Basic Theorems Regarding Sets of the First Category in a Topo. Sp.

# Basic Theorems Regarding Sets of the First Category in a Topological Space

Recall from the Sets of the First and Second Categories in a Topological Space page that if $(X, \tau)$ is a topological space then a set $A \subseteq X$ is said to be of the first category (or meager) if $A$ can be expressed as a countable union of nowhere dense subsets.

Furthermore, we said that $A$ is of the second category (or nonmeager) if $A$ cannot be expressed in such a manner.

We will now look at some basic theorems regarding sets of the first category in a topological space.

Theorem 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$ be of the first category. Then for any $B \subseteq A$ we have that $B$ is of the first category. |

**Proof:**Let $A \subseteq X$ be of the first category. Then $\displaystyle{A = \bigcup_{i \in I} A_i}$ where each $A_i$ is nowhere dense in $X$ and $I$ is a countable indexing set. If $B \subseteq A$, then $B \cap A_i$ is nowhere dense for each $i \in I$ since $B \cap A_i \subseteq A_i$. Moreover, we see that:

\begin{align} \quad B = \bigcup_{i \in I} (B \cap A_i) \end{align}

- This shows that $B$ is equal to a countable union of nowhere dense sets. So, $B$ is of the first category.

Theorem 2: Let $(X, \tau)$ be a topological space. If $\{ A_i \}_{i \in I}$ is a countable collection of sets of the first category ($I$ is a countable indexing set) then $\displaystyle{A = \bigcup_{i \in I} A_i}$ is of the first category. |

**Proof:**Let $\{ A_i \}_{i \in I}$ be a countable collection of sets of the first category. Then for each $i \in I$, $A_i$ is equal to a countable union of nowhere dense sets. Say $\displaystyle{A_i \bigcup_{j \in J_i} A_{i,j}}$ where $J_i$ is a countable indexing set for all $i \in I$, and each $A_{i, j}$ is nowhere dense. Then:

\begin{align} \quad A = \bigcup_{i \in I} A_i = \bigcup_{i \in I} \left ( \bigcup_{j \in J_i} A_{i, j} \right ) \end{align}

- So $A$ is a countable union of a countable union of nowhere dense sets, which is countable. This shows that $A$ is of the first category. $\blacksquare$