topological spaces 5

Given two distance spaces $D,E$ . If the function

$f:x\in D\rightarrow f(x)=y\in E$

is continuous, for an $x_n\rightarrow x\quad (n=1,2,\cdots)$

$f(x_n)\rightarrow f(x)\quad (n\rightarrow \infty)$ .

On the other hand, on topological spaces $S,T$,
if the function $f(S\rightarrow T)$ is continuous,

for an arbitrary open set $C_1\in T$ , $f^{-1}(C_1)$  becomes the open set
in $S$ .

This definition is equivalent to next two definitions.

For an arbitrary closed set $D_1\in T$ , if the function is continuous,
$f^{-1}(D_1)$  becomes  the closed set in $S$ .

For $x\in S$ , let $f(x)=y\in T$ be. For the arbitrary neighbourhood $B_{\epsilon}(y)\in T$ ,
$f^{-1}(B_{\epsilon}(y))$ becomes the neighbourhood of $x$ in $S$ .

As no one will have the questions for the equality, the important thing is a proof.