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 $P$.