Finding topologies of sets in complex space

[complexnumber]

Homework Statement

Consider the following subsets of $$\mathbb{C}$$, whose
descriptions are given in polar coordinates. (Take $$r \geq 0$$ in
this question.)
$$\begin{align*}<br /> X_1 =& \{ (r,\theta) | r = 1 \} \\<br /> X_2 =& \{ (r,\theta) | r < 1 \} \\<br /> X_3 =& \{ (r,\theta) | 0 < \theta < \pi, r > 0 \} \\<br /> X_4 =& \{ (r,\theta) | r = \cos 2\theta \}<br /> \end{align*}$$
Give each set the usual topology inherited from $$\mathcal{C}$$.
Which, if any, of these sets are homeomorphic?

Homework Equations

The Attempt at a Solution

$$\tau_1 = \varnothing$$. $$\tau_2 = \{ B(z,r') \cap X_2 | r'<br /> > 0 \}$$. $$\tau_3 = \{ B(z,r') \cap X_3 | r' > 0 \}$$. $$\tau_4 =<br /> \varnothing$$.

$$X_2$$ is homeomorphic.

Are my answers correct? I am not sure if the topologies I wrote make sense at all.

 

[HallsofIvy]

Do you even understand what the words you are using mean?

It makes no sense to say that "$X_2$ is homeomorphic". The word "homeomorphic" does not apply to a single set. Two sets are "homeomorphic" is one can be continuously changed into the other. That is, we can shrink, stretch, or otherwise deform a set but we cannot cut, tear, or otherwise break it. In particular, you cannot change the dimension of a set and you cannot change the number of times it intersects itself. While you can change the size of a set with a "continuous" transformation, you cannot change whether it is bounded or not.

Do you mean $\tau_1, \tau_2, \tau_3, \tau_4$ to be the topologies on each of these sets? If so, $\tau_1$ and $\tau_4$ are certainly incorrect! A topology must include the entire set and the empty set at least so a toplogy is never empty. In any case, it is not necessary to write out the topologies- they are just the usual topology on $R^2$ restricted to the given set.

$X_1$ is the circle with center at (0, 0) and radius 1. It is a one dimensional closed path.

$X_2$ is the disk with center at (0, 0) and radius 1. it is a two dimensional, bounded, set.

$X_3$ is the upper half plane. It is two dimsensional and unbounded.

$X_4$ is a "four petal rose". It is one dimensional and intersects itself.

Which, if any, of those sets are homeomorphic to each other?

 

[complexnumber]

Do you even understand what the words you are using mean?

It makes no sense to say that "$X_2$ is homeomorphic". The word "homeomorphic" does not apply to a single set. Two sets are "homeomorphic" is one can be continuously changed into the other. That is, we can shrink, stretch, or otherwise deform a set but we cannot cut, tear, or otherwise break it. In particular, you cannot change the dimension of a set and you cannot change the number of times it intersects itself. While you can change the size of a set with a "continuous" transformation, you cannot change whether it is bounded or not.

Do you mean $\tau_1, \tau_2, \tau_3, \tau_4$ to be the topologies on each of these sets? If so, $\tau_1$ and $\tau_4$ are certainly incorrect! A topology must include the entire set and the empty set at least so a toplogy is never empty. In any case, it is not necessary to write out the topologies- they are just the usual topology on $R^2$ restricted to the given set.

$X_1$ is the circle with center at (0, 0) and radius 1. It is a one dimensional closed path.

$X_2$ is the disk with center at (0, 0) and radius 1. it is a two dimensional, bounded, set.

$X_3$ is the upper half plane. It is two dimsensional and unbounded.

$X_4$ is a "four petal rose". It is one dimensional and intersects itself.

Which, if any, of those sets are homeomorphic to each other?

Thank you very much for detailed explanation about homeomorphic sets. I did not know those rules and thought the question is asking which set is homeomorphic to the complex space $$\mathbb{C}$$.

None of these sets are homeomorphic to each other. Is it correct?

 

[HallsofIvy]

Yes, that is correct.

 

[alabastertard]

I think X2 and X3 are homeomorphic.