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

[Delong66]

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!

 

[andrewkirk]

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.

 

[WWGD]

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 .

 

[jbergman]

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.

 

[WWGD]

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.