Definition of a continuous function (through open sets) Why the definition of a continuous function defined in terms of inverse function? 
$f$ - continuous function, if for every open $V$, $f^{-1}(V)$ is open. 
 A: It's not defined in terms of an inverse function! If $f : X \to Y$ is not bijective, then an inverse function does not exist! The set $f^{-1}(V)$ is actually defined as
$$ f^{-1}(V) = \{ x \in X : f(x) \in V \}.$$
(And if $f$ is bijective, then this is the same as the image of $V$ under the inverse function $f^{-1}$.)
To give you some intuition, consider the example:
$$ f: \mathbb R \to \mathbb R, \ \ \ \ \ f(x) = \begin{cases} 0 & \ \  x < 0 \\ 1 & \ \ x \geq 0\end{cases}$$
This function is NOT continuous - it has a jump at $x = 0$. And indeed, if we take $V = (\tfrac 1 2, \infty)$ (which is open), then $f^{-1}(V) = [0, \infty)$ (which is NOT open).
It is possible to show that this definition with open sets is the same as the standard $\epsilon-\delta$ definition, as well as the sequential definition. The advantage of working with the open sets definition is that it generalises to arbitrary topological spaces, whereas the $\epsilon-\delta$ definition only generalises to metric spaces.
A: Because it is easier to use than the regular caculus like definition,
f is continuous iff for all x and open V nhood f(x), there be an open U nhood of x with f(U) subset V.   
The two defintions can be proven equivalent.
The later for real to real functions can be be shown equivalent to the usual epsilon delta definition.  Likewise, for R^n to R functions.
