Definition: Let (X,\mathcal T) be a topology. A basis for \mathcal T is a subset \mathcal B of \mathcal T such that the following hold:

  • The union of all elements of \mathcal B is X.
  • If B_1, B_2 \in \mathcal B then for every x\in B_1 \cap B_2, there is a B_x \in \mathcal B such that x\in B_x and B_x \subseteq B_1 \cap B_2.

Remark: If any of the two conditions mentioned above is not satisfied, then the collection \mathcal B is NOT a basis with respect to any topology on X. If however, both the conditions are met, then there is a unique topology on X whose basis is \mathcal B. It is called the topology generated by \mathcal B.

Definition: A collection \mathcal S of subsets of X whose union is X is said to be a subbasis of X. A topology generated by a subbasis \mathcal S is an arbitrary union of finite intersections of elements of \mathcal S.

Examples:

  • The collection of all (finite) open intervals of the real line form a basis as do the open intervals with rational endpoints. The former set is uncountable whereas the latter is countable. This emphasizes that two bases for a topology, unlike bases for a vector space need not have the same cardinality.
  • For readers familiar with measure theory, it is noteworthy that the collection of infinite open intervals (\infty,\alpha) : \alpha \in \mathbb R generate the Borel \sigma - algebra but is not a basis. (This might be attributed to the fact that allowing arbitrary unions and finite intersections is not so strong a condition as allowing countable unions and complements! )

Important Remark: The phrase “Verify that \mathcal T indeed defines a topology on X ” which is standard in mathematical literature can be misguiding. It means that the subbasis T is actually a topology; the topology generated by the subbasis T is the same as T.

Here are a few examples that may clarify the aforementioned point. I start with a legendary proof of the infinitude of primes due to Furstenberg.

    Theorem: There are infinitely many primes.
Proof: Define a topology \mathcal T on \mathbb Z as follows:

Define the basic open sets to be

S(a,b) := \{ a n + b : n \in \mathbb Z\} = a \mathbb Z + b where a,b \in \mathbb Z and a \neq 0.

I claim that the topology on \mathbb Z generated by these basic open sets is the set of these basic open sets and their unions (including the empty set as well).

For, they clearly cover \mathbb Z; every integer n lies in S(1,0). Also, if x \in S(a,b) \cap S(c,d) \neq \phi, that is,

x \equiv b(\mod a) \equiv d(\mod c)

then the Chinese Remainder Theorem will yield a solution x \equiv r (\mod ac). Thus, x \in S(ac,r) and the set \{ S(a,b): a,b \in \mathbb Z \text{ and } a \neq 0 \} of basic open sets is closed under finite intersection. (\mathbb Z is countable and in our case, countable union further reduces to finite union because \cup_{b=0}^{a-1} S(a,b) = \mathbb Z . This fact will be used very soon now.)

We now note the following two facts:

  • Every (non-empty) open set is infinite. (Proof obvious.)
  • Every basic open set S(a,b) is closed as well. This is because

S(a,b) = \mathbb Z \backslash \bigcup_{c\not\equiv b(\mod a)} S(a,c) .

The previous statement can perhaps be better written as:

S(a,b) = \mathbb Z \backslash \bigcup_{i=1}^{a-1} S(a,b+i).

Having made these two observations, we investigate the set \mathbb Z \backslash \bigcup_{p \text{ prime}} S(p,0).
From the set \mathbb Z of integers, we strike out multiples of 2, 3, 5, 7, 11, \ldots . What remain now is just \{-1,1\}. That is,
\{-1,1\} = \mathbb Z \backslash \bigcup_{p \text{ prime}} S(p,0)
Were primes finite, then the union of the basic open (hence closed) sets on the right would be closed and so its complement, \{ -1,1\} would be open. But no non-empty open set can be finite. What a way to get a contradiction!
\blacksquare
And now here is another one from Algebra:
The Zariski topology on the Spectrum of a ring

