The Differential Geometry of $E$

Let's take off from where we were last time...

$E$ is a manifold. Therefore, there exists a local chart with $p\in\mathcal{U}_\alpha$ such that the transition functions $\varphi_\beta\circ\varphi_\alpha^{-1}\circ\varphi^{-1}_\alpha$ is a differentiable function on $\varphi_\alpha(\mathcal{U}_\alpha)$. Consider $E$. For this section, we will disregard the proof shown in the last post, and assume that $E\subseteq\mathcal{E}$. (Note that taking the equality part of the $\subseteq$ gives back the result in the last post.) Let $V$ be a compact subset of $\mathcal{E}$ such that $E\subseteq V$. Then there exists a $\varrho\in C^\infty(\mathcal{E})$ which is identically $1$ on $E$, identically $0$ outside $E$. $\varrho$ is a metric on $E$ since $\forall x\in E,y\in\mathcal{E}\backslash E,\varrho(x,y)\geq0,\varrho(x,y)=\varrho(y,x),\mbox{ if and only if }x=y,\varrho(x,y)=0$. The triangle equality can be shown to be saturated. $\varrho$ can be the Ricci flow, therefore define the local Ricci curvature of $\mathcal{E}$. In fact, the Ricci flow states that $\mathcal{E}$ is locally Ricci flat.

Consider a derivation (in the general case written as $\mathcal{D}$) $X$ on $C^\infty(E)$. Let $X$ be such that $X\varrho=0$. $X$ is therefore the Killing vector field defined locally on $E$. On $E$, let us define an evolution parameter $s$ (physically interpreted as time). Then, upto a constant, $X=\partial/\partial s$. Let $\{X_i\}$ be the set of all Killing vector fields on $E$. Let us write $X_p:\varrho\to(X\varrho)(p)$. Then, $\{X_i\}\subset\bigcup_p\{X_p|X\in\{\mathcal{D}\}\}$. Consider a mapping $\iota:E\to E$. Write $\mathcal{G}_\iota=\mathcal{G}\circ\iota^{-1}\forall\mathcal{G}\in C^\infty(E)$. Then, $\forall X$, $(\mathcal{G}X)_\iota=\mathcal{G}_\iota X_\iota, (X\mathcal{G})_\iota=0=X_\iota\mathcal{G}_\iota$. Let $\mathcal{G}=\varrho$. Then, $(\varrho X)_\iota=\varrho(\iota^{-1}(X(\iota^{-1})))$. $\iota$ is a homeomorphism, therefore, $\iota^{-1}\in C^\infty(E)$. Therefore, $X(\iota^{-1})=0$. Since $\iota$ is the inclusion map, $\iota^{-1}(0)=0$, and therefore, $(\varrho X)_\iota=\varrho(0)$. This shows us that using just the Killing vector field $X$, the metric at $s=0$ can be found. By means of differential equations and functional analysis, it is possible to determine $\varrho(s)$.

Let us define an affine connection on $E$ as $\Delta:\bigcup_p\{X_p|X\in\{\mathcal{D}\}\}\to\bigcup_p\{X_p|X\in\{\mathcal{D}\}\},\Delta_{\mathcal{F}X+\mathcal{G}Y}=\mathcal{F}\Delta_X+\mathcal{G}
\Delta_Y,\Delta_{X}(\mathcal{F}Y)=\mathcal{F}\Delta_X(Y)$
for $X,Y\in \bigcup_p\{X_p|X\in\{\mathcal{D}\}\},\mathcal{F,G}\in C^\infty(E)$. Another affine connection can defined on $\mathcal{E}$ as $(\Delta^{-1}_X(Y))^\iota=\Delta_{X\circ\iota^{-1}}(Y\circ\iota^{-1})$. $\iota^{-1}$ is continuous and thus, $Y\circ\iota^{-1}=0$. Therefore, at a point $p\in E\exists x\in\bigcup_p\{X_p|X\in\{\mathcal{D}\}\}$ such that $X(p)=0$. There also exists an open neighborhood $N$ of $0$ (defined on a coordinate system on $E$) in $E$ and another open neighborhood $M$ of $p\in E$ such that for some maximal geodesic $\gamma$ along $X$, $X\to\gamma_X$ is a diffeomorphism of $N$ onto $M$ such that $\gamma(0)=p,\dot{\gamma}(0)=Y\neq0$. Thus, $\exists$ neighborhood $\mathfrak{N}$ of $X\in\bigcup_p\{X_p|X\in\{\mathcal{D}\}\}$ and interval $\mathfrak{I}$ about $0\in\mathbb{R}$ such that $(\mathfrak{n},t)\mapsto\gamma_\mathfrak{n}(t)$ is smooth on $\mathfrak{N}\times\mathfrak{I}$. Therefore, $M=\exp(N)$. If the evolutionary parameter $s$ is continuous, then we should be able to take $M$ arbitrarily close to $N$, so $M=N\cup\partial N$. Also, there exists $\mathcal{M}=M\cup\partial M,\mathcal{M}=\exp(\exp(N))$. Substituting this equation into $\mathcal{M}=M\cup\partial M$ gives $\exp(\exp(N))=\exp(N)\cup\partial\exp(N)$. Using the methods of boundary operators provides $N=\emptyset$. To resolve this, the evolutionary parameter $s$ must not be continuous. Therefore, if one considers a local tangent vector $\omega$ (Ricci flatness implies same direction) on $\mathcal{E}$, then since $s=||\omega||$, $\omega$ provides a direction of $s$ and is therefore quantized. This can therefore be generalized to conclude that \emph{all elements of $\bigcup_p\{X_p|X\in\{\mathcal{D}\}\}$ are quantized.} This provides some method of quantizing the metric $g$ defined on $E$ in the following manner. $X$ is quantized. Therefore, $\exists$ a smooth vector field $Y$ such that $X=nY$. Since $X\varrho=0$, $nY\varrho=Y(n\varrho)=0$. Therefore, the Killing vector field can be made smooth for the price of quantizing the metric ($\varrho$ is a semi-classical metric and therefore satisfies the flat-space Einstein equations. Solving provides one with the solution: $\forall s,\varrho(s)=\varrho(0)$. This means that the ``graviton'' is constant through general relativistic ``time''.).

