**Definition: **Let be a topology. A **basis** for is a subset of such that the following hold:

- The union of all elements of is .
- If then for every , there is a such that and .

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

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

**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 generate the Borel 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 indeed defines a topology on ” which is standard in mathematical literature can be misguiding. It means that the subbasis is actually a topology; the topology generated by the subbasis is the same as .

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 on as follows:

Define the basic open sets to be

where and .

I claim that the topology on 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 ; every integer lies in . Also, if , that is,

then the Chinese Remainder Theorem will yield a solution . Thus, and the set of basic open sets is closed under finite intersection. ( is countable and in our case, countable union further reduces to finite union because . 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 is closed as well. This is because

.

The previous statement can perhaps be better written as:

.

The Zariski topology on the Spectrum of a ring

Consider the set of all prime ideals of a ring ,

One defines a topology on this set of prime ideals (which is our now) as follows:

For any , define

Without loss in generality, we can assume that the subset is an ideal. (This can be seen by noting that the intersection of ideals is an ideal: Just intersect all elements of and get the prime ideal, say, ). So we shall now denote **points** of by for prime ideal rather than .

Consider now the topology generated by letting these 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 generated by these basic closed sets is itself, that is,

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

- Finite union:
- Arbitrary intersection: If are all prime ideals for some (possibly infinite) index set , then

**Remark: **A special interesting case of this is, that all open sets around 0 of a local ring are of the form . In particular, open sets around 0 of , the -adic integers are all of the form for some natural number .

**Remark: **A point , that is, a prime ideal is closed if and only if is not contained in other prime ideals, if and only if is a maximal ideal. For this reason, maximal ideals of are called closed points of the Zariski topology.

One point compactification:

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

**Proof of uniqueness: Suppose are two such spaces and are the special points of and respectively, then define the map as and . Verify that open sets in are mapped to open sets in .**

**Proof of existence: **** **Let . Topologize by calling those sets to be open in such that

- and open in OR
- but is compact in .

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

**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 ) 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 is the same as the collection of sets satisfying .

## 2 comments

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December 19, 2009 at 17:19

arungood post! I have a correction: chinese remainder works on relatively prime integers! ( there are easy counter examples with no solution)

If i remember correctly Bakul( it was their teams pet topology) just used good old euclid’s division algo in class to reduce x = r mod ac.

Unless of course, the Chinese have laid claim to the euclid’s division algo as well and call this the generalized chinese remainder whatever…

December 19, 2009 at 19:29

abhishekparabThe Chinese remainder needs to be applied suitably. After factorizing etc.