homeomorphic $\sin \frac{1}{x}$ and $\mathbb{R}$ Is it really true that graph of the function
$$f(x)=\begin{cases}\sin \frac{1}{x}& x\neq0\\0&x=0\end{cases}$$
homeomorphic to $\mathbb{R}$?
This function is not continuous, else it is true.
What's the main ways of prooving non-homeomorphic we have?
 A: This is false. The problem is that the graph of $f(x)$ is not path-connected, whereas $\mathbb{R}$ is, and path-connectedness is a topological invariant. You can prove the graph of $f(x)$ is not path-connected directly, or you can get it from the fact that the "topologist's sine curve" is not path-connected.
In response to the second question, this is the whole point of topological invariants. The easiest way to show two spaces are not homeomorphic is to show that a certain topological invariant differs between them. Ones I've used frequently in undergraduate topology courses are compactness (in its various forms), connectedness (including path and simple), Hausdorff property, Euler characteristic; these are worth keeping in mind.
You should almost never try to show directly from the definition that "there exists no continuous map with continuous inverse between $X$ and $Y$." Just by virtue of being a nonexistence statement, it's hard to prove without some simplification, and topological invariants are precisely that simplification --- they make this proof an easy (easier, anyway) verification.
