Proving Continuity of f: [0,1) to S^1

[latentcorpse]

how do i show that $f: [0,1) \rightarrow S^1; t \rightarrow (\cos{2 \pi t}, \sin{2 \pi t})$ is a continuous bijection but not a homeomorphism.

so i know i have to
(i) show f is a continuous bijection
(ii) show $f^{-1}$ is not continuous

so for (i),
surjectivity:
if i take a point $(a,b) \in S^1$ then f is surjective if $\exists t \in [0,1)$ such that $\cos{2 \pi t}=a$ and $\sin{2 \pi t}=b$

solving these for t i get $t=\frac{1}{2 \pi} \arccos{a}$ and $t=\frac{1}{2 \pi} \arcsin{b}$

how does this help prove surjectivity though? surely i am trying to find a unique value for $t \in [0,1)$?

then injectivity:
say i take $t_1 \neq t_2$, i need to show $f(t_1) \neq f(t_2)$ but i keep getting confused...how do you show this?

then for continuity:
can i just say that since both cos and sin are continuous functions over the given domain then f will be continuous or do i have to do more work, an $\epsilon - \delta$ proof perhaps?

thanks guys.

 

[Office_Shredder]

For surjectivity, you don't need a unique value of t (that's injectivity) but since t is between 0 and 1 anyway, you will get one.

For injectivity work backwards. Prove that f(t₁)=f(t₂) implies t₁=t₂. Remember that if (a,b) = (c,d) then you get that a=c and b=d

To show that f^-1 is not continuous, you want to find an open set in your domain that maps to a non-open set in your codomain. Hint: The endpoints of your domain were chosen very specifically.

 

[latentcorpse]

ok surely i do need a unique t for surjectivity because if

$t=\frac{1}{2 \pi} \arccos{a}$ and $t=\frac{1}{2 \pi} \arcsin{b}$ gave different values of t then there would be no value of $t \in [0,1)$ that mapped to $(a,b) \in s^1$?

is the following sufficient for injectivity:

let $f(t_1)=(a,b),f(t_2)=(c,d)$
assume $f(t_1)=f(t_2)$ then
$a=c,b=d$
then $\cos{2 \pi t_1}=\cos{2 \pi t_2}$ and $\sin{2 \pi t_1}=\sin{2 \pi t_2}$
$\Rightarrow t_1=t_2$

there is a hint for the continuity part : use the fact that open arcs are a basis for the topology on $S^1$.
i guess I'm supposed to make use of that...

so the domain is [0,1). i guess if i want to pick an open subset of that i could use something like $(t,1), t \neq 0$. is that what you meant?

 

[Office_Shredder]

For surjectivity: It's OK if more than one value of t maps to the same point (a,b) on your circle, you just need that there is at least one

 

[latentcorpse]

no i meant that if say t1 mapped to a with the cos (2 pi t) and t2 mapped to b with the sin (2 pi t) then what value of t maps to (a,b)? none surely? then it wouldn't be surjective?

 

[latentcorpse]

ok. so I've done the bijectivity. what do i do to show its continuous. can i just say sin and cos are continuous?

 

[Landau]

then $\cos{2 \pi t_1}=\cos{2 \pi t_2}$ and $\sin{2 \pi t_1}=\sin{2 \pi t_2}$
$\Rightarrow t_1=t_2$

This is correct, but I hope you did not conclude t1=t2 from $\cos{2 \pi t_1}=\cos{2 \pi t_2}$ separately, and from $\sin{2 \pi t_1}=\sin{2 \pi t_2}$ separately. After all, $$\sin(2\pi t_1)=\sin(2\pi t_2)$$ implies $$t_1=t_2$$ or $$t_1=\pi-t_2$$.

For continuity, yes: sine and cosine are continuous, from which it follows that f is continuous.

 

[madness]

How can you prove anything without specifying the topology on the domain or codomain?

