Let $X,Y \subset \mathbb{R}^n$ and prove that if $\dim X < \dim Y$ then; \begin{equation} \dim(Y \setminus X) = \dim Y \end{equation} The trivial case is easy (where $Y \cap X = \emptyset)$, since then we would simply have $Y \setminus X = Y$.
When $Y \cap X \neq \emptyset$, I thought the best way to show this case would be by proving the two inequalities separately. For the inequality $\dim(Y\setminus X) \leq \dim Y$:
We know that $\dim(Y\setminus X) = \dim(Y \cap X^c)$, so intuitively, I would like to be able to show that: \begin{equation} \dim(Y \cap X^c) = \inf\{\dim Y, \dim X^c\} \end{equation}
Since then this inequality would follow immediately. Not too sure how to go about this though, or how to begin with the other inequality. Any help and thoughts are appreciated!