Let the affine connection be the Levi-Civita connection $\Delta$. Obviously, $X=\Delta$, therefore $\Delta\varrho=0$. By the definition of the Levi-Civita connection, $\Delta:\mathrm{Riem}(E)\to\mathrm{Conn}(E)$, where $\mathrm{Conn}(E)$ is the space of all torsionless connections on $E$. All $\varrho$ are in $C^\infty(E)$, so $X\varrho=\Delta\varrho=0$. Thus, $\mathrm{Conn}(E)=\{0\}$, so the only torsionless connection on $E$ is $\Delta=X=0$. The Ricci curvature $\mathrm{Ric}$ is therefore $0$. Let $A$ be a torsionless tensor field so that $\Delta^{'}=\Delta+A=0$. Therefore, $\mathrm{d}\mathrm{Ric}=0$. Let $\hat{\varrho}$ be the scalar product induced by $\varrho$ on $T^{*}E$. Then, $\int_E\hat{\varrho}\mathrm{d}\mathrm{Ric}g=0$. Let $\mathcal{S}=-\int_E\hat{\varrho}\mathrm{Ric}g$. Then, evaluating and using Hamilton's least action principle gives $\hat{\varrho}\circ\hat{\varrho}=-2$. Writing as an integral gives $\int_E\mathrm{d}\hat{\varrho}\circ\varrho=-1$. If $\hat{\varrho}$ was a Grassmann variable, then this integral becomes $\int_E\varrho\circ\mathrm{d}\hat{\varrho}=1$, showing that $\hat{\varrho}$ is indeed a Grassmann variable. This is plausible since $\varrho$ is a $C^\infty$-multilinear function on $E$.

$\varrho$ is the creation operator of a classical fermionic field, and its complex conjugate is the annihilation operator of the same classical fermionic field. Since $\varrho$ is a creation operator, it must have only positive eigenvalues, so $\varrho$ is the $++++$ metric. I.e., $E$ is a homogenous Riemannian manifold. Therefore, $E$ can be embedded in $\mathbb{R}^n$. Hence, it is locally homeomorphic to $\mathbb{C}^{n/2}$ via charts $\varphi_i:U_i \to V_i,U_i \subset E, V_i \subset \mathbb{C}^{n/2}$ open and such that $\varphi_j \circ \varphi_i^{-1}: V_i \cap V_j \to V_i \cap V_j$ is holomorphic. $E$ is therefore
homeomorphic via charts on $\mathbb{C}^{n/2}$ to $\underbrace{\mathcal{R}\times\mathcal{R}\times\mathcal{R}\times\ldots\times\mathcal{R}}_{n/2\mbox{-times}}$. This space is unique in the sense explained above. Therefore, the Tychonoff topology induced on $E$ via $\mathcal{R}$ is unique upto a homeomorphism. We ask - what is the genus of $\mathcal{R}$? The only Riemann surface that is unique for low genus is the torus (genus $1$). Therefore, $E=\mathrm{T}^n$. This is compact (as we wanted) and is also an Abelian Lie group. Therefore, there exists a finite extension $1\to E\to\pi_0(E)\to 1$. If $n=4$, as in conventional general relativity, then $E$ is hyperK\"{a}hler \footnote{An algebraic geometric intervention: A Hilbert scheme of $E$ has a natural morphism to the infinite symmetric product of $E$.} (and thus symplectic) and has its holonomy group contained in $SU(2)$. Since $E$ is hyperK\"{a}hler, it is the Pontryagin dual of a discrete torsion-free Abelian group. An $SU(2)$ holonomy would produce $\mathcal{N}=2$ supersymmetry on $E$. We will not look into supersymmetric spacetime in this post, but we have shown that supersymmetry must be an intrinsic property of spacetime.

