# Prove that a closed interval [0,1] is nonhomogeneous

• sunjin09
In summary, the closed interval [0,1] cannot be a homogeneous topology because there is no bijective, open, and continuous mapping from [0,1] to itself such that h(1/2)=0. This can be proven by assuming such a mapping exists and showing a contradiction using the intermediate value theorem, which requires the completeness of real numbers. Alternatively, the topological property of connectedness can be used to show that the two points 0 and 1 have different properties in terms of having points on only one side, making it impossible for a bi-continuous function to map 1 to 1/2.
sunjin09

## Homework Statement

Prove that the closed interval [0,1] is not a homogeneous topology by showing that there's no bijective, open and continuous (bi-continuous) mapping h: [0,1]→[0,1] such that h(1/2)=0.

## Homework Equations

The closed interval is equipped with the usually metric. If the mapping mentioned above exists for any x, y such that h(x)=y, then the topology is called homogeneous.

## The Attempt at a Solution

The book says it is easy to verify this; I can kind of feel that it doesn't exist, since if h(1/2)=0 then h(1/2+)=h(1/2-) which violates the bijection, but this is obviously not a proof. Any suggestions?

If h(1/2)=0 then a small neighborhood around 1/2 and a small neighborhood around 0 have to be homeomorphic. I.e. we have a homeomorphism between an interval of the form [a,b) and one of the form (c,d). Can you prove that these are not homeomorphic?

I really have no idea. I'll need to assume such homeomorphism mapping exists and then come up with a contradiction, but with no luck. I'm at earlier stage of my study, I don't know what compactness is.

Here is my attempt: suppose h exists, h(1/2)=0, let h(1/2-ε)=a>0, h(1/2+ε)=b>0, suppose a<b (a≠b from bijectivity), then by continuity and bijectivity of h there exists x in (1/2-ε,1/2), such that h(x)=a/2, similarly there exists y in (1/2,1/2+ε) such that h(y)=a/2=h(x), where x≠y.

Somehow there seem to be no use of the continuity of h^{-1}, maybe I'm missing something? e.g, the existence of x and y may be less trivial than it looks?

sunjin09 said:
Here is my attempt: suppose h exists, h(1/2)=0, let h(1/2-ε)=a>0, h(1/2+ε)=b>0, suppose a<b (a≠b from bijectivity), then by continuity and bijectivity of h there exists x in (1/2-ε,1/2), such that h(x)=a/2, similarly there exists y in (1/2,1/2+ε) such that h(y)=a/2=h(x), where x≠y.

Somehow there seem to be no use of the continuity of h^{-1}, maybe I'm missing something? e.g, the existence of x and y may be less trivial than it looks?

It turns out that the intermediate value theorem I used requires the completeness of real numbers, thus cannot be argued using elementary topological arguments.

sunjin09 said:
It turns out that the intermediate value theorem I used requires the completeness of real numbers, thus cannot be argued using elementary topological arguments.

Ok, you don't know compact. Do you know connected? That's a topological property. You can use that.

"Homogeneous" basically means "the same throughout". The two points 0 and 1 have the property that there are points of the set only on one side while every other point has other points of the set on both sides. That's the idea of trying to map 1 to 1/2. That can't be done with bi-continuous function precisely for the reason that 1 has only those points that are less in the set while 1/2 has points on both sides.

HallsofIvy said:
"Homogeneous" basically means "the same throughout". The two points 0 and 1 have the property that there are points of the set only on one side while every other point has other points of the set on both sides. That's the idea of trying to map 1 to 1/2. That can't be done with bi-continuous function precisely for the reason that 1 has only those points that are less in the set while 1/2 has points on both sides.

That's pretty superficial. Having "points on one side" is not a topological property. This is a topology problem, not a real analysis problem as sunjin09 has already very correctly figured out. You have to abstract the notion of "having points on one side" to something that makes sense in topology. I'm suggesting connectedness.

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## 1. What does it mean for a closed interval [0,1] to be nonhomogeneous?

Nonhomogeneous means that the elements within the interval are not all the same or equal. In this case, the elements in the interval [0,1] are not all identical.

## 2. How can you prove that a closed interval [0,1] is nonhomogeneous?

To prove that a closed interval [0,1] is nonhomogeneous, you can provide a counterexample. For example, you can show that the elements 0 and 1 are not the same, therefore proving that the interval is nonhomogeneous.

## 3. Can you provide a mathematical proof for the nonhomogeneity of [0,1]?

Yes, we can prove the nonhomogeneity of [0,1] using mathematical notation and properties. We can show that there exists at least one pair of elements, such as 0 and 1, that are not equal within the interval [0,1]. This proves that the interval is nonhomogeneous.

## 4. Are there any other ways to prove the nonhomogeneity of a closed interval [0,1]?

Yes, another way to prove the nonhomogeneity of [0,1] is by using the definition of nonhomogeneous, which states that the elements within a set or interval must be distinct. By showing that 0 and 1 are distinct elements within the interval [0,1], we can prove its nonhomogeneity.

## 5. Why is it important to prove the nonhomogeneity of a closed interval [0,1]?

Proving the nonhomogeneity of a closed interval [0,1] is important because it helps us understand the properties and characteristics of the interval. It also allows us to make accurate mathematical statements and conclusions about the elements within the interval. Additionally, proving nonhomogeneity can be a crucial step in solving certain mathematical problems or proofs.

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