# How do I prove that $S^1 \not \approx D$?

• snesnerd
In summary, the author proves that the real line is not homeomorphic to the circle by contradiction and uses the fact that two points removed from a circle do not disconnect it.
snesnerd

## Homework Statement

I recently read and proved that the real line is not homeomorphic to the circle. It was very interesting and made sense to me. I started pondering to myself and thought of another interesting problem. What if I considered the circle $S^1$ and $D$ where $D$ is the open disk. The question I asked myself is the following: Is $S^1 \approx D$ where $\approx$ signifies a homeomorphism. I thought about it and think that they are not.

## The Attempt at a Solution

My idea goes as follows. Do a proof by contradiction and use the idea of what it means to be connected and not connected. Suppose they were homeomorphic. Clearly if I remove a point from $S^1$ then it is still connected. Same goes for $D$. It does not seem helpful at first, but what if I remove another point from each? I think I get one of them to be connected and one of them to be not connected which is a contradiction. Of course this is my idea. I would like to write it out nicely, so if anyone could help me out it would be greatly appreciated! I wasn't sure if this was the way to go.

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snesnerd said:

## Homework Statement

I recently read and proved that the real line is not homeomorphic to the circle. It was very interesting and made sense to me. I started pondering to myself and thought of another interesting problem. What if I considered the circle $S^1$ and $D$ where $D$ is the open disk. The question I asked myself is the following: Is $S^1 \approx D$ where $\approx$ signifies a homeomorphism. I thought about it and think that they are not.

## The Attempt at a Solution

My idea goes as follows. Do a proof by contradiction and use the idea of what it means to be connected and not connected. Suppose they were homeomorphic. Clearly if I remove a point from $S^1$ then it is still connected. Same goes for $D$. It does not seem helpful at first, but what if I remove another point from each? I think I get one of them to be connected and one of them to be not connected which is a contradiction. Of course this is my idea. I would like to write it out nicely, so if anyone could help me out it would be greatly appreciated! I wasn't sure if this was the way to go.

That's a fine idea. I'm not sure even how much more you need to write out. What do you think you would need?

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Well to be honest, the idea sounds good, but I would like to mathematically write it out nicely. As you know in mathematics it is easy to say something is true, but it is only true if you can prove it. Granted the wording still sort of proves it well, but I would like to translate what I said into math if you know what I mean. I was hoping I could get assistance on that. It would be greatly appreciated.

snesnerd said:
Well to be honest, the idea sounds good, but I would like to mathematically write it out nicely. As you know in mathematics it is easy to say something is true, but it is only true if you can prove it. Granted the wording still sort of proves it well, but I would like to translate what I said into math if you know what I mean. I was hoping I could get assistance on that. It would be greatly appreciated.

The idea is perfect. Removing two points from ##S^1## disconnects it. Removing two points from ##D## does not disconnect it. So they are not homeomorphic. For me, just saying that is enough. If you want to actually prove each of those two statements, which I'm not sure is necessary, then start with one of them. How would you prove removing two points from ##S^1## disconnects it?

Proof (Attempt) :

Suppose $S^1 \approx D$. Then $\exists$ a map $f : S^1 \rightarrow D$ that is $\approx$. Let $x \in (0,1)$ and $y = f(x)$. Then $f : S^1 - \{x\} \rightarrow D - \{y\}$. Here $S^1 - \{x\}$ and $D - \{y\}$ are connected. Suppose we remove another point $x' \in (0,1),$ and $y' = f(x')$. Now from here you can make the same argument except the last part of the proof would rely on telling the reader why $S^1$ becomes disconnected while $D$ is connected. Maybe you can help me fill the details in for this proof.

snesnerd, try to be more careful when you use LaTeX. You had several bits where you started with an itex tag, but finished the expression with a tex tag (including in the thread title).

You could also use the fact that the circle is compact while the open disk is not; for continuous functions (and hence homeomorphisms) maps compact sets to compact sets. But the argument you give is much more intuitive.

To finish off the proof, you can use your intuition. You can draw the straight line from the two removed points and extend that line forever. Then the section of the plane above the line and the section below are open sets that separate the two parts of the circle. To show the open disk is still connected, you can use the fact that path-connectedness implies connectedness. So pick any two points and then you draw a line from one point to the other. If it does not intersect any removed points, you have a path. If it intersects a removed point, you can 'delete' a piece of the path around the point and connect the two pieces with a half circle that avoids the point.

! said:
You could also use the fact that the circle is compact while the open disk is not; for continuous functions (and hence homeomorphisms) maps compact sets to compact sets. But the argument you give is much more intuitive.

To finish off the proof, you can use your intuition. You can draw the straight line from the two removed points and extend that line forever. Then the section of the plane above the line and the section below are open sets that separate the two parts of the circle. To show the open disk is still connected, you can use the fact that path-connectedness implies connectedness. So pick any two points and then you draw a line from one point to the other. If it does not intersect any removed points, you have a path. If it intersects a removed point, you can 'delete' a piece of the path around the point and connect the two pieces with a half circle that avoids the point.

Good advice. Before you try to write something super formal mathematically you should at least have an argument you can put into words.

## Question 1: What is the difference between $S^1$ and $D$?

The notation $S^1$ represents the one-dimensional unit circle, while $D$ represents the two-dimensional unit disk. The key difference is that $S^1$ has only one dimension (length), while $D$ has two dimensions (length and width).

## Question 2: Why is it important to prove that $S^1 \not \approx D$?

This proof is important in topology, the branch of mathematics concerned with the properties of geometric objects that are preserved through continuous deformations. Showing that $S^1$ and $D$ are not topologically equivalent provides insight into their distinct topological properties.

## Question 3: What is the definition of topological equivalence?

Two objects are topologically equivalent if there is a continuous and invertible mapping (homeomorphism) between them. This means that the objects can be continuously deformed into each other without any tearing or gluing.

## Question 4: How can I prove that $S^1 \not \approx D$?

This can be proven by showing that there is no homeomorphism between $S^1$ and $D$. One approach is to use the concept of connectedness, which is preserved under homeomorphisms. $S^1$ is connected (any two points can be connected by a continuous curve on the circle), while $D$ is not (the center point cannot be connected to any point on the boundary without intersecting the interior).

## Question 5: Can $S^1$ and $D$ be homeomorphic in higher dimensions?

No. The proof that $S^1$ and $D$ are not topologically equivalent extends to any higher dimension. This is because the fundamental properties of connectedness and dimensionality hold in any number of dimensions and cannot be altered by homeomorphisms.

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