$S_C$ is nonseparable. To prove this, consider the set $B$ of all sequences consisting only of $0$ and/or $1$. There is a bijection between $B$ and the discrete topology of $\mathbb{N}$. Write $\forall I\subsetneq \mathbb{N}$ $x_I=x_n$ such that if $n\in I$, $x_n=1$, and $0$ otherwise. The distance between any two elements of $B$ is $1$. Let us surround each point in $B$ by an open sphere of radius $0.5$. There is thus an uncountably infinite family of pairwise disjoint spheres. If some set $X$ is dense in $S_C$, $\exists \mbox{ at least one }x\in E_C\backslash\partial E_C$ in each of the spheres. $X$ is not countable. Therefore, $S_C$ is nonseparable.
$S_C$ is separable if it has a nonempty countable dense subset. Then $S_C$ would be empty, which is a contradiction. Therefore, each element of $S_C$ must be countable, and so $E_C\backslash\partial E_C$ should have a finite number of elements (since $S_C$ is the trivial topology). But since $S_C$ is nonseparable, $E_C-
\partial E_C$ must have an infinite number of elements. \emph{If $E_C\backslash\partial E_C$ has an infinite number of elements, then there is no predetermined ``origin'' of $E_C\backslash\partial E_C$.} This is a very important statement. If there is no origin, then there is no predetermined frame of reference. Therefore, the dynamics of objects, and their positions, e.t.c., depend on the position of the observer. This is the most fundamental concept in Relativity.
Uniqueness of $E_C(\backslash\partial E_C)$: $(E_C\backslash\partial E_C,T_t)=(E_C\backslash\partial E_C,S_C)$ is not metrizable. Therefore, globally, $(E_C\backslash\partial E_C,S_C)$ is not homeomorphic to any other metric space.
Therefore, there exists no continuous bijection $f:(E_C\backslash\partial E_C,S_C)\to (Y,g)$, with $g$ the metric on $Y$. Upto a homeomorphism, $(E_C\backslash\partial E_C,S_C)$ is therefore unique. We will now generalize. to $E_C$. Let us write $E_C\backslash\partial E_C=D$. Then, by means of a Taylor expansion, $E_C$ is just $\left(\sum^\infty_{k=0}\partial^k\right)D=\bigcup^\infty_{k=0}(\partial^kD)$. It is well known that the boundary operator $\partial$ can be considered to be the exterior differential for spaces. Therefore, $\forall n\geq2, \partial^n\mathcal{A}=\emptyset$. The equation for $E_C$ simplifies to $E_C=D\cup\partial D$. $D$, and therefore $\partial D$ too, is unique. Hence, upto a homeomorphism, $E_C$ is unique.
Uniqueness of $E$: $E_C$ is the compactification of $E$. Therefore, it has a metric. There exists an embedding of $E$ as a dense subset of a compact space. $E_C$ is embedded in $E$ be some map $f$. Since $E_C$ is a subspace of $E$, this map is surjective. $f$ is therefore a bijection. The same holds for $f^{-1}$, which, like $f$, is continuous. $E$ is therefore homeomorphic (isometric, in fact) by a volume preserving map to $E_C$. Since $E_C$ is unique, so is $E$.
We had shown earlier that $E_C$ is generated from $S_C$, which does not contain any unbounded or incomplete sequences. The same holds for $E$ with the topology $S$. Since $E$ is isometric to $E_C$, $S_C$ is the trivial and $S$ is the discrete topology of $E$. Therefore, all sequences of $E$ except for $E\backslash\partial E$ itself and $\emptyset$ are unbounded and incomplete. What seems ambiguous, though, is that the topology contains $E\backslash\partial E$, and not $E$ itself. The reason is that from $E\backslash\partial E$, we can construct $E$ by means of a Taylor expansion.
Locally, $\forall x\in E,\exists f(x)\mbox{ such that }f(x)\in\mathbb{R}^n$ if $E$ is $n$-dimensional. Let us generalize the definition of the Taylor neighborhood to an $n$-Taylor neighborhood - an $n$-Taylor neighborhood is an $D^n$ with radius $\epsilon$ embedded in $E$ of dimension $n$ such that $\forall x\in E$, functions at any point in $D^n$ can be accurately determined by means of a Taylor expansion. Consider a point $p\in\mathbb{R}^n$. The closed set of all points spaced a distance $\epsilon$ from $p$ is $D^n$. By the definition of an $n$-Taylor neighborhood, this implies that $E$ is locally homeomorphic to $\mathbb{R}^n$, and therefore is a manifold. $E$ is therefore a finite $n$-dimensional complex, which at every point has the same homology groups as the $(n-1)$-dimensional sphere. Again, consider some sequence in $E$. Since the trivial topology of $E$ consists of only bounded and complete sequences, the limit of any sequence in $E$ is come $l\in E$. All ($n$)-Taylor neighborhoods of $l$ therefore contain an infinite number of points of the sequence. This means that each sequence must itself have infinite points. There are no subsequences with a finite number of points. Any topology of $E$ therefore must contain only sequences with an infinite number of points. $E$ is also simply-connected.
Consider a cover $\{\mathcal{U}_\alpha\}$ of $E$. By the definition of a cover, $E\subseteq\bigcup_\alpha\mathcal{U}_\alpha$. We will use the fact that the two sets can be equal. Then, $\{\mathcal{U}_\alpha\}$ can be considered to be a topology on $E$, i.e., $\{\mathcal{U}_\alpha\}=S\mbox{ or }S_C$. $S\mbox{ and }S_C$ are therefore open covers of $E$. There exists some subset of $\{\mathcal{U}_\alpha\}$ that is finite and is a subcover of $E$. We will now proceed to construct $\{\mathcal{U}_\alpha\}$ from the set of all open neighborhoods of $E$.
