Dense and Nowhere Dense Sets

Found a neat properties for dense and nowhere dense sets in functional analysis assignments recently. Suppose $(X,\tau)$ is a topological space, and $A \subseteq X$ is an arbitrary subset. We say that $A$ is dense if the closure is the whole space itself, i.e. $\bar{A} = X$; we say that $A$ is nowhere dense if the interior of its closure is empty, i.e. $\mathring{\bar{A}} = \emptyset$.

Now, we can also think about closures and interiors in terms of their complement. We shall denote by $X \setminus A$ to be the complement of $A$ instead of the usual $A^C$; the reason is obvious once we get to the main part, I promise.

Usually in topology, the interior and closure of a set $A$, are defined to be the union of all open sets contained in $A$, and the intersection of all closed sets containing $A$ respectively; in other words, they are the largest open set contained in $A$, and the smallest closed set that contains $A$ respectively. In terms of complement, we have the interior $\mathring{A} = X \setminus \bar{A}$ and $\bar{A} = (X \setminus A)^{\circ}$, this is not hard to prove.

If $A$ is dense, then the complement has empty interior. We have

$\bar{A} = X \iff X \setminus \bar{A} = \emptyset \iff (X \setminus A)^{\circ} = \emptyset$

This is not surprising and is pretty standard. On the other hand, we claim that $A$ is nowhere dense if and only if its complement contains an open dense subset of $X$. This is slightly more work, we have

$\mathring{\bar{A}} = \emptyset \iff \bar{X \setminus \bar{A}} = X \iff \bar{(X \setminus A)^{\circ}} = X \iff (X \setminus A)^{\circ} = X$

And since the interior itself is an open set, hence $(X \setminus A)^{\circ}$ is the open dense subset we desired.

This is pretty useful result when trying to prove Baire Category’s Theorem, or for an application of the Baire’s Category Theorem. All the best doing your assignments!

Cocountable Topology

Given a space $X$, there are many ways to define a topology. More common methods are one of the following:

• explicitly list out all the open sets
• define the rules for a subset $U \subseteq X$ to be open
• define the basis/subbasis of the topology (Note: The basis here refers to topological basis, this is different than a vector space basis that we have learnt in linear algebra.)

More often than not, we will work with topological spaces that has infinitely many open sets. In this case it will be hard to explicitly list out all the open sets in the topological space. In this post, we will look at a topology call cocountable topology, and luckily, we can describe the open set of this topology, thus we do not need to describe a topological basis.

Given a ground set $X$, the cocountable topology $\tau$ on $X$ is the empty set together with the collection of subsets $U \subseteq X$ such that $X \setminus U$ is countable, namely the complement of $U$ is countable. For example, if $X = \{1,2,3\}$, then under the cocountable topology, none of the subsets of $X$ is open.

Now let’s assume that $X$ is an uncountable set.  By definition, $\emptyset, X \in \tau$. Then, let $\{U_{\lambda}\}_{\lambda \in \Lambda} \subseteq \tau$ be a collection of open sets in $X$ for some index set $\Lambda$. Thus $X \setminus U_{\lambda}$ is countable for all $\lambda \in \Lambda$. Now consider

$X \setminus \bigcup_{\lambda \in \Lambda} U_{\lambda} = \bigcap_{\lambda \in \Lambda} [X \setminus U_{\lambda}] \subseteq X \setminus U_{\lambda_0}$

by De Morgan’s law, where $\lambda_0 \in \Lambda$ is arbitrary. Since $X \setminus U_{\lambda}$ is countable for all $\lambda \in \Lambda$ and intersections will only result in smaller set, therefore $X \setminus \bigcup_{\lambda \in \Lambda} U_{\lambda}$ must be countable. Thus we may conclude that $\bigcup_{\lambda \in \Lambda} U_{\lambda} \in \tau$.

