TITLE: $(X \setminus F) \cap Y = Y \setminus A \Leftrightarrow A = F \cap Y$
QUESTION [0 upvotes]: Let $(X, \mathcal{T}), (Y, \mathcal{T}_Y)$ be topological spaces, where $Y \subset X$ and $\mathcal{T}_Y$ is the subset topology. Let $A \subset Y$. During a proof of something (it's not really important so I won't mention it), a particular comment has thrown me.

Let $F \subset X$. Then $(X \setminus F) \cap Y = Y \setminus A \Leftrightarrow A = F \cap Y$.

"This is set theoretical and easy to check."
How would I go about proving this? I don't really know where to start (I've tried considering complements of the complements, but I didn't really know what to do with it).

REPLY [2 votes]: Write $F = G \sqcup H$ where $\sqcup$ means disjoint union and
$$G = F\cap Y\\H = F\setminus (F\cap Y)$$
Then $X\setminus F = X\setminus (G\sqcup H) = (X\setminus G) \cap (X\setminus H)$.

Indeed, $x \in X\setminus (G\sqcup H)$ if and only if $x \in X$ and $x \notin G\sqcup Y$, that is, if and only if $x \in X$ and $x \notin G$ and $x \notin H$.

Now, $H \cap Y = \emptyset$, and since $Y\subset X$ we have that $Y \subset X\setminus H$. It follows that $(X\setminus H)\cap Y = Y$. Hence, $(X\setminus F) \cap Y = (X\setminus G) \cap Y$. In other words, that's the set of all $x \in X$ with $x \notin G$ and $x \in Y$. Once again, since $Y \subset X$ this reduces to $x \in Y$ and $x\notin G$, so $$(X\setminus F) \cap Y = Y\setminus G = Y \setminus (F\cap Y)$$
Finally, since $A \subset Y$, we have that $Y \setminus A = Y \setminus (F\cap Y)$ if and only if $A = F\cap Y$, which completes the proof.

REPLY [2 votes]: You can write taking $X$ as reference to any complementation $X\setminus F=F^\complement$ and $Y\setminus A=Y\cap A^\complement$ then finally you have that $F^\complement\cap Y=Y\cap A^\complement$.
The last equality can be re-written as $Y\setminus F=Y\setminus A$, then $Y\cap F=Y\cap A$, and cause $A\subset Y$ then finally $Y\cap F=A$.

The "hidden" double implication is $Y\setminus F=Y\setminus A \iff Y\cap F=Y\cap A$, and this can be proven seeing that $A\setminus B =A\setminus (A\cap B)$

Note: this is not a proof in the more elemental sense i.e. using logic operators and showing that some $x$ belongs to some set $A$. This is just using algebraic manipulations, in the same sense that when you solve any equation, to show the "iff" statement.
{"set_name": "stack_exchange", "score": 0, "question_id": 1791005}