Some questions relating topology and manifolds

[Hymne]

Hello there!
I just started reading Topological manifolds by John Lee and got one questions regarding the material.
I am thankful for any advice or answer!

The criteria for being a topological manifold is that the space is second countantable ( = there exists a countable neighborhood basis?).
I can't see why R^n fulfills this criteria.. the neighborhood basis for q is all the open sets that contain q. And if we view these as all the open balls with a radius varying on the reell line, these are not countable due to the incountability of R.. right?

In which way is R^n second countable?

 

[micromass]

Take only the balls with rational centers. Thus your basis should be
[tex]\mathcal{B}=\{B(q,1/n)~\vert~q\in \mathbb{Q},~n\in \mathbb{N}_0\}[/tex]

In fact, if your space in metrizable, then the space is second countable if and only if it is separable. And [tex]\mathbb{R}[/tex] is separable, since it contains [tex]\mathbb{Q}[/tex] as countable dense subset...

 

[Bacle]

Just to pick up some low-hanging fruit from Micromass' response, you can see how
every metric space is also 1st-countable, by using { B(q,1/n): n in Z+ }.

 

[mathwonk]

this is exactly why in calculus, in order to show "for every epsilon there is a delta", it suffices to take epsilon equal to 1/n for all positive integers n. I.e. there are uncountably many real epsilons, but it suffices to look only at the countable set of 1/n's

 

[blue_raver22]

Hi there! Great to hear that you are reading John Lee's book on topological manifolds. It's a great resource for understanding this topic.

To answer your question, yes, you are correct that the neighborhood basis for a point q in R^n would be all the open balls with varying radii along the real line. However, keep in mind that the definition of a countable neighborhood basis is that for any point in the space, there exists a countable collection of open sets that contain that point.

In the case of R^n, we can construct a countable neighborhood basis by considering all open balls with rational radii and centers at points with rational coordinates. Since the set of rational numbers is countable, this collection of open sets would also be countable.

Therefore, R^n does fulfill the criteria of being second countable and can be classified as a topological manifold. I hope this helps clarify things for you. Happy reading!