## Bases

Recall that though a subring or ideal of a ring may be rather huge, it often suffices to specify just a few elements which will generate the subring or ideal. Likewise, in a topology,  one can specify a few open sets and generate the rest via unions and finite intersections. We’ll expound upon that in what follows.

First, note that when (Xd) is a metric space, a subset U of X is open if and only if it is a union of (possibly infinitely many) open balls. Indeed, since every open ball is open, so is a union of multiple open balls. Conversely, if U is open, each element $x\in U$ is contained in some open ball $N(x,\epsilon) \subseteq U$ for some ε>0 which depends on x; hence U is the union of all these N(x, ε).

In other words, consider B = {N(x, ε) : x in X, ε>0}, the collection of all open balls; a subset of X is open if and only if it is a union of elements of B. Generalising this concept for topological spaces gives us the following definition.

Definition. A basis for a topology (X, T) is a collection of open sets, $B\subseteq T$ such that every open subset U of X is a union $U=\cup_i V_i$ of elements $V_i\in B$. We shall call an element $V\in B$ a basic open set.

Examples

1. As we saw above, the set B of open balls in a metric space (Xd) forms a basis of the induced topology.
2. In particular, the set of open intervals (ab) in R forms a basis.
3. In the discrete topology, the collection of singleton sets {x} forms a basis. In fact, if you take the collection of open balls {N(x, 1/2)} for the discrete metric, you get the same basis.
4. Since the metrics dd1 and d on Rn give rise to the same topology, we can pick the set of open balls under any one metric to produce a basis.
5. Exercise: find all possible bases of N under the right-order topology. [ Answer: there’s no non-trivial basis, i.e. any basis must include all open sets. ] Our next question is: suppose $B\subseteq \mathbf{P}(X)$ is some collection of subsets of X. Is it always the basis of some topology?

If it were, then clearly the resulting topology would be: $T=\{\cup_i V_i : \{V_i\}\subseteq B\}.$

Let’s check if T satisfies the axioms of a topology.

• Empty set and X : the empty set can be written as a union of an empty collection of $V_i$, so no problem there; for X, we need to specify that $\cup_i V_i = X$.
• Arbitrary union : this is obvious, since the union of sets $U_i := \cup_j V_{ij}$ (where each $V_{ij} \in B$) is also of the form $\cup_{i, j} V_{ij}$ which lies in T.
• Finite intersection : write $U_1 = \cup_i V_{1i}$ and $U_2 = \cup_j V_{2j}$, with each $V_{1i}, V_{2j}\in B$. Since intersection is distributive over union, we get: $U_1\cap U_2 = (\cup_i V_{1i})\cap (\cup_j V_{2j}) = \cup_{i,j} (V_{1i} \cap V_{2j}).$

For this to be in T, a sufficient condition is that $V_1\cap V_2\in T$ for all $V_1, V_2\in B$. On the other hand, this condition is obviously necessary since if $V_1, V_2\in B$, we have $V_1, V_2\in T\implies V_1\cap V_2 \in T$. Thus we have proven:

Theorem. A subset $B\subseteq \mathbf{P}(X)$ is a basis for some topology if and only if:

• the union of all $V\in B$ is the whole X; and
• for any $V_1, V_2\in B$, the intersection $V_1 \cap V_2$ is a union of elements from B.

## Application: Furstenberg’s Proof of the Infinitude of Primes

While he was an undergraduate, Hillel Furstenberg found an innovative proof that there’re infinitely many prime numbers, via the concept of topology.

Theorem. There are infinitely many prime numbers.

Proof (Furstenberg)

Define a topology on set of integers Z as follows. For integers a, b let V(a, b) be the set of all integers am+b for integer m. Then any intersection $V(a,b)\cap V(c,d)$ corresponds to solutions of simultaneous linear congruences $x\equiv b\pmod a$, $x\equiv d\pmod c$. From elementary number theory, this intersection is either empty or V(e, f) for some integers e, f (where e = lcm(ac)). Since V(1, 0) = Z, {V(a, b)} forms a basis for some topology T on Z.

Since the complement ZV(a, b) is a union of V(a, b’) for various b’, it is open, i.e. V(a, b) is both open and closed. Now, since any integer other than ±1 is divisible by some prime p, we have: $\mathbf{Z} -\{-1, +1\} = \cup_{p \text{ prime}} V(p, 0).$

If there’re finitely many primes, then the RHS is a union of finitely many closed sets and is hence closed. Thus, {-1, +1} is open, which is ridiculous: since every basic open set V(a, b) is infinite, every open set must be infinite too. ♦ ## Subbases

Let’s fix an underlying set X and consider various topologies on it. If $\{T_i\}$ is a collection of topologies on X, one easily checks that so is $T:=\cap_i T_i$ where a subset of X is open in T if and only if it’s open in all Ti. In short, an intersection of topologies is still a topology; so given any subset $S\subseteq \mathbf{P}(X)$, one can consider the collection of all topologies containing S (this is a non-empty collection since it includes the discrete topology)  and take their intersection.

Definition. Under the above definition, the resulting topology is called the topology generated by S. One also says that S is a subbasis for the topology T.

Put in another way, T is the “smallest” topology on X containing S, in the following sense:

• $S\subseteq T$;
• if T’ is any topology containing S, then $T\subseteq T'$.

[ This is similar to case where an arbitrary subset of a group or ring can be used to generate a subgroup or subring. ]

Theorem. The topology T generated by subbasis S has, as basis, $B(S) := \{W_1 \cap W_2 \cap \ldots \cap W_k : W_i\in S\} \cup \{X\}.$

[ Note: the additional term {X} may be superfluous if one interprets X as being the intersection of no Wi‘s. The more terms we intersect, the smaller the set becomes, so if we intersect no terms at all, we get the universal set X.]

Proof.

• The intersection of any two elements of B(S) still lies in it, so B(S) is indeed a basis for some topology T’.
• Since $S\subseteq B(S)\subseteq T'$, we have $T \subseteq T'$ too, since T is the smallest topology containing S.
• Finally, if $V=W_1\cap \ldots \cap W_k \in B(S)$, for some $W_i\in S$, then since $W_i\in T$, we have $V\in T$ as well. Thus, $T'\subseteq T$. ♦

In general, bases and subbases can be useful tools in topological proofs since to verify a property, it often only suffices to do it on basic open sets, or even subbasic open sets.

Summary. We have: $\text{Subbasis S} \stackrel{\footnotesize \begin{matrix}\text{take}\\ \text{finite} \\ \text{intersections}\end{matrix}}{\implies} \text{Basis }B \stackrel{\footnotesize\begin{matrix}\text{take}\\ \text{unions}\end{matrix}}{\implies} \text{Topology }T.$

Examples

1. On R, the set of intervals of the form (-∞, b) and (a, ∞) forms a subbasis; indeed, the intersection gives (-∞, b) ∩ (a, ∞) = (ab) or the empty set.
2. If |X|>2, then the collection of two-element subsets {ab} forms a subbasis since if abc are distinct, then {ab} ∩ {bc} = {a}.
3. For the topology on Z in Furstenberg’s proof, the set of V(ab) for prime power a forms a subbasis by Chinese Remainder Theorem.
This entry was posted in Notes and tagged , , , , , , , . Bookmark the permalink.