Continuity of a function between two topological spaces What is the motivation behind the definition of continuity of a function between two topological spaces (i.e, $f : X \to Y$ is said to be continuous if $f^{-1}(v)$ is open for all open sets $v$ in $Y$)?
 A: Question concerning the motivation of something are often not easy to answer. Yes, the definition of continuity of $f$ by "preimages of open sets are open" is short and elegant, and this is what good definitions should be. But I am not sure whether it is optimal in a motivational sense.
In any textbook you will find various properties of a function $f : X \to Y$ which are equivalent to the continuity of $f$. One of these properties is

For all $M \subset X$, $f(\overline M) \subset \overline{f(M)}$.

Here $\overline{\phantom X}$ denotes closure. In my opinion this explains very nicely what continuity means. Given a set $M \subset X$, the closure $\overline M$ is the set of all points of $X$ which can be arbitrarily closely approximated by points of $M$. Now consider $x \in \overline M$. Then $f(x)$ is contained in the closure $\overline{f(M)}$, i.e. can be arbitrarily closely approximated by points of $f(M)$. In other words, $f$ preserves the approximability relation and does not tear up something.
A: This is only a comment to give details to the above comment of J. W. Tanner. You can prove easily the following:
Let $f:X\to Y$ be a map of topological spaces. The following are equivalent:
(i) For any open subset $V$ of $Y$, $f^{-1}(V)$ is open in $X$.
(ii) For any $x\in X$, and any open subset $V$ of $Y$ containing $f(x)$, there exists an open subset $U$ of $X$ containing $x$ such that $f(U)\subset V$ (continuity at $x$).
