I'm focusing on $\frac 1 {x^2}$, but any function with a "jump" or going to infinity at one point works.
I was trying to convince myself about the definition of continuity using the topological definition.
It says that for any open set, the pre image has to be open. So i was thinking that looking at the graph of an uncontinuous function should show me a problem. Nevertheless, visualy, I do not see anything.
Here, with $\frac 1 {x^2} $, if I take an open set (for example an open interval) on the y-axis, I can always get an open set on the x-axis…
what am I missing ?
I have the same problem for cases with jumps. So I found out that in the case where one side of the jump is closed (so the limit on one side of the gap exists, meaning the function is continuous on the left for example), I get the contradiction I was expecting ! If I take an open set where the upper bound lies right beneath the second side of the function which is left continuous, I get a closed interval ! (contradiction).
But in the case (like on the graph beneath), where the middle point is in a totally different place and so the function is neither continuous on the left nor on the right, I, again, can't find a problem…
Please, what's going on ?
I tried to be as clear as I could. If, nevertheless, I haven't been able to make it consistent, please feel free to ask question in the comment section.
Best Answer
The function $f: x \mapsto \frac{1}{x^2}$ is actually continuous as $0$ is not in the domain. If you set $f(0) = 0$, you get for example $$f^{-1}((-1,1)) = (-\infty,-1) \cup (1,\infty) \cup \{0\},$$ which is not open.