Bases of a Topological Space

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Bases and Subbases of a Topological Space

The most straightforward way to specify a topological space is to list all open sets and thereby give the topology $\mathcal{T}$, but the entire collection $\mathcal{T}$ is not needed for this. For instance, if an open set $U=\bigcup U_i$ is given, and for every $i$ we have $U\neq U_i$ and $U_i\in\mathcal{T}$, then the information that $U\in\mathcal{T}$ may be regarded as redundant.

Definition 1 Let $(X,\mathcal{T})$ be a topological space. A subset $\mathcal{B}$ of $\mathcal{T}$ is called a base for $\mathcal{T}$ if for every $U\in\mathcal{T}$, there exists a family $(B_i)_{i\in I}$ of elements of $\mathcal{B}$ such that $U=\bigcup_{i\in I} B_i$.

In particular, since $X\in\mathcal{T}$, the collection $\mathcal{B}$ is also a covering of $X$. ([Set Theory] §Sum of Sets, ⁋Definition 1) Since the elements of $\mathcal{B}$ are all open sets by definition, it is appropriate to say that $\mathcal{B}$ is an open covering of $X$.

Proposition 2 Let $(X,\mathcal{T})$ be a topological space. Then $\mathcal{B}$ is a base for $\mathcal{T}$ if and only if the following three conditions are satisfied:

For each $x\in X$, there exists at least one $B\in\mathcal{B}$ such that $x\in B$;
If there exist two elements $B_1,B_2\in\mathcal{B}$ containing $x$, then there exists another $B_3\in\mathcal{B}$ such that $x\in B_3$ and $B_3\subseteq B_1\cap B_2$;
For each $U\in\mathcal{T}$ and $x\in U$, there exists $B\in\mathcal{B}$ such that $x\in B\subseteq U$.

Proof

First, assume that $\mathcal{B}$ is a base for $\mathcal{T}$. Then $\mathcal{B}$ is an open covering of $X$, so condition 1 is trivially satisfied.

On the other hand, if $B_1$ and $B_2$ are given as in condition 2, then $B_1\cap B_2$ is also an open set, so there exists a family $(B_i)_{i\in I}$ of elements of $\mathcal{B}$ such that $B_1\cap B_2=\bigcup_{i\in I} B_i$. Since $(B_i)_{i\in I}$ is an open covering of $B_1\cap B_2$, condition 2 is also trivially satisfied, just as condition 1. Replacing $B_1\cap B_2$ with an arbitrary open set $U$ yields condition 3.

Conversely, assume that the three conditions are satisfied and pick an arbitrary open set $U$. Then for each $x\in U$, condition 3 implies that there exists $B_x\in\mathcal{B}$ such that $x\in B_x\subseteq U$. Now $U=\bigcup_{x\in U} B_x$, which completes the proof.

From the above proposition, the following question arises naturally.

Let $X$ be an arbitrary set, and let $\mathcal{B}$ be a subset of $\mathcal{P}(X)$ satisfying the above conditions. Can we endow $X$ with a topology $\mathcal{T}$ so that $(X,\mathcal{T})$ is a topological space having $\mathcal{B}$ as a base?

The answer to this question is affirmative. [Mun] gives a direct proof of this, but we will first define a local base and then prove it in a simpler way.

Meanwhile, by refining Definition 1, we can define a subbase as follows.

Definition 3 A subbase for a topological space $(X,\mathcal{T})$ is an open covering $\mathcal{S}$ of $X$ such that for every $U\in\mathcal{T}$, there exists $S\in\mathcal{S}$ with $S\subseteq U$.

Collecting all finite intersections of elements of $\mathcal{S}$, we obtain a new collection $\mathcal{B}$. To check whether this collection is a base, we only need condition 2; but since the elements of $\mathcal{B}$ are obtained as finite intersections of elements of $\mathcal{S}$, the intersection $B_1\cap B_2$ is again a finite intersection of elements of $\mathcal{S}$, and hence $\mathcal{B}$ is a base.

Local Bases of a Topological Space

§Open Sets, ⁋Proposition 6 shows that if we can describe the neighborhood filter $\mathcal{N}(x)$ centered at a point $x$ in a topological space $X$, then we can completely recover the topology of $X$. On the other hand, since $\mathcal{N}(x)$ satisfies the conditions of §Open Sets, ⁋Proposition 6, in particular the first condition, we do not need all of $\mathcal{N}(x)$ to describe it.

Definition 4 Let $X$ be a topological space and let $A$ be a subset of $X$. A local base at $A$ is a coinitial subset of $(\mathcal{N}(A),\subseteq)$ consisting of open sets. ([Set Theory] §Elements of Ordered Sets)

As in §Open Sets, ⁋Definition 4, when $A$ is a singleton $\{x\}$, we call a local base at $A$ a local base at the point $x$. Then the following holds.

