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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. $\blacksquare$

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 $(X, \tau)$ be a topological space and $A\subseteq X$. 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 ${X}$ is called Noetherian if it satisfies the descending chain condition for closed subsets: for any sequence ${Y_1 \supseteq Y_2 \supseteq \ldots}$ of closed subsets of ${X}$, there is an integer ${r}$ such that ${Y_r=Y_{r+1}=\ldots}$.

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 ${X}$ be the Noetherian space and ${U_\alpha}$ be an open cover of ${X}$; ${\alpha \in J}$. Define ${\mathcal S}$ to be the family of finite union of ${U_\alpha}$‘s. Then ${\mathcal S}$ is a family of open subsets of ${X}$ and hence must have a maximal element, ${U}$. I claim, ${U=X}$ because if there is an element ${x\in X\backslash U}$, then since ${U_\alpha}$ is an open cover of ${x}$, I can find an open set ${U_x}$ containing ${x}$. Now, ${U \cup U_x \in \mathcal S}$ since it is the finite union of elements of the cover of X, and contains ${U}$ properly. This is a contradiction to the maximality of ${U}$. Hence ${X=U}$, a finite union of elements of the open cover.$\blacksquare$

Lemma: (Proof easy) If ${Y}$ is any subset of a topological space ${X}$, then dim ${Y \leq }$ dim ${X}$.

Theorem: If ${X}$ is a topological space covered by a family of open subsets ${\{U_i\}}$, then dim ${X}$ = ${\sup}$ dim ${U_i}$.

Proof:

We only prove the case that dim ${X<\infty}$.

Using the above lemma, dim ${U_i \leq }$ dim ${X}$ is clear, and so taking supremum gives one side of the equality, ${\sup}$ dim ${U_i\leq}$ dim ${X}$. Consider the chain

$\displaystyle X_1 \subsetneq X_2 \subsetneq \ldots \subsetneq X_n$

of closed subsets of ${X}$, where ${X_1}$ is a singleton. We show that this chain is preserved when interescted with ${U=U_i}$, for some ${i}$. This would mean that dim ${X \leq}$ dim ${U_i}$. Consider
$\displaystyle (X_1 \cap U) \subseteq (X_2 \cap U) \subseteq \ldots \subseteq (X_n \cap U)$

and assume that ${X_j\cap U = X_{j+1}\cap U}$ for some ${j}$. Now,

$\displaystyle X_{j+1} = (X_{j+1} \cap U) \cup (X_{j+1} \cap U^c)$

$\displaystyle = (X_j \cap U) \cup (X_{j+1} \cap U^c)$

$\displaystyle \subseteq X_j \cup (X_{j+1} \cap U^c) \subseteq X_{j+1}$

Hence, ${X_{j+1} \subseteq U^c}$ or, ${X_{j+1} \cap U = \phi}$ and thus, ${X_1 \cap U = \ldots = X_{j+1} \cap U = \phi}$. In particular, ${X_1 \cap U = \phi}$ for every ${U\in \{U_i\}}$. But since ${\{U_i\}}$ is a cover of ${X}$, some ${U_i}$ must contain the singleton ${X_1}$, thus giving us a contradiction. Hence there exists a ${U_i}$ such that
$\displaystyle (X_1 \cap U_i) \subsetneq (X_2 \cap U_i) \subsetneq \ldots \subsetneq (X_n \cap U_i)$

thus proving that dim ${X\leq }$ dim ${U_i}$.   $\blacksquare$

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 $P$.

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!

* Reference:
The ordered pair (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 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.

Abhishek Parab

I? An Indian. A mathematics student. A former engineer. A rubik's cube addict. A nature photographer. A Pink Floyd fan. An ardent lover of Chess & Counter-Strike.

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