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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.

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    http://math.stackexchange.com/questions/43096/is-it-true-that-a-space-filling-curve-cannot-be-injective-everywhere2012-11-03

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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.

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    @user43901: Thank you; glad it helped!2012-11-03