# Thread: How can I show

1. ## How can I show

That subset of a set of 1st category (can write as countable union of nowhere dense sets) is also 1st category? Intuitively it is clear, but I am not sure how to argue formally

Thank you for helping

2. Originally Posted by MMath09
That subset of a set of 1st category (can write as countable union of nowhere dense sets) is also 1st category?
If X is of category I then $\displaystyle X = \bigcup\limits_{n \in \mathbb{N}} {A_n }$ where each $\displaystyle {A_n }$ is nowhere dense.

$\displaystyle B \subseteq X\; \Rightarrow \;B \subseteq \bigcup\limits_{n \in \mathbb{N}} {A_n } \; \Rightarrow \;\bigcup\limits_{n \in \mathbb{N}} {\left( {B \cup A_n } \right)}$

3. Originally Posted by Plato
If X is of category I then $\displaystyle X = \bigcup\limits_{n \in \mathbb{N}} {A_n }$ where each $\displaystyle {A_n }$ is nowhere dense.

$\displaystyle B \subseteq X\; \Rightarrow \;B \subseteq \bigcup\limits_{n \in \mathbb{N}} {A_n } \; \Rightarrow \;\bigcup\limits_{n \in \mathbb{N}} {\left( {B \cup A_n } \right)}$
Is this last bit supposed to be $\displaystyle X = \;\bigcup\limits_{n \in \mathbb{N}} {\left( {B \cup A_n } \right)}$? So you write the set X as the union of the set X and the subset B

So if B was made up of somewhere dense sets or an uncountable number of sets then you could write X like that too & that would be a contradiction

OK thanks