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Let $(M,d)$ be a metric topological manifold (without boundary). We know that for any $m\in M$ there is an open neighbourhood $U\subseteq M$ of $m$ such that $U$ is homeomorphic to $\mathbb{R}^n$. Can we always choose $U$ to be an open ball in $M$ with respect to the metric $d$ around $m$?

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  • $\begingroup$ Do you want the chartmap to be an isometry in this case? Or still just a homeomophism? $\endgroup$
    – Babelfish
    Feb 20 '19 at 12:27
  • $\begingroup$ @Babelfish homeomorfism is enough. I'm only interested in the neighborhood. But you'll get bonus points for isometry. :) $\endgroup$
    – freakish
    Feb 20 '19 at 12:28
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Ok, this is not always possible. Don't mess with topology, it will mess with you!

Wolframalpha plots you the following image with the function $f\colon [0,\infty)\to \mathbb R, x\mapsto5x \sin(10 \log(x))$:

Plot

Now we can take the graph $X$ of this function, which is homeomorphic to $[0,\infty)$. The metric from $\mathbb R^2$ yields a restricted metric on $X$ which induces the same topology. But if we look at a ball $B$ around $(0,0)$, there propably is a point $p$ on the $x$-axis with $p\in B$, but the two "spikes" next to $p$ do not lie in $B$.

If we now add the line $\{(-t,0) \mid t \leq 0\}\subset \mathbb R^2$, we obtain a manifold without boundary.

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If we assume that the metric balls of the manifold $M$ are connected, then we are able to generate a covering of metric balls, which are homeomorphic to connected subsets of $\mathbb R^n$. This is an alternative definition of an atlas.

For every $m\in M$, there is an open neighbourhood $U_m\subseteq M$ of $m$, such that $U_m$ is homeomorphic to an open subset of $\mathbb R^n$. The open balls $B(x,r)=\{y\in M \mid d(x,y)<r\}$ form a base of the topology. Therefore, for each such $U_m$, we have a set of such balls $B_i$ such that $\bigcup_{i\in I_m} B_i = U$. Since $m\in U$, there is $j_m\in I_m$ such that $x\in B_{j_m}$. The restriction of the homeomorphism $U_m \to U\subset \mathbb R^n$ to the ball $B_{j_m}$ is still a homeomorphism onto its image, therefore we have an atlas $\mathcal A = \{(U_m ,\text{restriction})\mid m\in M\}$ of $M$, consisting of metric balls, which are connected by assumption, so the image is a connected open subset of $\mathbb R^n$.

If you want the charts to be homeomorphic to $\mathbb R^n$, you are in bad shape. Normally, this is an equivalent definition for manifolds. If you have charts which are homeomorphic to connected open subsets of $\mathbb R^n$, you can build charts which are homeomorphic to $\mathbb R^n$ by just refining the charts. You decompose the chart in such a way, that you get simply-connected open regions. Then those are homeomorphic to $\mathbb R^n$. In our case, this is not possible in general, since we usually lose the property charts are metric balls, if we refine the charts.

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  • $\begingroup$ But $[0,\infty)$ is not a topological manifold. At least not a manifold without boundary which is my case. $\endgroup$
    – freakish
    Feb 20 '19 at 16:13
  • $\begingroup$ Yeah, but we can just add a line to the other side to get a manifold. $\endgroup$
    – Babelfish
    Feb 20 '19 at 16:14
  • $\begingroup$ Yeahh, this looks good. Unfortunately. $\endgroup$
    – freakish
    Feb 20 '19 at 16:16
  • $\begingroup$ Correct me if I'm wrong but your previous answer seems to be correct if we additionally assume that balls are connected? $\endgroup$
    – freakish
    Feb 20 '19 at 17:16
  • $\begingroup$ Hmm, according to the wiki the condition "being homeomorphic to a connected subset of $\mathbb{R}^n$" is equivalent to "being homeomorphic to $\mathbb{R}^n$". What do you think about it? $\endgroup$
    – freakish
    Feb 21 '19 at 8:29

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