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Prove that the subspace $S^{1}=\{{(x,y)\in \mathbb{R^{2}} : d((x, y),(0,0))=1}\}$ with $d$ the euclidien metric on $\mathbb{R^{2}}$ is locally euclidean.. Let $(x,y)\in S^{1}$ such that $(x,y)\neq (1,0)$.

Then $(x,y)\in S^{1}\setminus{\{(1,0)\}}=\mathbb{R^{2}}\setminus\{(1,0)\}\bigcap S^{1}$ open in $S^{1}$ and we know that $S^{1}\setminus{\{(1,0)\}}$ is homeomorphic to $]0,2\pi[$.

Now for $\{(1,0)\}$, I can't seem to find an open subset of $S^{1}$ that contains $(1,0)$ and homeomorphic to some open subset of $\mathbb{R}$..any help please.

Paul Frost
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Abdellatif Ouhaddou
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    I"m guessing for $S^1\setminus{(1,0)}$ being homeomorphic to $(0,2\pi)$, you used the polar coordinates/ complex exponential $t\mapsto e^{it}$? Well, now try to do a similar thing with $S^1\setminus {(-1,0)}$ and $(-\pi,\pi)$. – peek-a-boo Apr 01 '21 at 15:01
  • Yes I used the function $(\cos(\theta),\sin(\theta))$ – Abdellatif Ouhaddou Apr 01 '21 at 15:04
  • ok, that's equivalent to what I wrote about $e^{it}$. Now use the exact same functions for the interval I specified (and draw a picture to convince yourself of why this works). – peek-a-boo Apr 01 '21 at 15:04
  • Well proving the bijection and continuity is trivial.. The problem is proving that the function is open – Abdellatif Ouhaddou Apr 01 '21 at 15:18
  • Then how did you prove it for the case of $S^1\setminus{(1,0)}$? – peek-a-boo Apr 01 '21 at 15:23
  • I didn't actually prove that part I only found a prove that uses the concept of compactness so I just assumed it because the book that I study the subject from has not presented compactness yet – Abdellatif Ouhaddou Apr 01 '21 at 15:26
  • well, I mean continuity of the inverse is also pretty easy because you can even write down an explicit formula. For example, in the first quadrant, the inverse is $\arctan\left(\frac{y}{x}\right)$ (i.e this maps the open first quadrant continuously onto $(0,\pi/2)$). In any other quadrants you should be able to find similar trigonometric formulae. – peek-a-boo Apr 01 '21 at 15:29
  • Yes I did calculate the inverse function but I did not prove the continuity because I think it's not easy to prove it using the topological definition of continuity – Abdellatif Ouhaddou Apr 01 '21 at 15:32
  • Don't use the definition directly. YOu should invoke all the theorems you know: products and compositions of continuous functions are continuous (such theorems shouldn't be too hard to prove directly from the topological definition). Also quotients are continuous (whenever the denominator is non-zero) etc. Use all these theorems (also, inverse trig functions are continuous wherever they're defined etc). – peek-a-boo Apr 01 '21 at 15:33
  • Okey thank you I'll give it a try – Abdellatif Ouhaddou Apr 01 '21 at 15:35

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You say you know that $S^{1}\setminus{\{(1,0)\}}$ is homeomorphic to $]0,2\pi[$. Let us assume that the existence of a homeomorphism $h : S^{1}\setminus{\{(1,0)\}} \to ]0,2\pi[$ is all we know. But now we have an obvious homeomomorphism $\phi : S^1 \to S^1, \phi(z) = -z$. It restricts to a homeomorphism $\psi : S^{1}\setminus{\{(-1,0)\}} \to S^{1}\setminus{\{(1,0)\}}$. Hence $h \circ \psi : S^{1}\setminus{\{(-1,0)\}} \to ]0,2\pi[$ is a homeomorphism.

See also my answers to Is it an open function? and Open sets on the unit circle $S^1$ .

Paul Frost
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