Looking at this theorem of continuous function by Bert Mendelson in "Introduction to Topology" ch 3, theorem 5.3:
A function $f: (X, T) \rightarrow (Y, T')$ is continuous if and only if for each open subset $O$ of $Y$, $f^{-1}(O)$ is an open subset of $X$.
This follows from the definition of continuous functions which is continuous at each point, after he defined continuous at a point.
However, if I just focus on the theorem:
Now let's take as an example the function $f(x) = 1/x$, and an open subset of $Y$ which I choose to be $(-1, 1)$, the preimage of this subset is $(-\infty, -1) \cup (1, \infty)$. Now, as far as I'm concerned this is an open subset of $\mathbb{R}$. Then for it not to be continuous I should be able to find an open subset of $Y$ such that the preimage is not an open subset of $X$. However, I cannot find such a subset.
I would like to understand why $1/x$ is not a continuous function on $\mathbb{R}$ using open subsets (not an epsilon and delta answer)
Best Answer
As pointed out in the comments, the function $f(x)=1/x$ is not a function on $\Bbb{R}$; it's a function on $\Bbb{R}\backslash\{0\}$, and in fact it is continuous on this domain.
When people say "$1/x$ is discontinuous", what they mean is that there is no continuous extension of this function to $\Bbb{R}$. For instance, let $g : \Bbb{R} \rightarrow \Bbb{R}$ be defined by $g(0)=0$ and $g(x)=1/x$ for $x \neq 0$. You should be able to show with the open sets definition that this function is discontinuous.