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  1. Does the dimension of a manifold depend on the topology? That is, can I endow a set with a topology $T$ and get an $n$-dimensional manifold, and endow the exact same set with another toplogy $T'$ and get an $m$-dimensional manifold, with $n\neq m$?

  2. Can a curve like this:

$\hskip1.3in$enter image description here

be given the topology of a 1-manifold? Or a curve like this cannot be a manifold with any topology?

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It seems hopeless to characterize what you can do with a bare set. $\mathbf R$ and $\mathbf R^2$ have the same cardinality, for example. –  Dylan Moreland Apr 15 '12 at 2:56
    
OK so you are saying that those questions are impossible to prove or disprove? –  kein Apr 15 '12 at 3:01
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@kein Exapnding on Dylan comment: For each $n$ you can give a topology to $\mathbb R$ so that $\mathbb R$ is an $n$-manifold. –  azarel Apr 15 '12 at 3:11
    
Yep, that is correct. If S is a bijection between R and R^n, you can give R the topology induced by S (the open sets of R are the images of S which are open on R^n in the usual sense), and with that S is an homeomorphism. Thanks for helping me clarify this topic. –  kein Apr 15 '12 at 3:49

1 Answer 1

(1) Yes, the dimension of the manifold depends on the topology. A $3$-manifold has neighborhoods which are homeomorphic to $\mathbb{R}^3$, a $2$-manifold has open neighborhoods homeomorphic to $\mathbb{R}^2$, and so forth.

As Dylan Moreland points out in the comments, $\mathbb{R}$ and $\mathbb{R}^2$ both have the same cardinality. We may regard them as in bijection, therefore as two representations of the same uncountable set. When we endow them with two different topologies, we have two manifolds of different dimension.

For another example, consider $\mathbb{Z}$ and $\mathbb{Q}$. Again, they are in bijection, but while $\mathbb{Z}$ (with its discrete topology) is a $0$-dimensional manifold, $\mathbb{Q}$ with the topology inherited from its inclusion in $\mathbb{R}$ is not a manifold at all.

(2) As a set, certainly; the curve can be put in bijection with $\mathbb{R}^2$ and given the same topology. If you are really asking whether the curve is a $1$-dimensional manifold, then the answer is "no." To be a manifold, about every point on the curve there must be an open neighborhood homeomorphic to $\mathbb{R}$. Observe that no connected neighborhood of its point of self-intersection is homeomorphic to $\mathbb{R}$ since removing any point of $\mathbb{R}$ results in two open components, but removing the point of self-intersection from a small connected neighborhood of the point of self-intersection results in four open components.


Here are some more general comments. You're conflating two levels of abstraction here: the basic machinery of point-set topology and the foundations of manifold theory. Manifold theory is more concrete than point-set topology, in the sense that all of manifold theory deals with very specific types of topological spaces. To admit the possibility of having the structure of a topological manifold, the topology of a set must already be Hausdorff, paracompact, and second-countable. This rules out entire classes of topological spaces, and even rules out many metric spaces.

Meanwhile, most of the topological spaces we encounter frequently are built from uncountable sets. It may be an understatement when I say that there are many, many topologies you can put on an uncountable set. Only a tiny, tiny fraction of them could have the faintest hope of ever possibly sitting under a manifold structure. So it seems strange to try to think about a manifold structure in terms of the topology of the underlying set; they're in different layers of abstraction.

Maybe a computing analogy would help. Working with a topology on a set is like designing a processor. Working with a manifold is like programming software. To program software, you need a working processor, but once your processor works, you generally forget about the processor design, unless it becomes relevant, and just deal with the programming language.


Edited in response to comments by Henry T. Horton and Mariano Suarez-Alvarez.

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Ey thanks for the elaborated answer! –  kein Apr 15 '12 at 3:16
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$\mathbb{Q}$ can be made into a $0$-dimensional manifold by equipping it with the discrete topology (since it is a countable set). You're probably thinking of it with the subspace topology inherited from $\mathbb{R}$, however. There is also the trivial example that $\varnothing$ can be a manifold of any dimension, since it satisfies the necessary conditions vacuously. –  Henry T. Horton Apr 15 '12 at 3:19
    
Just for precision's sake: it is not true that «removing the point of self-intersection from a small neighborhood of the point of self-intersection results in four open components», because you can start with a neighborhood of the point which is itself already not connected! (the components need not be open, either) –  Mariano Suárez-Alvarez Apr 15 '12 at 3:34
    
R^2 in fact is homeomorphic to R, with the topology of R^2 where, if S is a bijection from R^2 to R, the open subsets of R^2 are the images of S which are open in R (in the usual sense). With those topologies, S is an homeomorphism. –  kein Apr 15 '12 at 3:44

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