Is the additive group of real numbers (R,+) compact?

Question

I have a very naive question about basic topology. My goal is to determine the conditions on a vector field in order for its flow to define a proper (R,+)-action. I understand that if the group G is compact, then the G-action is proper so I have to determine if (R,+) is compact. It seems to me that being homomorph to the (compact) circle group via the exponential function which is continuous, then (R,+) must be compact as well but I didn't find any explicit proof of this assertion. Can someone help me with that ? Thanks very much

More precisely, I want to use the following property: Let X and Y be two topological spaces, if $f:X\mapsto Y$ is continuous and X is compact, then $f(X)$ is compact. The circle Lie group $T=\{z\in \mathbb C|\, |z|=1\}$ is compact and there is an homomorphism (the exponential function) $exp:T\mapsto (R,+)$ which is continuous. Therefore, identifying X=T and Y=(R,+), (R,+) must be compact ?

Ok, now I see my (rookie) mistake. Thanks to everyone !

Answer 1

No, $\Bbb{R}$ is not compact (with the standard topology). The cover by open intervals $(-n,n)$ has no finite subcover.

Answer 2

As Chris Eagle pointed out, the group $G = (\mathbb{R},+)$ isn't compact. Your argument regarding the function $\varphi: \mathbb{R} \to S^1$ defined by $\varphi(x) = e^{ix}$ (or $\varphi(x) = (\cos x,\sin x)$) doesn't show that $G$ is compact, since the function $\varphi$ isn't a bijection (and hence isn't a homeomorphism).

Note, in your edit, the spaces $X$ and $Y$ are backwards; $\exp: \mathbb{R} \to S^1$ not the other way around. Also when you just say "the exponential function", it leads me (and most others, I assume) to think of either the map $x \mapsto e^x$ or the exponential map from a Lie algebra to its Lie group. So unless you're referring to one of those maps (which you're not in this case, since the former is a map $\mathbb{R} \to \mathbb{R}^{>0}$ and the latter is $\mathbb{R} \to \mathbb{R}$), you should probably be more specific.

On the other hand, the function $\varphi$ is locally injective and therefore a local homeomorphism. So if you wanted to, you could use it to argue that $G$ is locally compact (every point has a compact neighborhood), although there are easier ways to see this.