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I'm trying to show that $f:\mathbb{R}\to S^{1}$ ($S^{1}$ being the unit sphere in $\mathbb{R}^{2})$ defined by $$f\left(t\right)=\left(\cos\left(2\pi t\right),\sin\left(2\pi t\right)\right)$$ is open but not closed. This when you take $\mathbb{R}$ with the standard metric topology and $S^{1}$ with the subspace topology induced by $\mathbb{R}^{2}$ with the standard metric topology. I tried a bunch of stuff and haven't really managed to get anything substantial...

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1 Answer 1

What you have to do to show non-closedness is to find a closed set in $\mathbb{R}$ mapping to a non-closed set in $S^1$. So we need some closed but unbounded set in the reals (as closed plus bounded would be compact, and compactness is preserved under $f$, and then the image would be closed). The set $A = \{ n + \frac{1}{n}: n=2,3,4,\ldots \}$ will do: its images get closer and closer to $(1,0)$ without ever reaching it, so the latter point is in $\overline{f[A]} \setminus f[A]$.

As to openness: it might help to know that every open set in the reals is a disjoint union of open intervals. Think about what the image of an open interval on the circle is.

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I know the image of an open interval under $f$ is some portion of the circle, if the interval contains two distinct integers then the image is the entire circle and if not then it's just a portion of it. The thing I'm having difficulty with is finding a ball $B\left(x,\varepsilon\right)\subseteq\mathbb{R}^{2}$ such that $f\left[\left(a,b\right)\right]=S^{1}\cap B\left(x,\varepsilon\right)$ (thus the image will be open in the subspace toplogy) –  Serpahimz Mar 14 '13 at 22:36

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