We shall give the proof by which following (1) is eqivalent to (2) .
Given two topological spaces $S,T$ and the function $f:S\rightarrow T$ .
If $f$ is continuous,
(1) for an arbitrary open set $D_1\in T$ , $f^{-1}(D_1)$ becomes open set in $S$ .
(2) for an $x\in S$ , let $f(x)=y$ be. For the arbitrary neighbourhood $B_{\epsilon}(y)\in T$
$f^{-1}(B_{\epsilon}(y))$ becomes the neighbourhood of $x$ in $S$ .
$(1)\rightarrow (2)$
If $B'(y)$ is a neighbourhood of $y$ , there is an $\epsilon$ such that
$y\in B_{\epsilon}(y)\subset B'(y)\in T$ .
Here, as we have accepted (1), $x\in f^{-1}(B_{\epsilon}(y))\subset f^{-1}(B'(y))$ ,
and both open sets are neighbourhoods.
$(2)\rightarrow (1)$
Let $Y\in T$ and $f^{-1}(Y)=X\subset S$ be. If $x\in X$ and $f(x)=y$ ,
as $y\in Y$ , there is a neighbourhood $B_{\epsilon}(y)$ which is in $Y$ .
Therefore, $f^{-1}(B_{\epsilon}(y))$ is a neighbourhood of $x$ and in $S$ .
As the neighbourhood is a open set, (1) is gotten.
[proposition]
A set $O$ is open if and only if, for an arbitrary element $x$ in $O$ ,
$O$ becomes a neighbourhood of $x$ .
