Suppose that we have a vector space $X$ and two vector subspaces $A, B$ such that $A \subset B$. Is the following generally true? $$ X / A \supset X/B $$
Can we show this via equivalence classes? By this I mean given some $x \in X$ can we show $[x] \in X/B \implies [x] \in X/A$ where $[x]$ is the equivalence class generated by $x$.
Edit: The motivation for this question was trying to understand a step in the proof in Lax - Functional Analysis on pg. 13. For some operators $M$ and $G$ it claims because $R_M \supset R_{I+G}$ this implies $$codim(R_M) \leq codim (R_{I + G})$$
Here $R_M$ denotes the range of $M$.
No. In the case that $A\subsetneq B$ we cannot say $X/A \supseteq X/B$, because they have different equivalence relations, hence different equivalence classes. This is a set-theoretic obstacle.
Unfortunately the assignment $i:X/B \hookrightarrow X/A$ sending the equivalence class $x+B$ to the equivalence class $x+A$ is not welldefined either. For an element $b \in B\setminus A$ we would have to have $$0+A = i(0+B) = i(b+B) =b +A \neq 0+A$$ where the last inequalities holds as $b\notin A$.