Are open sets in R^n always homeomorphic to R^n?

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In summary, open intervals in R are homeomorphic to R, but this does not extend to any dimension of Euclidean space. For example, an open 4-ball is not homeomorphic to R^4. Additionally, while an open annulus is open and connected, it is not simply-connected and therefore not homeomorphic to R^n. However, any connected open set in R^n is homeomorphic to R^n. An open set in R^n can also be homeomorphic to the disjoint union of equally many R^n's as connected parts of the set. While R^n is contractible, this is not enough to determine if an open set in R^n is homeomorphic to R^n.
  • #1
iLoveTopology
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I know that open intervals in R are homeomorphic to R. But does this extend to any dimension of Euclidean space? (Like an open 4-ball is it homeomorphic to R^4?)

My book doesn't talk about anything general like that and only gives examples from R^2.
 
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  • #2
No for general open sets; look at, e.g., an open annulus, or any disconnected open set. But an open n-disk D:={x in R^n : ||x||<1 } (or any translation of it) is homeomorphic to R^n.
 
  • #3
Any connected open set in R^n is homeomorphic to R^n, for any n. An open set in R^n is homeomorphic to the disjoint union of equally many R^n's as connected parts of your open set.
 
  • #4
Actually, an open annulus is open and connected, but not homeomorphic to R^n, since it is not simply-connected.
 
  • #5
Bacle2 said:
Actually, an open annulus is open and connected, but not homeomorphic to R^n, since it is not simply-connected.

thanks for the correction, i meant simply connected :)
 
  • #6
disregardthat said:
thanks for the correction, i meant simply connected :)

Also not true, then open set [itex]B(0,1)\setminus\{0\}[/itex] of [itex]\mathbb{R}^3[/itex] is simply connected but not homeomorphic to [itex]\mathbb{R}^3[/itex].
 

1. What is an open set in R^n?

An open set in R^n is a subset of the n-dimensional real numbers in which every point within the set has a neighborhood entirely contained within the set. In other words, for every point in the set, there exists a small "open ball" around that point which is also contained within the set.

2. What does it mean for two sets to be homeomorphic?

Two sets are homeomorphic if there exists a continuous function between them that is one-to-one, onto, and has a continuous inverse. In simpler terms, two sets are homeomorphic if they have the same shape and structure, even if the elements within them are different.

3. Are all open sets in R^n homeomorphic to R^n?

No, not all open sets in R^n are homeomorphic to R^n. While it is true that open sets in R^n are homeomorphic to each other, they are not necessarily homeomorphic to the entire space R^n. Some open sets may have a different shape or structure than R^n and therefore cannot be mapped onto it.

4. How do you prove that open sets in R^n are homeomorphic to R^n?

There are several ways to prove that open sets in R^n are homeomorphic to R^n. One way is to construct a specific homeomorphism between the two sets, showing that there exists a continuous function that is one-to-one, onto, and has a continuous inverse. Another way is to use topological invariants, such as the number of holes or connected components, to show that the two sets have the same structure.

5. Why is it important to study homeomorphism between open sets in R^n and R^n?

Studying homeomorphism between open sets in R^n and R^n allows us to better understand the topological properties and structures of these sets. It also helps us identify which sets are equivalent in terms of shape and structure, which can be useful in solving problems in mathematics and other fields such as physics and engineering.

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