I've been trying to show that $f:\mathbb R \rightarrow\mathbb R^2$; $x\mapsto(\cos(2\pi x),\sin(2\pi x))$ is continuous under the standard topologies on $\mathbb R$ and $\mathbb R^2$ and have had no luck doing so.
Any help proving this would be much appreciated.
Hint: The euclidian topology on $\mathbb R^2$ is equivalent to the product topology, which is continuous, iff $f$ is continuous in each factor. Therefore, it suffices to prove that $x \mapsto \cos(2 \pi x)$ is continuous, since $x \mapsto \sin(2 \pi(x))$ is just the composition $x \mapsto \cos(2 \pi(x)) \mapsto 1-\cos(x)^2$. The power series definition would suffice to prove the claim, but it is actually uniformly continuous
Since this is kind of a weird verification, you an instead use the map $x \mapsto e^{ix}$ which identifies $\mathbb R^2 \cong \mathbb C$, and show that this is continuous (either by definition, power series, etc.)
one other approach might be: this claim is equivalent to showing that the map $f:\mathbb R^2 \to S^1$, where $S^1$ is the unit circle is continuous with respect to the subspace topology. Remove the point $(1,0) \in S^1$ and argue geometrically that this is now a homeomorphism from the open interval $(0,1) \to S^1 \setminus\{(1,0)\}$, and the map extends continuously to the closure (this is in fact a quotient map for $\mathbb R/\sim$ where $x \sim y \iff x-y \in \mathbb Z$.)
