Why is this picture an example of a $1$-dimensional manifold?
My thought process is: the circle must have a point removed from it because otherwise it would be self-intersecting, and self-intersection isn't a property of manifolds? Is that correct reasoning? So a circle with all its points isn't a manifold?
Is this property true for all manifolds, or for those only in $\mathbb{R}^n$?
On
The basic condition (for memorization) for a set to be a n-dimensional manifold is that it locally "looks" like a Euclidean $R^n$ i.e. locally, you can define a map from the neighborhood on the manifold to some open set in $R^n$ and that map is a homeomorphism (i.e. continuous, bijective, inverse is continuous too). Both the circle, and circle with one point removed are manifolds. You may in general need to define more than one map to cover the whole manifold.
A circle with a point removed from it, as in your picture, is actually homeomorphic to $\mathbb{R}$: for each point $S'$ on this "punctured circle", there is a corresponding point $S$ on the real line $\mathbb{R}$, and vice versa, as shown in the following picture:
This correspondence, called stereographic projection, is a homeomorphism. So the punctured circle is equivalent to the simplest 1-dimensional manifold, which is $\mathbb{R}$.
The (complete) circle $S^1$ can be seen to be a 1-dimensional manifold because you can write $S^1 = U\cup V$ where $U$ and $V$ are two open subsets each homeomorphic to $\mathbb{R}$: just choose two different points $x$ and $y$ and let $U = S^1\setminus\{x\}$ and $V = S^1\setminus\{y\}$, which are homemorphic to $\mathbb{R}$ using the above pictorial argument.