Separability of $E$ and $E_C$ and a few more interesting concepts (Uniqueness of $E$!)...

Let's continue from the last post!

$S_C$ 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.

Introducing Myself and some technical formalisms...

Hey Guys! I'm Sanath Devalapurkar and I am a differential topologist, with an emphasis on the topological aspects. I am interested in mathematical physics and will be using this blog as a forum to publish my ideas on spacetime, relativity, quantum mechanics, and of course - Quantum Gravity!

Shall we begin?

Let us define a set $E$ as spacetime, with metric $g$ (All spacetimes have a Lorentzian metric, and so we shall endow $E$ with one.). Consider an infinitesimal $(n-1)$ dimensional sphere with radius $\epsilon$ and center $x\in E$, embedded in a spacelike $(n-1)$ dimensional slice of $E$. Let $\epsilon$ be small enough so that the metric at the point $x+\delta, |\delta|\leq|\epsilon|$, can be approximated by a Taylor expansion $g(x+\delta)=\sum_{k=0}\frac{\delta^k}{k!}\frac{\mathrm{d}^kg(x)}{\mathrm{d}x^k}$. Define this particular (non-unique) $(n-1)$ dimensional disk $D^{n-1}$ with radius $\epsilon$ as the \emph{Taylor neighborhood} $\mathcal{O}_x$ of the point $x$. This defines a topology on $E$ by $T=\{\mathcal{O}\subsetneq E|\forall x\in\mathcal{O}\exists\epsilon>0, \{y\in E|g(x,y)<\epsilon\}\subsetneq\mathcal{O}\}$. This provides a few local topological properties of spacetime:

• $\forall x,y\in E,\exists\mathcal{O}_x,\mathcal{O}_y\mbox{ such that }x\not\in\mathcal{O}_y$;
• $\forall x,y\in E,\exists\mathcal{O}_x,\mathcal{O}_y\mbox{ such that }\mathcal{O}_x\cap\mathcal{O}_y=\emptyset$;
• $\forall \mathcal{O}_x,\mathcal{O}_y,\exists\mbox{ closed neighborhoods }M,P\subset E\mbox{ such that }\mathcal{O}_x\cap\mathcal{O}_y=\emptyset$.

Consider the set $S$ of all possible sequences in $E$. Let $l$ be some finite number. Then, a bounded sequence is defined as a converging sequence $(x_n)$: $\forall U\in\{\mbox{set of all (Taylor) neighborhoods of $x$}\}=\mathcal{C}(x)\exists N\in\mathbb{N}\mbox{ such that }\forall n\geq N\Rightarrow x_n\in U$. We will consider $E$ to have the property that all sequences are complete, i.e., the sequence converges to some $l\in E$. Write $S_C$ as the set of all unbounded sequences factored out from $E$. Let $E_C$ be the set that results from this factorization. $E_C$ is the compactification of $E$. Let us define $\partial E$ as the set of all points in $E$ such that for some $P\in E$, the neighborhood of $P$ contains points which lie in $E$ and also do not lie in $E$.

We would now like to clarify what was meant by ``results from this factorization'' in the previous paragraph. A sequence is an ordered open set. The union of sequences is also a sequence, and the same hold for intersections. There also exists the ``empty'' sequence, and the sequence containing all the points of $E\backslash\partial E$ (we say $E\backslash\partial E$ and not $E$ because sequences are open). Therefore, the set of all sequences is a topology. In fact, by grouping together elements of $E$ to form ordered sequences, it is seen that the topology generated by $S$ is the ``ordered'' discrete topology.

We now attempt to find if $S_C$ is a topology. Consider two disjoint sequences in $S_C$. These sequences are complete and bounded. Therefore, there exists some finite $l\in E$ to which the sequences converge. (The $l$ for each sequence may be different.) The union of these sequences is also a sequence, and the intersection is the empty sequence. The sequence containing all the points of $E_C$ is complete if $E_C$ is compact. By the construction of $E_C$, $E_C$ is compact, and therefore since complete subspaces (if these subspaces are open, they are sequences) of $E_C$ are closed (since the compactification of a metric space is a metric space), and sequences as defined above are open, complete sequences in $E_C$ do not exist, with the exception of the empty sequence and $E_C\backslash\partial E_C$ itself. $S_C$ is then just the set containing the empty sequence and $E_C\backslash\partial E_C$, i.e., $S_C$ provides the trivial topology on $E_C\backslash\partial E_C$.

This is interesting... What does this mean for the separability of $E$ and $E_C$? We'll find out next time! So long, friends!