I'm studying analysis and I've ran into this proposition saying that a function from a metric space X to a metric space Y, is continuous if and only if for every open set O in Y, the inverse image of O with respect to f is open in X. Similar statement with closed subsets of Y.
My question is - does f need to be onto for this to work? Should we only work with sets inside f(X)? I mean, consider f(x) = sinx from (-pi,pi) to [-1,1]. If we take the inverse of the closed interval [-1,1] do we get the open (-pi,pi)? What am I missing here?
Sorry for the very newbie way of expressing but its the first time I am asking a question here.
You do not need surjectivity of $f$. I assume what bothers you about the example of $\sin(x)$ is that the preimage of the closed interval $[-1, 1]$ is the open interval $(-\pi, \pi)$. However, you've chosen to restrict the domain of your function to the space $X = (-\pi, \pi)$: but this is a closed subset of $X$! ($X$ has the subspace topology from $\mathbb{R}$).
Clarification: Just to clarify the seemingly different answers of Nick and myself, we interpreted your problem two different ways. I assumed you were explicitly taking $(-\pi, \pi)$ to be the domain of $f$, and Nick was assuming you wanted the domain to be all of $\mathbb{R}$ and had a mistake in calculating the preimage of $[-1, 1]$.
As an even simpler example, let's consider the map $f : \mathbb{R} \to \mathbb{R}$ which just sends everything to $0$. This is a continuous map because the preimage of any open set is still open: If your open set you take the preimage of does not contain $0$, then the preimage is empty (which is open), but if you look at an open set which does contain $0$, the preimage is all of $\mathbb{R}$ (which is also open).