# $f: \mathbb{R} \to S^1$ where $f(x) = (\cos x, \sin x)$ open and closed mapping?

Show that $f: \mathbb{R} \to S^1$ where $f(x) = (\cos x, \sin x)$ is both an open and closed mapping, or provide counter-examples if one or both are not true.

Well, my hypothesis is that they are both true, that open sets in $\mathbb{R}$ map to open sets, and closed sets in $\mathbb{R}$ map to closed sets. I could not think of any counter-examples since yesterday, unless I am missing something.

Anyway, it is a trivial case if $f(X) = S^1$ since $S^1$ is both and open and closed in itself.

I feel intuitively that an open or closed interval will map to an open or closed arc segment on the given unit circle, respectively. But I don't know how to write this in mathematical form. Or this not true in general?

• This map is indeed an open and closed mapping. – Omnomnomnom Sep 2 '15 at 18:53
• Consider the set $f(\mathbb{Z})$. – JHance Sep 2 '15 at 19:01
• I take that back. $f$ is an open mapping, but not a closed mapping. – Omnomnomnom Sep 2 '15 at 19:03
• If $A$ is closed and bounded, then $f(A)$ is, indeed, closed. All counterexamples are unbounded. – Akiva Weinberger Sep 2 '15 at 19:07
• @JHance isn't the image of $f(\mathbb{Z})$ a closed set of isolated points on the unit circle? – mr eyeglasses Sep 2 '15 at 19:07

This map is not closed. Consider for example $F = \{2n\pi + \frac 1n: n = 1,2,\ldots\}$. It is open though. To prove it note that every open subset of $\mathbb{R}$ is a union of open intervals. Since an open interval is mapped onto an open arc (which is open if we think about $S_1$ with arc metric) or the whole circle, the image is a union of open sets which is open.

• How do I prove that an open interval is mapped onto an open arc? – mr eyeglasses Sep 2 '15 at 19:20
• In order to prove that it's an arc you can use the fact that the continuous image of a connected set is connected (as long as we believe that arcs are the only connected subsets of $S_1$). Now, we can check the cases where the arc is closed, open, or mixed to reach the conclusion. The other way to prove it would be by using the continuity of $\cos x$ and $\sin x$ to find a ball contained in the image for every its point - here we don't really need the fact that the image is an arc. – ajr Sep 2 '15 at 19:32
• I'm not sure why f(F) is not closed? – IntegrateThis Sep 22 '18 at 18:19
• $F$ is a sequence of points with the limit $(1,0)$. Then it is not closed, because $(1,0) \notin F$. – ajr Sep 24 '18 at 15:40

$f$ is indeed open.

To show that $f$ is open, it is sufficient to show that $f$ takes open intervals to open sets (since the open intervals form a topological basis).

In order to show that an open arc is an open set, it suffices to show that there is a continuous onto map from the arc to an open interval.

• Take the endpoints of the arc. Any circle that passes through those endpoints that is not the circle itself does not intersect the circle in any other point (a circle is determined by three points). Take an open disc bounded by such a circle; this is open in the plane and its intersection with the circle is the open arc, so the arc is open. – Matt Samuel Sep 2 '15 at 19:20
• To make it completely sound you'd have to figure out exactly why the intersection is the arc. You may have to bound the radius (can two circles be tangent at two points? I don't know.). You also have to restrict it to the smaller arc. In any case, I'm sure it can be made to work. The moral of the story is that math is hard. – Matt Samuel Sep 2 '15 at 19:35
• Ah! Easy. You can just insist that they aren't tangent. Then it works. Anyway, I'll stop posting tons of comments now. – Matt Samuel Sep 2 '15 at 19:43
• Arcs are parameterized by angles. An open arc is all points that are at angles strictly in between two angles. That help? – Matt Samuel Sep 2 '15 at 20:11
• $f(]-3\pi,3\pi[)$ seems very closed to me. – Michael Hoppe Sep 2 '15 at 20:12