Subset of Meager Set is Meager Set
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be meager in $T$.
Let $B \subseteq A$.
Then $B$ is meager in $T$.
Proof
Since $A$ is meager in $T$:
- there exists a countable collection of sets $\set {U_n: n \in \N}$ nowhere dense in $T$ such that $\ds A = \bigcup_{n \mathop \in \N} U_n$.
Then, we have:
\(\ds B\) | \(=\) | \(\ds A \cap B\) | Intersection with Subset is Subset | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\bigcup_{n \mathop \in \N} U_n} \cap B\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \bigcup_{n \mathop \in \N} \paren {U_n \cap B}\) | Union Distributes over Intersection |
From Intersection is Subset:
- $U_n \cap B \subseteq U_n$
From Subset of Nowhere Dense Subset is Nowhere Dense:
- $U_n \cap B$ is nowhere dense in $T$.
Then, we see that:
- $B$ can be written as the union of nowhere dense sets in $T$.
That is:
- $B$ is meager in $T$.
$\blacksquare$