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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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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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  • $\begingroup$ :Thank you so much for responding so quickly. I don't know why I am struggling with the concept of embedding. Everyone says it is easy, which frustrates me even more. Can you either explain an intuitive notion of embedding (the definitions confuse me) or point me towards a reference that has an intuitive explanation? Real quick: in your example, the y-cordinate of the image of f is always zero? Since arctan x takes all real numbers and spits them out to (-pi/2, pi/2), you used this. I am just wondering whether you can explain the motivation. Sorry if this is too basic. $\endgroup$ – user43901 Nov 3 '12 at 3:44
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    $\begingroup$ An embedding is a homeomorphism onto its image, so you have to look for homeomorphisms of the real line onto a bounded interval. Any strictly monotonic continuous function on the real line is a homeomorphism, so you only have to find a bounded one. And then just use any standard embedding of this interval into the plane. $\endgroup$ – Lukas Geyer Nov 3 '12 at 3:49
  • $\begingroup$ @user43901: I think the most intuitive way of looking at an embedding is as the reverse operation of taking a subspace. $\endgroup$ – Niels J. Diepeveen Nov 3 '12 at 3:58
  • $\begingroup$ 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. $\endgroup$ – user43901 Nov 3 '12 at 3:58
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    $\begingroup$ @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. $\endgroup$ – Brian M. Scott Nov 3 '12 at 16:59

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