# Homework Help: Finding topologies of sets in complex space

1. May 12, 2010

### complexnumber

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

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

$$X_2$$ is homeomorphic.

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

2. May 12, 2010

### 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?

3. May 13, 2010

### complexnumber

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?

4. May 13, 2010

### HallsofIvy

Yes, that is correct.

5. Jun 7, 2010

### alabastertard

I think X2 and X3 are homeomorphic.