Let $\mathcal{O}$ be the set if all neighborhoods of $E$. To $\mathcal{O}$ add $\emptyset,E$. Add to $\mathcal{O}$ also all finite intersections of elements of $E$ to obtain $\mathcal{X}$. To $\mathcal{X}$ add all (in)finite unions of elements of $\mathcal{X}$. $\mathcal{X}$ is then a cover of $X$. Thus, all elements of the cover along with some $N\in\mathbb{N}$ defines some $n\geq N$ such that any sequence in $E$ converges to some $l\in E$. This shows that $E$ is complete. Suppose we had begun with an arbitrary family of sets. $\{\mathcal{U}_\alpha\}$ defines another space in which $E$ is a subspace. Any compact Hausdorff space is a closed subset of another compact Hausdorff space \footnote{It is well known that every locally compact Hausdorff space with countable basis is paracompact, and that paracompactness implies the existence of a Lorentzian metric. Since $E$ is a compact (and thus locally compact) Hausdorff space, $E$ has a Lorentzian metric on it. As was mentioned in the beginning of the paper, general relativity \emph{requires} a Lorentzian metric, and we have shown that $E$ has one.}. $E$ is therefore a closed subset of some compact Hausdorff space space $\mathcal{E}$. since $E$ is a metric space, $\mathcal{E}$ is also a metric space. Since $E$ has infinite elements, $\mathcal{E}$ is not separable. It can be observed that $E$ is dense in $\mathcal{E}$. Therefore, $\mathcal{E}$ is the completion of $E$. $\mathcal{E}$ is unique. All Cauchy sequences therefore converge in $\mathcal{E}$.
Consider a convergent Cauchy sequence consisting of elements from both the sets $E$ and $\mathcal{E}\backslash E$. The subsequence of elements from $E$ only converges to some $l\in E$ due to completeness of sequences. The same holds for $\mathcal{E}\backslash E$. What does a convergent Cauchy sequence consisting of elements from both the sets $E$ and $\mathcal{E}\backslash E$ therefore converge to? We will now attempt to answer this question. For any point $p\in\mathcal{E}$, the ($n$)-Taylor neighborhood $\mathcal{U}$ of $p$ with
- $p\in\mathcal{E}\backslash E$ is $\mathcal{U}\backslash(\mathcal{U}\cup E)$;
- $p\in E$ is $E\backslash(\mathcal{U}\cup E)$.
This ensures that if a sequence $(p_n)\in\mathcal{E}\backslash E$, then $l\in\mathcal{E}\backslash E$, and if $(p_n)\in E$, then $l\in E$. This ensures completeness of both $E$ and $\mathcal{E}\backslash E$. This means that every decreasing sequence of non-empty closed subsets of $\mathcal{E}\backslash E$, with diameters going to $0$, have a non-empty intersections. Since $E$ is closed, $\mathcal{E}\backslash E$ is closed. This supports the fact that a complete subset of a metric space is closed. It can be shown that $\mathcal{E}\backslash E$ is compact. Every compact Hausdorff space is H-closed. Therefore, every open cover of $E\mbox{ and } \mathcal{E}\backslash E$ has a finite subfamily with a dense union. $E$ is Hausdorff. Therefore, $E$ is a neighborhood space. Consider the following: $E\mbox{ and } \mathcal{E}\backslash E$ are closed in $\mathcal{E}$. Therefore, every sequence in both $E\mbox{ and } \mathcal{E}\backslash E$ that converges to a limit $l\in\mathcal{E}$ has its limit in $E\mbox{ or } \mathcal{E}\backslash E$ respectively. Also, every compact metric space is second countable. Therefore, $\exists\beta\subset S$ such that $\forall\mathcal{O}\in S,\exists B\in\beta$ such that $x\in B\subset\mathcal{O}$. There also exists some countable collection $\mathcal{U} = \{U_i\}_{i=1}^\infty$ of open subsets of $E\mbox{ and } \mathcal{E}\backslash E$ such that any open subset of $E\mbox{ or } \mathcal{E}\backslash E$ respectively can be written as a union of elements of some subfamily of $\mathcal{U}$. By Urysohn's metrization theorem, it can therefore be said that both $E\mbox{ and } \mathcal{E}\backslash E$ are not regular, and so every non-empty closed subset $C$ of $E\mbox{ and } \mathcal{E}\backslash E$ and a point $p\not\in C$ admit non-overlapping open neighborhoods.
Let $U$ be a subset of $\mathbb{R}^n$. Locally, $\exists$ a diffeomorphism $\phi:E\to U$, since $\mathcal{E}$ is a manifold. $\phi^{-1}$ defines a coordinate system on $E$. The uniqueness of $E$ implies that upto a diffeomorphism, $E$ is an open set of $\mathbb{R}^n$. $E$ is embedded in $\mathcal{E}$. Therefore, consider the map $f:E\to\mathcal{E}$. If $E$ is embedded in $\mathcal{E}$, then $E$ is a manifold if and only if $\forall \mathcal{Z}\in\mathcal{E}$, $x\in f^{-1}(\mathcal{Z})$ has a neighborhood $U\in E$ such that $f^{-1}(\mathcal{Z})\cap U$ is a manifold. It is possible to take $\mathcal{Z}=E$, and $E$ as a sufficiently large neighbor hoods of $x\in f^{-1}(\mathcal{Z})$. Then $f^{-1}(E)\cap E=E$, since $E$ is a manifold. Therefore, $f:E\to \mathcal{E}$ can be taken to be the inclusion map. Therefore, $\mathcal{E}=E$. This shows that $E$ is unique both as a subset and as a compact Hausdorff space.
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