Can there be a embedding from the real number line to the plane such it is both closed and bounded?

Just a little clarification on the question: when I say line I am essentially referring to the real number line and the plane being $\mathbb R^2$.

I am really new to the notion of embedding; hence, I am having difficulty envisioning the process. I know that $f$ embeds a (compact) metric space M onto N if $f$ is a homeomorphism, which means both $f \text{ and } f^{-1}$ exists and are continuous.

I mean it is easy to see that the real number line is definitely a closed subset of $\mathbb R^2$ and but can anyone give me an example of a homeomorphic $h$ that will embedd the real line onto the plane in a bounded way? I know the unit circle does not work.

If someone could kindly explain the concept first and then provide me with some handle on the problem, I will be grateful

SO the final point I want to reach to prove that there cannot be a embedding from the real line to the plane such it is both closed and bounded.

But there are understanding in intermediate steps, which I am lacking. If someone can point me to some resource that explains the concept, that will be great too.

-

A topological embedding is easy, e.g., $f(x) = (\arctan x, 0)$ embeds $\mathbb{R}$ into $\mathbb{R}^2$ with image $(-\pi/2,\pi/2) \times \{0\}$. If you want to preserve the metric (or preserve it up to multiplicative constant) then the image can obviously not be bounded because $\mathbb{R}$ is not bounded. Similarly, if you want a topological embedding with a closed and bounded image, that won't work either because it would be compact, and $\mathbb{R}$ is not.
I see, the point of the monotonic cont. function hit home with me. I can see why now you used the arctan function. What is the significance of the singleton {0} in your example. Will it not suffice to say that $f(x)$ is just what you gave instead showing the Cartesian product? The online resources on embedding is very jargon-ish, which turns me off. Hence, if you can direct me towards any resource from where you learned these notions, let me know. – user43901 Nov 3 '12 at 3:58
@user43901: Suppose that there were a homeomorphism $f:\Bbb R\to K$ such that $K$ is a compact set. Then $f^{-1}:K\to\Bbb R$ would also be a homeomorphism, and hence continuous. The continuous image of a compact set is compact, and $\Bbb R=f^{-1}[K]$, so $\Bbb R$ would be compact. Bur $\Bbb R$ isn’t compact, so there can be no such homeomorphism. More succinctly, compactness is a topological property, meaning that it is perserved by homeomorphisms. Thus, if $X$ and $Y$ are homeomorphic, either both are compact, or neither is compact. – Brian M. Scott Nov 3 '12 at 16:59