Topology of a circle

Question

I'm currently trying to learn topology by myself, before I started I knew topology had something to do with shaped stretched but not torn or glued. Now that I have started to learn I see talk about open and closed sets, neighborhoods, limits and such, but I don't understand how this relates to my previous understanding of topology.

I would be interested to know what is the topology, or the topological space of a circle (or the equivalent of a circle in topology, as a circle is the same as a square and many other things).

Thanks.

Answer 1

Boy, I had the exact same experience.

When you talk about "shapes" "stretched" and "not torn" and "glued" in math you have to formally define what they are.

Okay. A "shape" is a bunch of points that are packed together solidly somehow. When we stretch it we are coming up with a way to map all the points, x, in the first shape, to points, y, in the second stretched shape so that if two points are near or connected to each other in the first shape, they are near or connected to each other in the second. What's more we can map every step of the way of the stretch with an "in between" mapping where near and connected points stay near and connected, and no points don't "jump" in and glue themselves to places they weren't before. So a stretch is a continuous mapping.

And this is where the concept of open sets, and neighborhoods, come in handy. A neighborhood can be thought of as all the points that are near a point x. Now we can stretch the neighborhood out so it is a million miles long but we can't "rip" it. Some some section of the neighborhood of a point will remain in a neighborhood of the point after stretching and that'd be true of all points. An open set is a "shape" where all points are in a neighborhood that is inside that shape. If we stretch the shape all the points will remain inside neighborhoods. No point will be cut or pushed to the edge. Open sets get mapped to open set.

So make sense of this a "topology" is a set of points and a set of rules that tell us which sets are open. A toplogy can be stretched any way but the stretching of open sets will result in open sets.

Okay, the circle. Known as $S_1$. Note: there are 3 different things that can be colloquially called a circle. We need to be precise about which ones we mean. It can mean: a) All the points that are equal distance from a center. This is edge of a disc. This is the correct and understood meaning of a circle. This is topologically equivalent to the perimeter of a square. Techically a circle is $\{(x,y)|x , y \in R, (x-a)^2 + (y-b)^2 = r^2\}$. It can be stretched and enlarged or shrunk but it can't be "cut". No matter how you contract it you can always find a path from a to b but there will always be a "hole" in the center. You can not contract a path "around" the circle to a single point.

b) Incorrectly it can refer to all the points on the edge and inside the circle. We call this a closed disc and it is topologically equivalent to a square or rectangle or triangle or whatever (A square technically includes the points inside it; a circle doesn't) with edge. This is "solid" and can be contracted down to a single point which a circle can not. But this will always have have an boundary of a bunch of points that will always have neighborhoods consisting of points in the disc and outside the disc. This boundary, no matter how stretched, will always be topologically equivalent to a circle.

c) Or we could incorrectly refer to all the points inside the disc but not the edge. This is an open disk. It can, surprisingly, be stretched infinitely into and is therefore topologically equivalent the entire $\mathbb R^2$ plane.

Answer 2

Topolgy deals with notions of convergence and the translation into other spaces of these properties.

Let $(X, \mathcal{O}_X)$ and $(Y, \mathcal{O}_Y)$ topological spaces. A function is called continuous, if the pre-image of every open set is open. More formally:

$f\colon (X,\mathcal{O}_X) \rightarrow (Y,\mathcal{O}_Y)$ is continous (in all $x \in X$) $\iff$ $\forall O \in \mathcal{O}_Y : f^{-1}(O) \in \mathcal{O}_X$

The cool thing here to notice: There is a map between sets and between structures defined by open sets!

In some exercises, you can see that this defintion of continous coincides with the ones seen in real analysis (as a special case). There are some more subtle defintions using filters or nets to the equivalent formulations, which you can have a look at when the time comes.

A homeomorphism is a function from $X \rightarrow Y$, which is continous and where the inverse function $f^{-1}$ is continous as well. Topological properties are properties which are preserved under homeomorphisms. Continuity forbids you to cut things, but to transform them by stretching.

Notice here that the definition for continuity totally depends on the topologies $\mathcal{O}_X$ and $\mathcal{O}_Y$, and therefore on open sets! For example:

Have a look at $f\colon (X,\mathcal{O}_X) \rightarrow (Y,\mathcal{O}_Y)$. If you use the $\mathcal{O}_X= \tau_{discrete}$, denoting the discrete topology, you can prove as an easy exercise that no matter what the topolgy $\mathcal{O}_Y$ is, the function is continous. Or: If $\mathcal{O}_Y = \tau_{trivial}$, then the function is continuous by the above defintion as well for all $\mathcal{O}_X$. Making the $Y$-topolgy smaller and the $X$-topology bigger will certainly make things continous. But the trivial topology is quite uninteressing. For instance if you like to have a concept like neighbourhood, this is way too rough. You need more sets. The question topologists ask now is: I want to have a look at two spaces with certain topological properties (like convergence, compactness, connectedness, separability, etc.). Can I still get a homeomorphism between the two?

So, if you like to equip the unit-ball with a topology so that it is homeomorphic to the square, you have to ask yourself: Which system of open sets do I have to take? Can I find a bicontinous function then? Have a look here, how that is done: Prove a square is homeomorphic to a circle

The idea here is that there are topologies induced by metrics. And there are plenty of metrics in $\mathbb{R}^2$ like the supremum metric or the euclidean metric.

Answer 3

At the heart of it all is the "topological idea of a circle", or of a loop for short. A loop does not have to look like the circle $x^2+y^2=1$ in the plane, but even a four year old child recognizes a loop when he sees one, and complains when it is fenced in by such a loop. In order to define precisely what a loop is we have to set up a standard model of a loop, as well as a procedure to decide whether any particular geometric object presented to us qualifies as a loop. As standard model we could take the unit circle $S^1$ in the $(x,y)$-plane mentioned above, or the quotient set ${\mathbb R}/{\mathbb Z}$ of real numbers modulo $1$. I think you could accept both of these as loops in an intuitive way.

Now it comes to a formalization of "shaped stretched but not torn or glued", necessary for a mathematically precise procedure to test the "loopness" of a given geometric object. That's where the notions of general topology: open, closed, continuous, homeomorphism, etc., come in. Note that mathematics needed some 2500 years to come up with a definitive version of these notions, even though already Euclid knew what a loop is.