0
$\begingroup$

Lemma: Let U be an open subset of $\mathbb{R}^n$ and $x \in U$ then there exists a ball with rational center $(\mathbb{Q}^n$ and rational radius containing x, that is contained in U.

What I want to show: Every open subset of $\mathbb{R}^n$ is the union of some some collection of elements in the following set:

$\{$ Open balls with rational center and rational radius $\}$

Let U be an arbitrary open set in $\mathbb{R}^n$. By the above lemma, for each $x \in U$ you can find a rational center $x'$ and rational radius $r'>0$ so that $x \in B(x',r')$ $\subset U$. Hence the following collection:

$B'=$ $\{$ B(a,r) : a $\in \mathbb{Q}^n$ , $r>0$ , B(a,r) $\subset U$ $\}$ gives us

$U=$ $\bigcup_{B \in B'}$ $B$

is the proof correct?

As this is part of the proof of showing that $\mathbb{R}^n$ is second countable, it follows that since every metric space is Hausdorff, $\mathbb{R}^n$ is Hausdorff, and trivially, it is locally euclidean of dimension n (take the identity map, which is a homeomorphism) and so $\mathbb{R}^n$ is a n- dimensional topological manifold, am I correct?

$\endgroup$
  • $\begingroup$ I assume that you require manifolds to be second countable? Then yes and yes. $\endgroup$ – freakish Aug 5 at 18:39
  • $\begingroup$ @freakish Yup, a topological space M is a n - dimensional manifold if (1) It is second countable (2) It is Hausdorff (3) Every point in M has a neighborhood homeomorphic to an open subset of $\mathbb{R}^n$ $\endgroup$ – topologicalmagician Aug 5 at 18:40
  • $\begingroup$ @topologicalmagician You prove correctly that $\mathbb{R}^n$ is second countable. Also, since $\mathbb{R}^n$ is Hausdorff and locally euclidean for the reasons you said above, it is a topological manifold too. $\endgroup$ – bing-nagata-smirnov Aug 5 at 19:03
0
$\begingroup$

This community wiki solution is intended to clear the question from the unanswered queue.

Yes, your proofs that $\mathbb R^n$ is cecond countable and that $\mathbb R^n$ is an $n$- dimensional topological manifold are correct.

$\endgroup$

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.