 

[latentcorpse]

ok so id have to say that $\sin{(2 \pi t_1)}=\sin{(2 \pi t_2)} \Rightarrow t_1=t_2$ or $t_1=\pi - t_2$
and $\cos{(2 \pi t_1)}=\cos{(2 \pi t_2)} \Rightarrow t_1=t_2$ or $t_1=\pi + t_2$
and so $t_1=t_2$ is the solution that satisfies both simultaneously.
how's that?

what about the need to specify a topology on either $[0,1)$ or $S^1$?

ok. as for finding the open set in [0,1) that maps to a non open set in S^1.
could i use something like (a,1)?

 

[Landau]

ok so id have to say that $\sin{(2 \pi t_1)}=\sin{(2 \pi t_2)} \Rightarrow t_1=t_2$ or $t_1=\pi - t_2$
and $\cos{(2 \pi t_1)}=\cos{(2 \pi t_2)} \Rightarrow t_1=t_2$ or $t_1=\pi + t_2$
and so $t_1=t_2$ is the solution that satisfies both simultaneously.
how's that?

That's ok. You could also divide the equations to obtain $$\tan{(2 \pi t_1)}=\tan{(2 \pi t_2)}$$. Since tan is bijective on our interval, this implies t1=t2.

 

[latentcorpse]

ok thanks. can i use the interval (a,1) to show that the inverse isn't continuous or did u have a different set in mind?

 

[mathstime]

could someone possibly go through the surjectivity again in more detail for the slower students amongst us (ie: myself??!)

 

[Office_Shredder]

what about the need to specify a topology

I think you're safe assuming that the topologies are the metric topologies that you're familiar with.

ok thanks. can i use the interval (a,1) to show that the inverse isn't continuous or did u have a different set in mind?

Wrong endpoint... the image of that is going to be everything in the circle on the arc between (1,0) (corresponding to your endpoint 1) to whatever point a is mapped to, and won't include either. So you can see you still get an open set in S¹

The key here is that a set of the form [0,a) is open even though it 'looks' like it isn't; this is because there aren't any points in your interval to the left of 0 so small open balls around 0 are in fact contained in [0,a)

Now consider what the image of [0,a) is

 

[latentcorpse]

well the image of [0,a) is the open arc from (1,0) to whatever a is mapped to.
but in this case it includes the point (1,0).
why does this make it not open? is it do with open balls again? i think I am getting confused with the defn of open in terms of balls and the defn of an open set.

thanks.

 

[Office_Shredder]

Basically, an open set is one in which if a point is in the set, then points nearby are near the set

An open ball around a point x is just all the points within some distance epsilon > 0, NOT including the boundary (points that are epsilon away).

For the circle case, it has what's called an induced topology. Basically, a set in S¹ is open if it's the intersection of S¹ with an open set in R². It happens to be that this reduces to something easy to work with... a set in S¹ is open if given any point in the set, points nearby it on the arc are also in the set.

So looking at the image of [0,a), (1,0) is in the image, but points below it (small negative y value) are not in the image, which means the image of that set is not open

 

[Office_Shredder]

could someone possibly go through the surjectivity again in more detail for the slower students amongst us (ie: myself??!)

Ok, so we want to find t between 0 and 1 so that given (a,b) a²+b²=1, then cos(2pit)=a, sin(2pit)=b. Taking arccosine and arcsine, we see that

2pi*t = arccos(a), 2pi*t=arcsin(b).

The way that I would solve this is that by the definition of cosine and sine, given a point on the unit circle there MUST be some value x so that (cos(x),sin(x)) is that point, for 0<=x<2pi. So pick t=x/2pi

 

[Landau]

The key here is that a set of the form [0,a) is open even though it 'looks' like it isn't; this is because there aren't any points in your interval to the left of 0 so small open balls around 0 are in fact contained in [0,a)

To be precise (and to avoid confusion), I would say that [0,a) is open in the domain [0,1).

Basically, an open set is one in which if a point is in the set, then points nearby are near the set

You mean in