Consider the set of all prime ideals of a ring A,

\text{ Spec } (A) = \{ \mathfrak p : \mathfrak p \text{ prime in } A\}

One defines a topology on this set of prime ideals \text{ Spec } (A) (which is our X now) as follows:

For any B \subseteq A , define

\mathcal Z (B) := \{ \mathfrak p \in \text{ Spec } (A), \text{ and } B \subseteq \mathfrak p \}

Without loss in generality, we can assume that the subset B is an ideal. (This can be seen by noting that the intersection of ideals is an ideal: Just intersect all elements of \mathcal Z(B) and get the prime ideal, say, \mathfrak b). So we shall now denote points of X = \text{Spec }(A) by \mathcal Z(\mathfrak b) for prime ideal \mathfrak b rather than \mathcal Z(B).

Consider now the topology generated by letting these \mathcal Z(\mathfrak b) as basic closed sets. (The understanding for a ‘closed basis’ is that the topology is generated by arbitrarily intersecting and taking finite unions of these ‘basic closed sets’).

The claim here is that the topology \mathcal T generated by these basic closed sets is itself, that is,

\mathcal T = \{ \mathcal Z(\mathfrak b) :\mathfrak b\in \text{Spec }(A) \}

This is, by observing the following facts, the proof of which is simple and is left to the reader:

  1. A = \text{Spec } (0) \Rightarrow A \in \mathcal T
  2. \phi = \text{Spec} (A) \Rightarrow \phi \in \mathcal T
  3. Finite union: \mathcal Z(\mathfrak a) \cup \mathcal Z(\mathfrak b) = \mathcal Z(\mathfrak a \mathfrak b)
  4. Arbitrary intersection: If \mathfrak a_j: j \in \Lambda are all prime ideals for some (possibly infinite) index set \Lambda, then \cap_{j \in \Lambda} \mathcal Z(\mathfrak a_j) = \mathcal Z(\cap_{j\in \Lambda} \mathfrak a_j)

Remark: A special interesting case of this is, that all open sets around 0 of a local ring (A,\mathfrak m) are of the form \mathfrak{m}^n; n\in \mathbb N. In particular, open sets around 0 of \mathbb{Z}_p, the p-adic integers are all of the form p^n \mathbb{Z}_p for some natural number n.

Remark: A point \mathfrak p \in X, that is, a prime ideal \mathfrak p \in \text{Spec }(A) is closed if and only if \mathfrak p is not contained in other prime ideals, if and only if \mathfrak p is a maximal ideal. For this reason, maximal ideals of A are called closed points of the Zariski topology.

One point compactification:

Theorem: Let X be a locally compact Hausdorff space. Then there is a compact Hausdorff space Y, unique upto homeomorphism such that X is a subspace of Y and $Y \backslash X$ is a singleton.

Proof of uniqueness: Suppose Y,Y' are two such spaces and p,p' are the special points of Y and Y' respectively, then define the map f: Y \to Y' as f(x)=x and f(p)=p'. Verify that open sets in Y are mapped to open sets in Y'.

Proof of existence: Let Y:= X\cup \{ \infty \}. Topologize Y by calling those sets U \subseteq Y to be open in Y such that

  • \infty \in U and U open in X OR
  • \infty \not\in U but X\backslash U is compact in X.

I leave to the reader, the amazing verification that these declared open sets actually form a topology, that X is a subspace of Y and that Y is compact and Hausdorff. It is amazing in the sense that the hypotheses X being locally compact and Hausdorff are nicely used. Clean work. \blacksquare

Remark: Actually, this compactification example is not suitable in this context since there is no basis involved. But the point that I am trying to make is that We would like to call a few subsets of a given set (satisfying a particular property say P) to be open and prove that they satisfy the requirements of open sets of a topology rather than consider the topology generated by them. Indeed, another roundabout way of saying this is that the topology generated by sets satisfying P is the same as the collection of sets satisfying P.

Advertisement