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Sohei YASUDA gave me a proof of a lemma whose proof is instructive and interesting, using many little facts from group theory and topology. I thought I should blog it.

**Lemma:** G is a compact topological group and H is an open subgroup. Then H is also closed (!) and of finite index.

**Proof.** G can be written as a union of cosets, each of which is open in G thanks to continuity of the group product. Thus the complement of H is a union of open (co)sets so is open. H is thus closed.

Consider now the quotient map from G to G/H. (Here, I’m not assuming H is normal in G but rather thinking of G/H as the quotient topological group with it’s topology coming from the quotient map). By definition of the quotient map, H cl-open in G implies it’s image, the identity coset of G/H too is cl-open.

Compactness of G inherits to the topology on G/H so the identity element is discrete on a compact set. The set G/H better be finite now, thus proving the second assertion.

I am really short of time, so let me summarize. I have joined the Institute of Mathematical Sciences for my PhD. Right now, I am desperately trying to avoid being succumbed to the pressure of grueling coursework here. But a mathematician is a masochist, so I am enjoying the mathematical agony!

A famous theorem in point-set topology due to Kuratowski was stated as an assignment problem in our topology class. With a hint from our Topology professor, I solved the problem which I regard as the best problem in point-set topology I ever solved yet. The problem seems too fascinating to be even true. It appeared ﬁrst in a paper due to Kuratowski and was made popular by Kelley in his book *General Topology* and can be found as a “starred” problem in Munkres’ *Topology*. I could solve it, not mainly because of the hint given by my professor but with the comfort that someone has already solved the problem. It does seem daunting to begin working on an open problem. Without further ado, I state the problem:

**Let be a topological space and . By iteratively applying operations of closure and complemention, one cannot produce more than 14 disjoint sets.**

The proof of this theorem is found in the adjoining section “Notes” or by clicking here.

Though the following was found in Hartshorne’s *Algebraic Geometry* (read, ‘terror maths’), this is really point-set topology.

**Definition: **A topological space is called Noetherian if it satisfies the descending chain condition for closed subsets: for any sequence of closed subsets of , there is an integer such that .

We have the following properties of Noetherian spaces:

**Proposition: **A Noetherian space is compact.

**Proof: **It is easy to show that the Noetherian property is equivalent to every nonempty family of open subsets having a maximal element. (The proof resembles a similar one in ring theory).

Let be the Noetherian space and be an open cover of ; . Define to be the family of finite union of ‘s. Then is a family of open subsets of and hence must have a maximal element, . I claim, because if there is an element , then since is an open cover of , I can find an open set containing . Now, since it is the finite union of elements of the cover of X, and contains properly. This is a contradiction to the maximality of . Hence , a finite union of elements of the open cover.

**Lemma:** (Proof easy) If is any subset of a topological space , then dim dim .

**Theorem:** If is a topological space covered by a family of open subsets , then dim = dim .

**Proof:**

We only prove the case that dim .

Using the above lemma, dim dim is clear, and so taking supremum gives one side of the equality, dim dim . Consider the chain

of closed subsets of , where is a singleton. We show that this chain is preserved when interescted with , for some . This would mean that dim dim . Consider

and assume that for some . Now,

Hence, or, and thus, . In particular, for every . But since is a cover of , some must contain the singleton , thus giving us a contradiction. Hence there exists a such that

thus proving that dim dim .

**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 .

Recently, I came upon an interesting topology*, the **VIP topology**. By interesting, I mean that it is useful for producing some pretty cool counterexamples.

It is defined as follows:* X is any (non-empty) set. a is a fixed element of X. The subset U of X is open if either it is empty, or if it contains a.*

It is easy to verify that the collection

**T(X)**of such open sets defined above is a Topology on X. (a is called the VIP for obvious reasons). The topology, called VIP topology satisfies the following properties:

- There always exists a
*basis of***T(X)***with cardinality*equal to**X**. Just take the basis to be**B(X)**= { {**a**,**x**} :**x**is an element of**X**}. Note that this topology is different from the Discrete Topology, which is generated by singletons as basic open sets.

- The VIP topology is
*T0*(i.e. Hausdorff) but not T½. Given any two distinct points**x**and**a**,there exists the open set {**a**} containing**a**but not**x**. A symmetric argument does not work if**x**and**a**are interchanged, which means that VIP is not T½.

- VIP is
*connected*. For, to exist separation into two disjoint open sets, each must contain**a**or one of them must be empty.

A similar topology is obtained if one replaces the necessity of ‘inclusion’ of **a** to ‘exclusion’. In other words, a subset **U** of **X** is open if it does not contain the (fixed) element **a** of **X** (unless **U** is the whole space **X**, just to avoid the triviality trap!) Such a topology (check!) is called the **Outcast Topology**. Just like the VIP, it has some interesting properties, one of them being that it is trivially a compact space!

**X**,

**T**(

**X**)) is a

*topological space*if

**T**(

**X**) is a subset of the power-set of

**X**containing the empty set, the whole set

**X**along with arbitrary unions and finite intersections of elements of

**T**(

**X**).

A (topological) space is *T0*, if for any two distinct points, there exists an open set containing one but not the other.

A space is *T½* if for two distinct points **x **and **y**, there exists an open set containing **x** but not **y**.

A space is *T1* if given two distinct points **x** and **y**, there exist disjoint open sets **U** and **V** containing **x** and **y** respectively. This is the same as Hausdorff space.

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