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The question I'm looking at is as follows:

Prove that there is an embedding of the line as a closed subset of the plane, and there is an embedding of the line as a bounded subset of the plane, but there is no embedding of the line as a closed and bounded subset of the plane.

My understanding of embedding is that it needs to be a homomorphism from $ \mathbb{R} \to f(x) \in \mathbb{R}^2$. I.e. the entire number line needs to be in $ \mathbb{R}^2$ in some shape or form after the transformation.

My thoughts for the closed subset are simply $ f(x) : x \to (1,x) $ as this is is effectively the identity function plus one dimension. It is closed as all 0 limits are contained, and unbounded as the Cauchy sequences do not converge as x approaches $ - \infty $ and $ \infty $.

For bounded $ f(x) : x \to (arctanh(x),x) $ on $ (-1,1) $ which encodes the entire number line, has Cauchy sequences converging at limits, but does not contain $ x = -1 $ or $ x = 1 $.

Are these intuitions correct for these parts of the question, or am I misinterpreting embedding as a concept? Are there significantly simpler answers? I feel like I'm missing something.

liqiudilk
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    Your intuition and examples are good. –  Sep 24 '15 at 04:36
  • On the second example, do you mean for $x\mapsto(\arctan x,0)$? The second coordinate being $x$ will go to $\infty$ as $x\to\infty$, hence the image wouldn't be bounded. – Clayton Sep 24 '15 at 04:53
  • Ah. I'd tried to avoid that with $ x \in (-1,1)$. I thought that restricting the domain was fine as long as the entire number line was translated anyway. Is $ x \to (arctanh(x),0) $ just a vertical line in the plane? Also, it's worth noting that I chose arctanh and not arctan. Is this a poor choice? If so, why? – liqiudilk Sep 24 '15 at 04:59

1 Answers1

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The first example is fine (essentially embed the reals as a straight line in the plane), which gives us an embedding $e$ of $\mathbb{R}$ into $\mathbb{R}^2$ where $e[\mathbb{R}]$ is closed.

To make it bounded, use the essentially same idea, using that $f(x) = \arctan(x)$ is a homeomorphism between $\mathbb{R}$ and $(-\frac{\pi}{2},\frac{\pi}{2})$ which is a bounded open interval. To get it into the plane, add a fixed coordinate: $e(x) = (\arctan(x),0)$, which has the bounded image $(-\frac{\pi}{2},\frac{\pi}{2}) \times \{0\}$ and is still a homeomorphism between $\mathbb{R}$ and its image. Adding $x$ as a second coordinate makes it unbounded again (it is a homeomorphism still).

If there were some embedding $e: \mathbb{R} \rightarrow \mathbb{R}^2$ where $e[\mathbb{R}]$ (which is homeomorphic to $\mathbb{R}$ by definition of an embedding!) would be closed and bounded, and $e[\mathbb{R}]$ would be compact by the Heine-Borel theorem, which $\mathbb{R}$ is not (and compactness is preserved by homeomorphism...).

Henno Brandsma
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