Proposition 5 Let $(X,\mathcal{T})$ be a topological space. Then a subset $\mathcal{B}$ of $\mathcal{T}$ is a base for $X$ if and only if for each $x\in X$, the elements of $\mathcal{B}$ containing $x$ form a local base at $x$.

Proof

For convenience, let us denote by $\mathcal{B}(x)$ the collection of elements of $\mathcal{B}$ containing $x$.

First, suppose that $\mathcal{B}$ is a base for $X$, and pick an arbitrary point $x\in X$ and a neighborhood $V$. Then there exists an open set $U$ such that $x\in U\subseteq V$. Since $\mathcal{B}$ is a base for $X$, there exist $U_i\in\mathcal{B}$ such that $U=\bigcup U_i$. Since $x\in U$, we have $x\in U_i$ for some $i$, and therefore $U_i\in\mathcal{B}(x)$.

Conversely, suppose that $\mathcal{B}$ satisfies the given condition and let $U$ be an arbitrary open set. Then for any $x\in U$, we have $U\in\mathcal{N}(x)$, so the given condition yields a suitable $V(x)\in\mathcal{B}(x)$ such that $x\in V(x)\subseteq U$. Now $U=\bigcup V(x)$, which gives the desired result.

Earlier we asked whether, given an arbitrary collection satisfying the first and second conditions of Proposition 2, there exists a topology on $X$ having it as a base. This becomes an easy corollary of the above proposition.

Corollary 6 Let $X$ be a set, and let $\mathcal{B}$ be a subset of $\mathcal{P}(X)$ satisfying the following two conditions.

For each $x$, there exists at least one $B\in\mathcal{B}$ such that $x\in B$.
If there exist $B_1,B_2\in\mathcal{B}$ containing $x$, then there exists another $B_3\in\mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2$.

Then there exists a unique topology $\mathcal{T}$ on $X$ such that $\mathcal{B}$ is a base for this topology $\mathcal{T}$.

Proof

Define $\mathcal{B}(x)$ to be the collection of elements of $\mathcal{B}$ containing $x$, just as in the proof of the previous proposition. Also define

\[\mathcal{N}(x)=\mathop{\uparrow}\mathcal{B}(x):=\bigcup_{B\in\mathcal{B}}\mathop{\uparrow}B\]

That is, $\mathcal{N}(x)$ is the collection containing every element of $\mathcal{B}$ that contains the given $x\in X$, together with all elements of $\mathcal{P}(X)$ larger than it.

For any $V\in\mathcal{N}(x)$, suppose $V\subseteq V'$. By the definition of $\mathcal{N}(x)$, there exists $U\in\mathcal{B}(x)$ such that $U\subseteq V$, and for such $U$ we have $U\subseteq V'$, so $V'\in\mathcal{N}(x)$.
Suppose that elements $V_1,\ldots,V_n$ of $\mathcal{N}(x)$ are given. Then there exist $U_i\in\mathcal{B}(x)$ such that $U_i\subseteq V_i$. Using the second condition of Proposition 2 inductively, we can find a suitable $U\in\mathcal{B}(x)$ such that $U\subseteq U_1\cap\cdots\cap U_n$. In particular, $U\subseteq V_1\cap\cdots\cap V_n$, so $V_1\cap\cdots\cap V_n\in\mathcal{N}(x)$.
For any element $V$ of $\mathcal{N}(x)$, there exists $W\in\mathcal{B}(x)$ such that $W\subseteq V$, so $x\in V$.
For any element $V$ of $\mathcal{N}(x)$, pick $W\in\mathcal{B}(x)$ such that $W\subseteq V$. Then $W\in\mathcal{B}$, and therefore $W\in\mathcal{B}(y)$ for any $y\in W$. Since $W\subseteq V$, it follows that $V\in\mathcal{N}(y)$ for all $y$.

Now we can apply §Open Sets, ⁋Proposition 6 to obtain a topology $\mathcal{T}$, and in this topological space $\mathcal{B}(x)$ is a local base at $x$, so by Proposition 5, $\mathcal{B}$ is a base for $\mathcal{T}$.

The topology obtained from $\mathcal{B}$ through this process is called the topology generated by $\mathcal{B}$. Similarly, we make a base from a subbase $\mathcal{S}$, and the topology generated by this base is called the topology obtained from $\mathcal{S}$.

References

[Bou] N. Bourbaki, General Topology. Elements of mathematics. Springer, 1995.
[Mun] J.R. Munkres, Topology. Featured Titles for Topology. Prentice Hall, Incorporated, 2000.

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Bases and Subbases of a Topological Space

Local Bases of a Topological Space

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