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) ?
Best Answer
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.