lol

1 files changed, 63 insertions, 3 deletions

diff --git a/mindmap/limit.org b/mindmap/limit.org
index d25409b..d0f6679 100644
--- a/mindmap/limit.org
+++ b/mindmap/limit.org
@@ -73,11 +73,42 @@ where the central "object of importance" (a common theme in math is one where yo
 Specifically, the limits of universal nets have a deep relation to compactness, but here we will explore the most informative and essential
 form of this idea and its algebraic properties. We will quickly go over the one-point compactification, and then introduce the stone-cech
 compactification after.
-*** One Point Compactification
+** One Point Compactification
 :PROPERTIES:
 :ID:       339b32e7-ad89-40d7-8b11-5b293bd1056f
 :END:
-*** Stone Cech Compcatification
+The one-point compactification is the simplest possible compactification of a topological space as you are adding only one point, and it does have a
+rather simple definition, although it is really only interesting in locally compact hausdorff spaces.
+
+Let $X$ be a locally compact Hausdorff space, then its one-point compactification is $X \cup \lbrace \infty \rbrace$, where the topology defined on
+this is as follows:
+1. if $U$ is open in $X$, $U$ is open in $X \cup \lbrace \infty \rbrace$.
+2. if $F \subset X$ is a compact subset and $\infty \in F^{c}$, then $F^{c}$ is open.
+The topology generated by these open sets it the topology associated with the one-point compactification of $X$. If $X$ is locally compact hausdorff,
+then in fact this topology is compact hausdorff, which is why it is the notable case. We shall see this in a proof.
+#+begin_theorem
+If $X$ is a locally compact Hausdorff space and $X^{\plus} = X \cup \lbrace \infty \rbrace$ is the one-point compactification of $X$, then $X^{\plus}$ is a
+compact Hausdorff space.
+#+end_theorem
+
+#+begin_proof
+In order to prove this, we must first prove it is compact, then we must prove it is Hausdorff. For the first we will use proof by
+contradiction. Let $\lbrace x_{\alpha}\rbrace$ be a universal net in
+$X^{\plus}$, then suppose $\lbrace  x_{\alpha}\rbrace$ does not converge in $X^{\plus}$. Then $\lbrace x_{\alpha} \rbrace$ also doesn't converge to $\infty$, and let $U_{\infty}$ be an open
+neighborhood of $\infty$ which $\lbrace x_{\alpha} \rbrace$ is not eventually in. Then the complement $U_{\infty}^{c}$ must be compact (the only way to define a
+neighborhood of $\infty$ is in terms of the complements of compact sets). But if $\lbrace x_{\alpha} \rbrace$ is eventually in $U_{\infty}^{c}$ it is eventually in a compact
+set and must converge. However $\lbrace x_{\alpha} \rbrace$ is universal and therefore must eventually be in either $U_{\infty}$ (impossible by construction) or $U_{\infty}^{c}$
+(also impossible). Contradiction!
+
+To prove that it is Hausdorff, it is enough to prove that $\infty$ is separated from the other points (this is because all points in $X$ already are
+separated by open sets in $X$). Let $x \in X$, then there exists an open neighborhood $U$ such that $\infty \not \in \overline{U}$ (choose $U$ such that
+$\overline{U}$ is compact, and this set exists due to locally compact property of $X$). Then $\overline{U}^{c}$ is a neighborhood of $\infty$ and is disjoint from $U$.
+#+end_proof
+
+Importantly, the one-point compactification can be thought of as a generalisation of the compactification of $\mathbb{R}^n$ via identification with
+$S^n$, and it can be thought of as undoing stereographic projection. It is also the smallest possible compactification as you are only adding one
+point. Note that it is possible for $X$ itself to be compact, and in that case $\infty$ is a disconnected component.
+** Stone-Cech Compactification
 :PROPERTIES:
 :ID:       14bebb09-2e38-4b55-adc0-97ba571331af
 :END:
@@ -85,7 +116,8 @@ We can construct the Stone Cech Compcatification on a completely regular topolog
 but will at least give us the Hausdorff property in the compactified space. To start, let $A$ be the set of all $f_{\alpha}: X \rightarrow [0, 1]_{\alpha}$  such that $f$ is
 continuous (with $\alpha$ being an arbitrary but consistent index), and let us define a Tychonoff space $Y = \prod_{\alpha \in A}[0, 1]_{\alpha}$  and an embedding $\phi: X \rightarrow Y$
 where the embedding $\phi$ is defined as $(\phi(x))_{\alpha }= f_{\alpha}(x)$. Then the idea is that the /closure/ of $\phi(X)$ in $Y$ is a compactification of $X$.
-In fact, this is sort of analogous to currying in the theory of computer science.
+In fact, this is sort of analogous to currying in the theory of computer science, or delayed or /lazy evaluation/, and as we shall see, it will share
+similar algebraic properties.
 
