Preimage of sets, complement of sets, continuity of functions I just got some simple questions in real analysis regarding preimage and complement of sets and continuity.


*

*Suppose $f:X\to Y$,
then does $f^{-1} (Y\setminus F)=f^{-1} (Y)\setminus f^{-1} (F)=X\setminus f^{-1} (F)$ always hold? But why the first equality is true?

*Can we rewrite the definition of continuity of a function in metric space as  $f:X\to Y$ is continuous at $a \in X \Longleftrightarrow $ given $\epsilon >0$, there exists $\delta >0$ such that$f(B^{0}(a,\delta))\subset B^0(f(a), \epsilon)$, where $B^0$ represents open balls.


Thank you.
 A: *

*The first equality is true:
\begin{align*}
f^{-1}(Y\setminus F)&=\{x\in X:f(x)\in Y\setminus F\}\\
&=\{x\in X:f(x)\in Y\text{ and }f(x)\not\in F\}\\
&=\{x\in X:f(x)\in Y\}\cap\{x\in X:f(x)\not\in F\}\\
&=\{x\in X:f(x)\in Y\}\cap\left(X\setminus\{x\in X:f(x)\in F\}\right)\\
&=f^{-1}(Y)\cap\left(X\setminus f^{-1}(F)\right)\\
&=f^{-1}(Y)\setminus f^{-1}(F).
\end{align*}
The second equality is also true because $f$ being a function implies $f^{-1}(Y)=X$. Note that in general the preimage acts as we would wish with most set operations.

*This is true. I guess your definition of continuity in a metric space goes like this: we say $f:X\to Y$ is continuous at $a\in X$ if
$$
\forall\epsilon>0,\exists\delta>0,\forall x\in X,d_X(x,a)<\delta\implies d_Y(f(x),f(a))<\epsilon.
$$
But the condition
$$
\forall x\in X,d_X(x,a)<\delta\implies d_Y(f(x),f(a))<\epsilon
$$
is readily seen to be equivalent to
$$
f(B^0(a,\delta))\subset B^0(f(a),\epsilon).
$$

A: Yes to both. To see why the first is true, observe that $x\in f^{-1}(Y\setminus F)$ if and only if $f(x)\in Y\setminus F$ if and only if $x\notin f^{-1}(F)$.
