# A topology problem (homeomorphisms)

1. Feb 10, 2007

### quasar987

1. The problem statement, all variables and given/known data
Our professor gave us a few exercises at the end of class the other day and it is possible that he wrote the problem wrong, or I copied it wrong, but in any case, something's fishy about it. It says

Show that there exists a topology on the set $\mathcal{D}$ of all lines of $\mathbb{R}^2$ that makes it homeomorphic to the circle $\mathbb{S}^1$.

3. The attempt at a solution

I think I found a proof that no such topology can exists on the simple ground that no bijection can exist btw $\mathcal{D}$ and $\mathbb{S}^1$. It goes like this...

1° Note that $\approx$ denoting isomorphisms of sets (i.e. $A\approx B$ means there exists a bijection btw A and B) is an equivalence relation on the set of all sets.

2° $\mathbb{S}^1$ as a set is isomorphic to $[1,0)$ via, say, $t\mapsto e^{2\pi it}$.

3° $\mathcal{D}\approx \mathbb{R}^2$ via $\{(x,mx+b): x\in\mathbb{R}\}\mapsto (m,b)$

4° $[[1,0)]_{\approx}\neq [\mathbb{R}^2]_{\approx}$. So because $\approx$ is an equivalence relation, $[\mathcal{D}]_{\approx} \neq [\mathbb{S}^1]_{\approx}$, i.e. there can exists no bijection btw these two sets, let alone homeomorphisms.

Last edited: Feb 10, 2007
2. Feb 10, 2007

### StatusX

There is a bijection between [0,1) and R^2. But I still think you're right, the problem was probably written wrong, and they meant the space of lines through the origin.

3. Feb 10, 2007

### quasar987

That's scary; what does it look like?

That's very possible. Our prof is particularly sloppy when it comes to distinguishing a space and the quotient space associated with some "natural" equivalence relation.

Last edited: Feb 10, 2007
4. Feb 10, 2007

### AKG

Clearly, there are at least as many lines as there are real numbers. On the other hand, any line can be determined by two points in the plane, so the number of lines is less than or equal to the size of R4. So

$$|\mathbb{R}| \leq |\mathcal{D}| \leq |\mathbb{R}^4| = |\mathbb{R}|$$

So $\mathcal{D} \approx [0, 1) \approx \mathbb{S}^1$. So there's a bijection from the set of lines to the circle. Therefore, it's obvious that there's a topology on the set of lines making the two homemorphic (choose a the topology induced by this bijection).

5. Feb 10, 2007

### quasar987

Hi AKG, thanks for the reply!

Your solution looks very simple and elegant but I can't follow it. In particular, what do you mean by "size"? And why do you evoke R^4 (why four)?

Please try to talk in the language I used in my OP if possible because I never had a proper exposure to the basic notions of set theory so there's a lot of holes in my knowledge of this area.

6. Feb 11, 2007

### quasar987

Ah, I see how to construct the bijection btw [0,1) and R². And now I have

$$\mathbb{S}^1\approx [0,1)\approx \mathbb{R}^2\approx \mathcal{D}$$

and the problem is over before it even began! :grumpy: (define a topology on $\mathcal{D}$ as {$\phi(U):U\in \tau_{\mathbb{S}^1}$} where phi is a bijection btw S^1 and D)

Btw - I'm still curious about AKG's R^4 argument and the |A| notation.

7. Feb 11, 2007

### d_leet

As to the notation |A| I'm fairly certain that that represents the cardinality of A.

8. Feb 11, 2007

### matt grime

The problem is most definitely not over before it began. You notional bijeciton is absolutely not continuous with continuous inverse until you prove it.

9. Feb 11, 2007

### quasar987

But this is automatic after one defines the topology on D as the images of the opens of S^1, is it not?

Does it make sense to compare the cardinality of two sets of infinite cardinality? I guess so; if a bijection can be found btw some subset of a set and another set, than the first can be said to have cardinality greater of equal to the second. But this does not explain $|\mathbb{R}^4| = |\mathbb{R}|$.

Last edited: Feb 11, 2007
10. Feb 11, 2007

### StatusX

You can prove that the cartesian product of a finite number of copies of a space with some infinite cardinality has the same infinite cardinality. There's even a continuous bijection from, say, R to R^2 (a space filling curve), however its inverse isn't continuous so it isn't a homeomorphism.

11. Feb 11, 2007

### AKG

|X| denotes the cardinality of the set X, which is the same thing as the size of X ("cardinality" is the technical term, "size" is the informal synonym of "cardinality"). From set theory, we have a very important fact which you will need many times in all math courses, so remember it:

$$|X| = |Y| \Leftrightarrow X \approx Y$$

where $X \approx Y$ means, by definition, that there exists a bijection from X to Y. It should be clear that, given a topological space (X,T) and a set Y:

"there exists a topology S for Y such that (X,T) is homeomorphic to (Y,S)"

iff

"|X| = |Y|"

The forward direction is obvious. For the reverse direction, we assume |X| = |Y|, so we know there's a bijection f from X to Y. We can use this f together with the topology T to induce a topology S such that f witnesses that (X,T) and (Y,S) are homoemorphic. This induced topology S = {f(U) : U is in T}.

So to prove what you're asked to prove, it suffices to show that $|\mathcal{D}| = |\mathbb{S}^1|$.

$\mathbb{R}^4 = \mathbb{R}^2 \times \mathbb{R}^2$, so $\mathbb{R}^4$ is the set of all pairs of points in the plane. And a pair of points in the plane determines a line in the plane. Each pair of points determines a unique line in the plane, but a line in the plane does not determine a unique pair of points in the plane (that is, if you pick two points in the plane, there is only one line that passes through both, but if you pick a line in the plane, you can pick many pairs of points through which that line passes).

Now why is $\mathbb{R} \approx \mathbb{R}^4$? You can in fact prove that if X is infinite and n is finite, then $X^n \approx X$. There might be a stronger result that one can prove, that allows n to be infinite in certain cases, but the above is enough for this problem. In particular, R is infinite and 4 is finite. Now if you're taking a course in topology, a fact like this should be something you can freely assume without proof. If not, I can sketch a proof:

We'll prove by induction that |Rn| = |R|. We need a key fact, which you should already know, and if not, be very sure to remember:

|A| < |B| iff there is an injection from A to B iff there's a surjection from B to A.

There's an obvious injection from R to R2. To get an injection going the other way, first let's agree that if x is a real number with two decimal expansions (meaning that it can end in all 0's or all 9's), we choose the one that ends in all 0's. Under this restriction, each real number has a unique decimal expansion. Now if (x,y) is an element in R2, then we can create a new real number by whose decimal expansion is made by taking alternate digits from x and y. For example, if we have:

x = 0.88888888..., y = 0.7373737373...

then the new number we make is:

0.8783878387838783...

It's not hard to prove that this is a well-defined injection, making sure we keep the restriction of only looking at the expansion that ends in all 0's when we have a number that gives us an option to look at an expansion ending with all 9's or all 0's. So we have an injection from R to [/b]R[/b]2 and an injection the other way. From the fact I mentioned, we get |R| < |R2| and |R2| < |R|. So |R| = |R2|. Next, assume that |Rk| = |R|. Then:

$$|\mathbb{R}^{k+1}| = |\mathbb{R}^k \times \mathbb{R}| = |\mathbb{R} \times \mathbb{R}| = |\mathbb{R}|$$

12. Feb 11, 2007

### quasar987

Awesome AKG, thank you very much.