Show that if $U$ is open and $A$ is closed, then $U\setminus A = \{ x\in U : x\notin A \}$ is open. What can be said about $A\setminus U$
I dont quite get why $U\setminus A = \{x\in U : x\notin A\}$ is open?
If $x\in U$ and $x\notin A$ then isn't $U\setminus A$ with just be $U$?
They dont even have common element in the set?
When they dont have common element, isnt $A\setminus U$ will just be the same???
Thanks
Hint: rewrite $U\setminus A=U\cap A^c$. Now if $A$ is closed, what does that mean for the complement $A^c$?
$U \setminus A$ is, in words, all those elements of $U$ that aren't element of $A$. So if $A$ and $U$ were disjoint then this would be equal to $U$, but certainly not in general. If $A = X$ (where $X$ is the whole space) then no elements of $U$ wouldn't be elements of $A$ and the set would be empty. So it does depend on $A$.
Now $U \setminus A = U \cap (X \setminus A)$ (this is immediate from my description in words) and then we have written it as the finite intersection of an open set $U$ and the complement of a closed set $A$, so another open set. So the result must be open.
A set is closed iff its complement is open.
The intersection of two open sets is open.
$A$ is closed in $X$ $\implies X - A$ is open in $X$
$U$ is open in $X$
So $(X- A) \cap U$ is open in $X$ ($\because$ finite intersection of open sets is open)
$\implies (X \cap U) - (A \cap U)$ is open in $X$
$\implies U - (A \cap U)$ = $ U - A$ (can be seen using the venn-diagram)
$\implies U - A$ is open in $X$
similarly for $A - U$
$X - U$ is closed in $X$
$\implies A \cap (X - U)$ is closed in $X$ ($\because$ arbitrary intersection of closed sets is closed)
$\implies$ $(A \cap X) - (A \cap U) = A - (A \cap U) = A - U$
Hence, $A - U$ is closed in $X$.