 How do we know the space is compact? We know that $Y$ is compact because $[0, 1]$ is compact, and we apply Tychonoff's theorem. How do we know that
 $\overline{\phi(X)}$ is compact? It is closed and a subset of a compact set. However, what we have /not/ shown thus far is that $\phi(X)$ is truly an embedding. To see
@@ -93,3 +125,31 @@ this, the completely regular property of $X$ saves the day; if we /didn't/ have
 separated by any function, and then you'd lose the one-to-one property of $\phi$. Also, $\phi$ is clearly always continuous; we use the property that
 $\pi_{\alpha}\circ \phi(x) = f_{\alpha}(x)$, and $\phi$ is continuous iff its projections $f_{\alpha}$ are continuous. Now all we need to show is that $\phi^{-1}$ is continuous, which we can
 also do with the completely regular property.
+
+Before this we will introduce some more standard notation that will make it seem much more like currying in programming. For example we can just drop
+the index $\alpha$ and index by the function $f$ instead. Of course in real programming this is terrible as you'd want to index by pointer, however in math
+we have infinite power so we're just going to index by the literal function. How this will work is that instead of writing
+$(\phi(x))_{\alpha} = f_{\alpha}(x)$, we will instead
+write $\phi(x)(f) = f(x)$, and we will index our space with the set $A$ directly. The standard way to write the space $\overline{\phi(X)}$ is actually $\beta X$ so we'll
+write it that way from now on as well. I just thought that the above would have been a more intuitive explanation for the concept for me in the past. We will
+call $\phi$ the /evaluation map/, for obvious reasons if you come from programming.
+#+begin_theorem
+if $X$ is a completely regular space and $\phi: X \rightarrow \beta X$ is the evaluation map of all continuous $f: X \rightarrow [0, 1]$, then $\phi$ is an open map on its image $\phi(X)$.
+#+end_theorem
+
+#+begin_proof
+In this proof we will use the net definition of continuity. Suppose $\phi(x_{\alpha}) \rightarrow \phi(x)$, yet $x_{\alpha} \not \rightarrow x$. Then there exists some open neighborhood  $U$ of
+$x$ such that $x_{\alpha}$ is not eventually in $U$. Because of complete regularity, there exists a map $f$ separating $x$ from $U^{c}$. If
+$\phi(x_{\alpha}) \rightarrow \phi(x)$, then $\pi_{f} \circ \phi(x_{\alpha}) \rightarrow \pi_{f} \circ \phi(x)$, but clearly this is equivalent to $f(x_{\alpha}) \rightarrow f(x)$. Any subnet of a convergent net converges to the same
+value, so we create a subnet $\lbrace y_{\alpha}\rbrace$ of $\lbrace x_{\alpha} \rbrace$ such that $\lbrace y_{\alpha}\rbrace$ is eventually in
+$U^{c}$ (this is possible because $\lbrace x_{\alpha}\rbrace$ is frequently in $U^{c}$). Then $f(y_{\alpha}) \rightarrow 1$ ($f(U^{c}) \equiv 1$ by construction), yet $f(x) = 0$. But this is clearly absurd, because $\lbrace y_{\alpha} \rbrace$
+converges uniquely, and the constant $1$ net cannot converge to $0$! Contradiction.
+#+end_proof
+*** Algebra on Limits
+Often times it is useful to think of limits as /objects in themselves/ rather than an object that you apply to, say, a sequence. Often times algebras on
+different /kinds/ of limits enables oneself to draw on connections between limits and many other fields of mathematics. For instance, the /closure/ of a
+set is exactly the same set with all its limit points included, and both closures, and as we will see, limits, are /idempotent/, which is to say,
+applying them once is the same thing as applying them twice. Note that if $f: X \rightarrow Y$ where $Y$ is any topological space and $f$ is any continuous
+function, then $\beta f(X) = f(\beta X)$, which one can represent with a commutative diagram, where $\beta f$ is the /unique extension/ of the mapping $f$. Actually, in a moment
+we will see that the funcor commuting is equivalent to the /limit/ commuting.
+