Non-contractibility of a set

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In summary, the conversation discusses the non-contractibility of the set BxB\D, which consists of points in BxB that are not on the diagonal D. It is shown that this set is path connected and a potential proof for its non-contractibility is suggested by computing a homotopy invariant. An idea is presented to show that the circle is not contractible by mapping it to BxB\D and showing that it would result in the circle being a retract of the unit disc, which is not possible.
  • #1
Goklayeh
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Consider the unit ball [itex]B:=B_1(0)\subset \mathbb{R}^2[/itex]. How can one prove that the set [itex]B\times B \setminus D[/itex], where [itex]D:=\left\{(x, x)\biggr| x \in B\right\}[/itex] is the diagonal, is non-contractible? Is it even disconnected? Thank you in advance.
 
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It is not hard to see that this set is path connected: take two points in it and find a path between them by varying only the first coordinate until you reach the desired value, and then varying only the second coordinate. The fact that this procedure can be carried out amounts to the fact that in B, you can find a path btw any two points that avoids a third one.

Now about your actual question, the usual way to prove non-contractibility of a space is to compute a homotopy invariant of it (like the fundamental group) that does not coincides with the value of that invariant for the one-point space. Have you tried doing this?
 
  • #3
Here is an idea.

If x and y are different points in the unit disc then one can draw a line segment from x through y and extend it until it touches the boundary circle. This defines a map,f(x,y), from BxB\D onto the circle. Is this map continuous?

If so then, the continuous map (x,y) ->(0,f(x,y)) maps is the identity on the circle, (0,e[itex]^{i\theta}[/itex]).

If BxB\D were contactible then the compositions

((0,e[itex]^{i\theta}[/itex]),t) -> BxB\D X I -> BxB\D -> (0,f(x,y)) where the first arrow is inclusion, and the second a contraction homotopy, would show that the circle is contractible.

But the circle is not contractible because a contraction homotopy would make the circle into a retract of the unit disc.
 
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1. What does it mean for a set to be non-contractible?

Non-contractibility of a set refers to the inability of the set to be continuously deformed into a single point. In other words, it is not possible to shrink or expand the set into a single point without tearing or breaking it.

2. How is non-contractibility related to topology?

Non-contractibility is a topological property, meaning it is a characteristic of a set that is preserved under continuous deformations. In topology, the concept of continuity is used to define and study different types of spaces, and non-contractibility is one of the important properties used to distinguish between different types of spaces.

3. Can a set be non-contractible in one dimension but not in another?

Yes, a set can be non-contractible in one dimension but contractible in another. For example, a circle is non-contractible in two dimensions, but it is contractible in three dimensions as it can be continuously deformed into a sphere.

4. What is the significance of non-contractibility in mathematics?

Non-contractibility is an important concept in topology, which is a branch of mathematics that studies the properties of spaces that are preserved under continuous deformations. It helps to classify different types of spaces and understand their topological properties.

5. Are there any real-world examples of non-contractible sets?

Yes, there are real-world examples of non-contractible sets. One example is a rubber band. It cannot be continuously deformed into a single point without breaking it. Another example is a Möbius strip, which is a continuous loop with only one side and one edge. It cannot be continuously deformed into a circle without tearing it.

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