The answer to the question of the thread title is yes, according to what I found on web. Now a manifold is by definition a topological space that (aside from other conditions) is locally Euclidean.(adsbygoogle = window.adsbygoogle || []).push({});

What does such condition means? Is it the same as saying that each of its points must have a homeomorphic neighborhood to ##\mathbb{R}^n## for a n-dimensional manifold?

If so, how can we prove that it's possible to find an open neighborhood around each point of the circle? Could we use open spheres to do that?

I have tried to prove this by considering the unit circle in ##\mathbb{R}^2## (with the usual metric) as

$$\mathbb{S}^1: \big[x \in \mathbb{R}^2 \ | \ d(x,0) = 1 \big]$$ where ##0## is the point where the circle is centered in. Let ##P_0## be the center of an open sphere on the circle.

We have ##d(0,x) = 1 = d(0,P_0)##

Using the "triangle inequality" we have

##d(0,P_0) + d(P_0, x) \leq d(0,x)##

The solution is ##d(P_0,x) = 0##. This says that ##P_0## and ##x## are the same point! That is, the open sphere contains only its center point. Let's call the open sphere ##A##.

Now as this open sphere is in ##\mathbb{R}^2## a parametrization for its points could be ##\gamma: [0,\delta) \times [0, 2 \pi) \longrightarrow A## (the inverse mapping is the usual spherical polar coordinates). We would need both intervals in ##[0,\delta) \times [0, 2 \pi)## to be open in order to it to be homeomorphic to ##\mathbb{R}^2##. But if we do that for our open sphere ##A##, we will exclude its unique point, namely the central point ##P_0##. The solution would be that it doesn't contain any points at all. But then our proof that ##\mathbb{S}^1## is a manifold fails. How do we solve the problem?

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# I Is the circle a manifold?

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