I am encountering some difficulties and would like to ask for a hint. I have a continuous function$$f:R\rightarrow R: f(x)=x^2$$ and am supposed to find preimages of the following intervals
$$1.(a,b) , 0\leq a< b $$ $$2.(a,b) , a < b \leq 0$$ $$3.(a,b) , a \leq 0 <b$$
I am wondering how the function could spit out a negative value as needed in 2. where a is supposed to be strictly smaller than 0.
I apologize, i should add that I do not believe equation 2 to have a preimage, however part 4 of the exercise is then the following:
With the previous results show that $f$ is (topologically) continuous. (Recall that any open set in R can be represented as a union of intervals of the form (a, b) with a < b.).
As I understand, $f(x)=x^2$ is continuous topologically and therefore i don't believe my answer of equation 2 not having a preimage to be correct. Would you be so kind to explain if my understanding is incorrect?
On the plus side, you are thinking correctly about the preimage. This is, roughly, why the definition of a topology requires $\varnothing$ to be open.
For your case (2), you have that $(a,b)$ consists of purely negative numbers. Hence, as you suspected, $f^{-1}(a,b) = \varnothing$. This is, of course, open.