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On page 86 of John Lee's Introduction to smooth manifolds there is an example of an injective immersion that is not a topological embedding:

$\beta : (-\pi, \pi) \to \mathbb{R}^2$, defined by $\beta(t) = (\sin{2t}, \sin{t})$, or pictorially:

Figure taken from the above-mentioned book

It is explained that, although $\beta$ is an injective immersion, it is not a smooth embedding since the image is compact while the domain is not. My understanding is that the image, while bounded in $\mathbb{R}^2$, is an open subset of the plane, whereas the statement claims that it is not.

Would anyone please explain why the image is compact? Thank you.

Balarka Sen
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Behrooz
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    Even if you think that the image is not closed, because you are confused bout what happens at the origin, you should see that it is certainly not open. – Carsten S Dec 21 '16 at 18:14

5 Answers5

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The image is literally a leminscate in $\Bbb R^2$.

It's clearly not open as if you take a point on the leminscate, any small neighborhood of it in $\Bbb R^2$ gets outside the curve (i.e. hits the complement). It's in fact closed, because a leminscate is a level curve which are closed because they are preimage of $0 \in \Bbb R$ by a continuous function.

As you noted, it's bounded, so that guarantees compactness.

Balarka Sen
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  • But don't we need the lemniscate to be open in the subspace topology? Here the open sets would be arcs on the figure eight. By this definition, it is open. Not as a subset of $\mathbb R^2 $ but as a subspace. Right? – gary Oct 29 '18 at 00:01
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First proof: If $\beta(t_n)$ is a sequence of points in the image, the sequence $t_n$ is bounded in $\mathbb{R}$, hence there is a subsequence $t_{n_p}$ that converges to a $t \in [-\pi, +\pi]$. By continuity of the sine, $\beta(t_{n_p})$ converges to $(\sin(2t), \sin(t))$, which is equal to $\beta(t)$ if $t\in (-\pi, \pi)$ and to $\beta(0)$ otherwise. Thus, every sequence in the image has a subsequence that converges in the image, which is the definition of compactness.

Second proof: Let $\gamma$ be the map $t \mapsto (\sin(2t), \sin(t))$ from $[-\pi, \pi]$ into $\mathbb{R}^2$ The image of $\gamma$ is the same as $\beta$, hence it is the image of a compact set by a continuous map.

Gribouillis
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  • Gribouillis, the domain of $\beta$ is the open interval $(-\pi, \pi)$. – Behrooz Dec 21 '16 at 18:55
  • Yes but a sequence in $(-\pi, \pi)$ has a limit point in $[-\pi, \pi]$. Similarly $\gamma$ as a different domain but the same image, because $\gamma(\pi) = \gamma(-\pi) = \beta(0)$ – Gribouillis Dec 21 '16 at 19:23
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It contains all its limit points, so it is a closed subset of $\mathbb{R}^2$. Since it is bounded as well, by the Heine-Borel theorem, it is a compact subset of the plane $\mathbb{R}^2$.

Behrooz
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The image is a solid figure eight: the only point of contention is the origin, but the "hole" between the abutting open ends is "plugged" by the middle of the curve.

Unit
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Note that interior of $\beta ((-\pi,\pi)) \subset \mathbb{R}^2$ is empty. So it cannot be an open subset of $\mathbb{R}^2$. Now take any open cover $\{U_i\}_{i \in I}$ of the image. For some $j \in I$, we must have $0 \in U_j$. Clearly, the rest of the image can be covered by finitely many $U_k$'s where $k \in I$. So we started with any open cover and find a finite subcover. That is, image is compact subset of $\mathbb{R}^2$.

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