1
$\begingroup$

Can we regard Hausdorff space as a manifold of class enter image description here?(p≥1)

And I want to know the relation among the concept Hausdorff space,metric space,vector space,tangent space and manifold.

What's the common ground between enter image description here-manifold and smooth manifold?

$\endgroup$
  • 1
    $\begingroup$ Do you mean the other way around? A differentiable manifold is a Hausdorff space, but many topological spaces certainly aren't manifolds. $\endgroup$ – Christopher A. Wong Feb 23 '14 at 21:31
3
$\begingroup$

A topological space is intuitively a space with a notion of "nearness". These spaces are very general, and some of them are called Hausdorff spaces because they satisfy a special property.

A metric space is a space with a notion of distance, and this distance will always imply a notion of nearness: In other words, every metric space is a topological space. In fact, every metric space is a Hausdorff space.

A manifold is a special kind of topological space which locally looks like a Euclidean space. Manifolds are by definition Hausdorff spaces. There are both manifolds which are not metric spaces and metric spaces which are not manifolds. However, if the manifold is smooth, then we can view it as a subspace of Euclidean space and inherit the Euclidean metric. This is not trivial however: See Whitney Embedding Theorem.

A vector space is a completely different object. In general it makes no sense to ask whether some topological/metric space is a vector space and vice versa. However, there are many mathematical objects which have both the structure of a topological space and a vector space, and such that these structures are compatible in a precise sense. Examples are Hilbert and Banach spaces from functional analysis, and of course Euclidean spaces.

$\endgroup$
  • $\begingroup$ Minor correction/comment: If your definition of manifold includes paracompactness, then every manifold is a metric space. $\endgroup$ – Espen Nielsen May 7 '14 at 17:54
5
$\begingroup$

No, the space $X:=\{(x,y)\in\mathbb{R}^2:xy=0\}$ is Hausdorff when endowed with the subspace topology from $\mathbb{R}^2$. Yet, $X$ does not support any type of manifold structure since it is not locally Euclidean near $(0,0)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.