Sets of the First and Second Categories in a Topological Space

# Sets of the First and Second Categories in a Topological Space

Recall from the Dense and Nowhere Dense Sets in a Topological Space page that if $(X, \tau)$ is a topological space then a set $A \subseteq X$ is said to be dense in $X$ if the intersection of $A$ with all open sets (except for the empty set) is nonempty, that is, for all $U \in \tau \setminus \{ \emptyset \}$ we have that:

(1)
\begin{align} \quad A \cap U \neq \emptyset \end{align}

Furthermore, $A$ is said to be nowhere dense if the interior of the closure of $A$ is empty, that is:

(2)
\begin{align} \quad \mathrm{int} (\bar{A}) = \emptyset \end{align}

We will now look at two very important definitions regarding whether an arbitrary set $A \subseteq X$ can either be written as the union of a countable collection of nowhere dense subsets of $X$ or not.

 Definition: Let $(X, \tau)$ be a topological space. A set $A \subseteq X$ is said to be of The First Category or Meager if $A$ can be expressed as the union of a countable number of nowhere dense subsets of $X$. If $A$ cannot be expressed as such a union, then $A$ is said to be of The Second Category or Nonmeager.

Note that in general it is much easier to show that a set $A \subseteq X$ of a topological space $(X, \tau)$ is of the first category since we only need to find a countable collection of nowhere dense subsets, say $\{ A_1, A_2, ... \}$ (possibly finite) where each $A_i$ is nowhere dense such that:

(3)
\begin{align} \quad A = \bigcup_{i=1}^{\infty} A_i \end{align}

Showing that $A \subseteq X$ is of the second category is much more difficult since we must show that no such union of a countable collection of nowhere dense subsets from $X$ equals $A$.

For an example of a set of the first category, consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology of open intervals and consider the set $\mathbb{Q} \subseteq \mathbb{R}$ of rational numbers. We already know that the set of rational numbers is countable, so the following union is a union of a countable collection of subsets of $X$:

(4)
\begin{align} \quad \mathbb{Q} = \bigcup_{q \in \mathbb{Q}} \{ q \} \end{align}

Each of the sets $\{ q \}$ is nowhere dense. Therefore $\mathbb{Q}$ can be expressed as the union of a countable collection of nowhere dense subsets of $X$, so $\mathbb{Q}$ is of the first category.