborhood of $x$ in the first topology, so it is a neighborhood of $x$ in the second. Consequently there is an open set $V_x\in\mathcal{T}_2$ with $x\in V_x\subseteq U$. Taking the union over all $x\in U$ gives $$ U=\bigcup_{x\in U}V_x. $$ If $U=\varnothing$, it belongs to both topologies. Otherwise, the displayed union is open in $\mathcal{T}_2$, so $U\in\mathcal{T}_2$. Hence $\mathcal{T}_1\subseteq\mathcal{T}_2$. $\square$

This result also shows why neighborhood descriptions are a useful comparison tool: instead of comparing entire collections of open sets, one can compare the neighborhoods at each point. The direction matters. More open sets produce more neighborhoods, not fewer, because a set qualifies as a neighborhood whenever it contains at least one open set around the point.

What the Axioms Accomplish

The open-set axioms and neighborhood axioms express the same structure at different scales. Arbitrary unions and finite intersections govern open sets globally. Neighborhood systems retain the same structure locally, with upward closure and finite intersections handling basic set operations, and local coherence ensuring that nearby points see compatible neighborhoods.

A common pitfall is to confuse a neighborhood with an open set. A neighborhood may properly contain an open set containing the point; it need not itself be open. Another is to omit local coherence when proposing neighborhood data. The first three neighborhood rules may look plausible, but without the fourth they need not recover the specified neighborhoods from the topology they generate. Checking every axiom, including its pointwise quantifiers, is what makes the equivalence reliable.

Check Your Understanding

Use the definitions and proofs above to answer the following questions.

How can a set fail to be open but still be a neighborhood of a point?
Why does upward closure imply that a union of open sets defined from a neighborhood system is open?
Where is the local coherence axiom used when recovering the original neighborhoods from the reconstructed topology?
For the indiscrete topology, which sets are neighborhoods of a fixed point, and why?
If $\mathcal{T}_1\subseteq\mathcal{T}_2$, which way do the corresponding neighborhood inclusions go?
In the three-point example, which required inclusion fails when testing local coherence at $a$?

What You'll Learn