Meagre Sets: Algebra Let meagre subsets be defined as:
$A\text{ meagre}\iff A=\bigcup_{\lvert K\rvert\leq\aleph_0} A_k\text{ with }\overline{A_k}°=\varnothing$
Then it satisfies:
$B\subseteq A\text{ meagre}\Rightarrow B\text{ meagre}$
$A_k\text{ meagre}\Rightarrow\bigcup_{\lvert K\rvert\leq\aleph_0} A_k\text{ meagre}$
Flipping this around, is it possible to define meagre subsets by these properties?
 A: Given a set $X$, we say that $I$ is a $\sigma$-ideal on $X$ if the following holds:


*

*$I\subseteq\mathcal P(X)$, is not empty and $X\notin I$.

*If $A\in I$ and $B\subseteq A$, then $B\in I$.

*If for $n\in\Bbb N$ we have $A_n\in I$ then $\bigcup A_n\in I$. (If we weaken this just for finitely many, we have an ideal, without the $\sigma$.)


The meager sets form a $\sigma$-ideal. But so do the zero measure sets (in the Lebesgue measure, for example). So do the countable subsets of $X$, if $X$ is uncountable.
The properties that you suggest are exactly those of a $\sigma$-ideal, but those are not unique at all to meager sets.

One can extend this question and ask, given a $\sigma$-ideal $I$ on an uncountable set $X$, is there a topology where $I$ is exactly the meager sets of the topology?
Well, there is, but it's not a very exciting topology. We declare $\tau$ to be the topology $\{A\subseteq X\mid X\setminus A\in I\lor A=\varnothing\}$. Then $\tau$ is a topology on $X$ as it includes $X$ and $\varnothing$, closed under finite (and countable) intersections and arbitrary unions (because it's closed under supersets).
Note now that given $A\subseteq X$ then $\overline{A}=X$ if and only if $A\notin I$ if and only if $\overline{A}\notin I$.
Moreover all the closed sets, except $X$, have an empty interior. Therefore a set is meager if and only if it is closed and not equal to $X$. 
A: The property "being a set" (not very demanding!) verifies:
$B\subset A$ is a set $\Rightarrow$ $B$ is a set.
$A_k\text{ is a set}\Rightarrow\bigcup_{\lvert K\rvert\leq\aleph_0} A_k\text{ is a set}$
A: If $A$ is meager, then there is a sequence of nowhere dense subsets $\{A_n\}$ such that $A=\bigcup_{n=1}^\infty A_n$. Now $B=B\cap A=\bigcup_{n=1}^\infty(B\cap A_n)$. We just need show that each $B\cap A_n$ is nowhere dense, and this will finish the proof for the first.
But we can show something a little more general. Let $A$ be nowhere dense and $B\subseteq A$. Then $B$ is nowhere dense. Proof: $B\subseteq A$ implies $\overline{B}\subseteq\overline{A}$. And this implies $\overline{B}^\circ\subseteq\overline{A}^\circ=\varnothing$. Thus
$\overline{B}^\circ=\varnothing$. Thus $B$ is nowhere dense.
