Are the coordinate axes a 1d- or 2d-differentiable manifold?

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Discussion Overview

The discussion centers on whether the coordinate axes in the plane, defined as the set $$ D=\{ (x,0) \in \mathbb{R}^2 : x \in \mathbb{R}\} \cup \{ (0,y) \in \mathbb{R}^2 : y \in \mathbb{R} \}$$, can be classified as a 1-dimensional or 2-dimensional differentiable manifold. The scope includes theoretical considerations of manifold properties and topology.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant asserts that the set D is not a manifold because every point must have a neighborhood homeomorphic to an open subset of Euclidean space, which fails at the origin.
  • Another participant argues that removing the origin would disconnect the set into four components, suggesting that it does not meet the criteria for a differentiable manifold.
  • A later reply proposes that if the origin is removed, the remaining set could be considered a topological manifold, as it consists of four lines, each being a 1-manifold globally homeomorphic to the Reals.

Areas of Agreement / Disagreement

Participants generally disagree on the classification of the set D as a manifold. While some argue it is not a manifold due to the properties at the origin, others suggest that removing the origin allows it to be considered a topological manifold.

Contextual Notes

Limitations include the dependence on the definitions of manifold and topological space, as well as the unresolved nature of the implications of removing the origin on the manifold structure.

Delong66
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Suppose $$ D=\{ (x,0) \in \mathbb{R}^2 : x \in \mathbb{R}\} \cup \{ (0,y) \in \mathbb{R}^2 : y \in \mathbb{R} \}$$ is a subset of $$\mathbb{R}^2 $$ with subspace topology. Can this be a 1d or 2d manifold?
Thank you!
 
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The set D is not a manifold. Every point in a manifold must have a neighbourhood that is homeomorphic to an open subset of a Euclidean space. The point (0,0) in set D has no such neighbourhood, as any open set containing (0,0) has an intersection of the two lines in it, and neither 1D nor 2D Euclidean space has an open subset consisting of such an intersection.
 
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And removal of a single point, the origin, would disconnect it into 4 components, unlike any surface or 1-manifold. It's clearly not a differentiable ( if it was a manifold ), as its tangent space is not defined at the origin. It's not even a manifold with boundary, as no neighborhood of the origin is homeomorphic to a (subspace) neighborhood of the upper half plane .
 
WWGD said:
And removal of a single point, the origin, would disconnect it into 4 components, unlike any surface or 1-manifold. It's clearly not a differentiable ( if it was a manifold ), as its tangent space is not defined at the origin. It's not even a manifold with boundary, as no neighborhood of the origin is homeomorphic to a (subspace) neighborhood of the upper half plane .
If you remove the origin, I believe it is a topological manifold.
 
jbergman said:
If you remove the origin, I believe it is a topological manifold.
Indeed, 4 lines, each a 1-manifold, globally homeomorphic to the Reals. A manifold with 4 connected components.
 

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