# Metric topological manifold and the choice of neighbourhood

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 $$m$$?

• Do you want the chartmap to be an isometry in this case? Or still just a homeomophism? Feb 20 '19 at 12:27
• @Babelfish homeomorfism is enough. I'm only interested in the neighborhood. But you'll get bonus points for isometry. :) Feb 20 '19 at 12:28

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 $$f\colon [0,\infty)\to \mathbb R, x\mapsto5x \sin(10 \log(x))$$:

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 $$B(x,r)=\{y\in M \mid d(x,y) form a base of the topology. Therefore, for each such $$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.

• But $[0,\infty)$ is not a topological manifold. At least not a manifold without boundary which is my case. Feb 20 '19 at 16:13
• Yeah, but we can just add a line to the other side to get a manifold. Feb 20 '19 at 16:14
• Yeahh, this looks good. Unfortunately. Feb 20 '19 at 16:16
• Correct me if I'm wrong but your previous answer seems to be correct if we additionally assume that balls are connected? Feb 20 '19 at 17:16
• 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? Feb 21 '19 at 8:29