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?

  • $\begingroup$ I assume that you require manifolds to be second countable? Then yes and yes. $\endgroup$
    – freakish
    Aug 5 '19 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 $\mathbb{R}^n$ $\endgroup$ Aug 5 '19 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. $\endgroup$ Aug 5 '19 at 19:03

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

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


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