Why does $\operatorname{Spec}(\mathbb{C}[[X]])-\langle X\rangle$ "have topological properties identical" to $C^1$? (Mumford's Red Book) In Example F of section II.2 of the Red Book of Varieties and Schemes by Mumford (page 82 in my copy) concerns taking Spec of a noetherian local ring $O$, and then considering the subscheme $X$ when the unique closed point is removed. Specifically, it says that taking $O$ to be $\mathbb{C}[[X_1, ..., X_n]]$, then the resulting $X$ turns out to "have topological properties identical to the ordinary $(2n-1)$-sphere."
So in the simple case of taking the univariate power series $\mathbb{C}[[T]]$ (switching to $T$ to not confuse it with the subscheme), then removing the maximal ideal generated by $T$, it sounds like the claim is such a subscheme $X$ is isomorphic to the circle.
Is that what it's saying? If so how?
I'm having a hard time figuring out what the points of $X$ even would be (with $\mathbb{C}$ being a field, I'm not seeing a lot of opportunities for prime ideals in the power series...), let alone why it would be Hausdorff, why the "ends would touch", etc. that a circle is. Maybe it's not saying it's "the same as" $C^1$ but rather shares some properties?
Clarification of whether I am understanding the point correctly, or help finding elements of $X$ would be appreciated!
 A: Being the complement $V(T)$, the scheme $X$ is $D(T)$ which is $\mathrm{Spec}(\Bbb C[[T]][T^{-1}])$ $\Bbb C[[T]][T^{-1}]=\Bbb C((T))$ is actually the quotient field of $\Bbb C[[T]]$ and is known as the field of formal Laurant series. Therefore, $X$ has only one underlying point and in particular can't be topologically equivalent to $S^1$. (Side remark: In fact, there's no scheme with underlying topological space homeomorphic to $S^1$)
But there's more to scheme than just the topology. The theory of finite étale morphisms is a nice analogue of the usual covering space theory for topological spaces. And there's even a fundamental group that controls this covering theory, called the étale fundamental group. It turns out that this fundamental group for $X$ is $\widehat{\Bbb Z}$, the profinite completion of $\Bbb Z=\pi_1(S^1)$. Because the étale fundamental group is always profinite, this is the closest the étale fundamental group can be to $\Bbb Z$ and we even get that the category of finite coverings of $S^1$ is equivalent to the category of finite étale coverings of $X$.
Here's a more intuitive explanation for this connection (beware, I'm switching between algebraic and analytic categories here, to make this precise you probably need analytification and the the étale topology). You know that the complex plane $\Bbb C$ can be regarded (modulo adding a generic point) as the affine line $\mathrm{Spec}(\Bbb C[T])$. Intuitively, passing to the completion with respect to the ideal $(T)$ corresponds to zooming in "really close" (in some sense, infinitely close) to the origin $T=0$. Now what happens in the analytic topology of $\Bbb C$ if you take a small open ball around the origin and remove the origin from that? You get a small punctured ball, which deformation retracts to $S^1$. So these contain the same topological information, because they are homotopy equivalent.
From this description, the aforementioned category equivalence between finite covers gets more concrete: every finite covering of $S^1$ is equivalent to one given by $S^1 \to S^1: z \mapsto z^n$ (using complex multiplication on $S^1 \subset \Bbb C$) for some nonzero $n \in \Bbb N$. Similarly, every finite étale covering of $X$ is induced by a map $\Bbb C[[T]] \to \Bbb C[[T]], T \mapsto T^n$. These are very similar, and you can also work out their groups of deck transformations to be induced by multiplying with $n$-th roots of unity (i.e. degree $2\pi/n$ rotations).
Of course, the fundamental group is not the only topological invariant, we also have e.g. cohomology. The (say, singular) cohomology of $S^1$ has a particularly simple form: for every abelian group $A$, we have $$H^i(S^1,A)=\begin{cases}A& \text{if $i=0,1$} \\ 0 &\text{else}\end{cases}$$
For technical reasons, not all coefficient groups work well for étale cohomology, but one can take for example any finite abelian group $A$ or $A=\Bbb Z_p$ or $A=\Bbb Q_p$, then there's a useful and well-behaved theory of étale cohomology and for these coefficients it turns out that
$$H^i(X,A)=\begin{cases}A& \text{if $i=0,1$} \\ 0 &\text{else}\end{cases}$$
(which is an exercise in (profinite) group cohomology).
Sorry that this is all kind of informal, I fear making this precise might involve high-level concepts such as the étale homotopy type.