Finally, let $\{U_i\}_{i=1}^{n} \subseteq X$ be a finite collection. Then by De Morgan’s law, we have $X \setminus \bigcap_{i=1}^{n} U_i = \bigcup_{i=1}^{n} (X \setminus U_i)$. Notice that finite union of countable sets is still countable (in fact, countable union of countable sets is countable), thus $X \setminus \bigcap_{i=1}^{n} U_i$ is countable and $\bigcap_{i=1}^{n} U_i \in \tau$.

Therefore $\tau$ is indeed a topology.

The Intersection of Topologies result in a Topology

Topology is the study of invariant properties of mathematical objects. Through the study of topology, mathematicians to translate properties between mathematical objects. There are many different subfields of topology, e.g. general topology (a.k.a. point-set topology), algebraic topology, etc. Unfortunately (or fortunately?), I have only encountered general topology in my undergraduate degree.

Perhaps one interesting (basic) result in topology is as follows

Given a space $X$ and a collection of topologies $\tau_i$, where $i \in I$ and $I$ is an index set, the intersection $\bigcap_{i \in I} \tau_i$ is also a topology on $X$.

Given a collection of objects that share property a property, let’s say property $a$, we do not expect the union and intersection of this collection to share the same properties. For instance, the set $\{n\}$, where $n$ is an integer, is finite. But when you take the union of all singleton set (namely $\{n\}$), we have an infinite set.

So why is the above statement true? To prove this, we need need at least the definition of a topology, so here it goes:

Let $X$ be a set, and $\mathcal{P}(X)$ be its power set. A subset $\tau \subseteq \mathcal{P}(X)$ is called a topology if the following conditions hold:

1. $\emptyset, X \in \tau$
2. Arbitrary union of elements of $\tau$ is also an element of $\tau$, i.e. $\bigcup_{\lambda \in \Lambda} U_{\lambda} \in \tau$ where $U_{\lambda} \in \tau$ is arbitrarily chosen.
3. Finite intersection of elements of $\tau$ is also an element of $\tau$, i.e. if $n \in \mathbb{Z}$, then $\bigcap_{i=1}^n U_{i} \in \tau$ where $U_{i} \in \tau$ is arbitrarily chosen

So far so good. Now that we know the definition of a topology, we can easily see that the union of topologies is NOT a topology; consider $X = \{a,b,c\}$, and we have two topologies $\tau_1 = \{\emptyset, \{a\}, X\}$ and $\tau_2 = \{\emptyset, \{b\}, X\}$, then the union is

$\tau_1 \cup \tau_2 = \{\emptyset, \{a\},\{b\},X\}$

This violates the arbitrary union property; $\{a\}$ and $\{b\}$ are elements of the union, but $\{a\} \cup \{b\} = \{a,b\}$ is not. Note that this is not to say that all unions of topologies does not result in a topology; it is only the counterexample to the claim that all unions of topologies gives a topology.

We have shown above that the union of topologies itself need not be a topology. So why is the intersection of of topologies is itself a topology? Clearly, by definition of topology, each topology includes the sets $\emptyset$ and $X$, thus the empty set $\emptyset$ and $X$ is in the intersection. Now let $I$ be an index set, and for $i \in I$, $U_{i} \in \bigcap_{\lambda \in \Lambda} \tau_{\lambda}$ be an arbitrary family of elements in the intersection of the topologies. Then by definition of topology, arbitrary unions of elements in $\{U_i\}_{i \in I}$ is also in $\tau_{\lambda}$ for all $\lambda \in \Lambda$. Thus the arbitrary union is also an element of $\bigcap_{\lambda \in \Lambda} \tau_{\lambda}$. By similar reasoning, we can also conclude that finite intersection of elements in $\bigcap_{\lambda \in \Lambda} \tau_{\lambda}$ is also an element of $\bigcap_{\lambda \in \Lambda} \tau_{\lambda}$ itself. Thus $\bigcap_{\lambda \in \Lambda} \tau_{\lambda}$ is a topology.