Is the function $f(x) = 1/x$ continuous? A function f is mapped from the non-zero reals to the reals . We assume the natural topology to be induced on the domain. Then is the function f(x) = 1/x continuous ?
EDIT Suppose I use this definition of continuity :
         The inverse image of any open set in the co-domain is an open set in the domain.
         Then what could I say about the inverse image of, let's say (-1,1) ?        
 A: Answer concerning your edit.
In full terms the preimage of $\left(-1,1\right)$ under function $f:\mathbb{R}\backslash\left\{ 0\right\} \rightarrow\mathbb{R}$,
prescribed by $x\mapsto\frac{1}{x}$, is set $\left(-\infty,-1\right)\cup\left(1,\infty\right)$.
This is evidently an open subset of $\mathbb{R}\backslash\left\{ 0\right\} $.
It can be shown that $f^{-1}\left(U\right)$ is open in $\mathbb{R}\backslash\left\{ 0\right\} $ if $U$ is open in $\mathbb{R}$ which means that $f$ is continuous.
A: For the topology, induced by a metric, a function $f$ is continuous if for every $x$ and every $\epsilon$, there exists such a $\delta $ that for all $y$, $d(x,y)<\delta$ implies $d(f(x)-f(y))<\epsilon$.
This is the standard definition of continuity known from mathematical analysis. Using tools  from analysis, it is not hard to show that $f$ is, indeed, continuous.
A: For example you can say for $x,y>1$ you have
$$
\left|\frac{1}{x}-\frac{1}{y}\right|=\left|\frac{x-y}{xy}\right|\leqslant|x-y|
$$
The general case follows by symmetry, since $1/x$ is its inverse.
