prelude - Topology

Topological space

A set $X$ and a collection of subsets of $X$ , (called open sets) $T$ is a topogical space if:

$\emptyset, X \in T$
if $A_{α} \in T$ is an arbitrary collection of open sets, for each $α \in I$ (where $I$ is any index set), $⋃_{α \in I} A_{α} \in T$ . That is, arbitrary union of open sets, is open.
Similarly, for a finite collection of open sets $A_{i} \in T$ , the intersection $⋂_{i = 1}^{n} A_{i} \in T$ . That is, finite intersection of open sets are open.

Note: for a topological space $(X, T)$ , Saying $A$ is open is equivalent to saying $A \in T$ . So we might simply use open sets to refer to the (sub)sets in the topology $T$ . It is important to note that the concept of "open sets" is therefore dependent on the choice of topology $T$ on $X$ .
Moreover, for a point $p \in X$ we say that $U$ is an open set around $p$ if $p \in U$ . This is just to help write more intuitive sentences.

Neighborhood

Given a topological space $(X, T)$ , and a point $p \in X$ , then $V$ is a neighborhood around $p$ if There is an open set $U$ around $p$ that is contained in $V$ . More formally, $V$ is a neighborhood if $p \in U \subset V \subset X$ , where $U \in T$ . The set of all neighborhoods of $p$ is denoted as $N (p)$

Limit point

let $(X, T)$ be a topological space and $S \subset X$ . then a point $p \in X$ is a limit point of $S$ if for all neighborhoods $V$ around $p$ , $V$ has a non empty intersection with $S$ , and contains points other than $p$ . That is, for all $V \in N (p)$ , $(V - p) \cap S \neq \emptyset$ .
The set of all limit points of $S$ is denoted as $L P (S)$

Closed Set

Let $(X, T)$ be a topological and $S \subset X$ . Then $S$ is closed if it contains all it's limit points. That is, $L P (S) \subset S$ .

CLOSED AND OPEN ARE NOT OPPOSITES!!!!

This is something important to keep in mind. If a set is not closed, that doesn't mean its open. And if a set is not open, it doesn't mean that it is closed. It's a weird case of English names not playing well with their co-responding math definitions.
For example, if $(X, T)$ is a topological space, then $X$ is open by definition ( $X \in T)$ , However, $X$ also contains all its limit points ( $X$ contains all points!) Hence $X$ is also closed. It is both closed and open, or "clopen" (this is an actual term).
Define the standard toplogy on $R$ where the open sets are unions of open intervals $(a, b)$ . consider the set $S = (0, 1]$ . Clearly this set is not open, as no union of open intervals or finite intersection of open intervals can result in $S$ . However, it is also not closed, as $0$ is a limit point of $S$ , but $0 \notin S$ . So this is an example of a set that neither close nor open in the standard topology on $R$ .

Henceforth we may omit mentioning $T$ and simply say $X$ is a topological space, of course, $T$ , a topology of $X$ is still there.

Closure of a set

let $X$ be a topological space and $S \subset X$ . Then the closure of $S$ written $\bar{S}$ is the union of $S$ with all the limit points of $S$ . That is, $\bar{S} = S \cup L P (S)$ .

The closure of a set is closed

Let X be a topological space, and $S \subset X$ . Then $\bar{S} = S \cup L P (S)$ is closed.

$* p r o o f * :$ Suppose $\overset{―}{S}$ is not closed. Then there exists a point $q$ which is a limit point of $\overset{―}{S}$ but not in $\overset{―}{S}$ . Clearly, $q$ is not a limit point of $S$ . Therefore, $q$ must be a limit point of $LP (S)$ . Take any open set $U$ around $q$ . There exists some $p \in LP (S)$ such that $(U - {q}) \cap {p} \neq \emptyset$ . This means that $p \in U$ , which implies that $U$ is an open set around $p$ . Since $p$ is a limit point of $S$ , this also means that $U$ intersects (non-emptily and without $p$ ) with $S$ . Therefore, $q$ must be a limit point of $S$ .

$◻$

The closure is the smallest containing closed set

Let $X$ be a topological space and $S \subset X$ . Then if $E$ is a closed set such that $S \subset E$ , then $\bar{S} \subset E$ . That is, any closed set that contains $S$ must also contain it's closure.

$* p r o o f * :$ Since $E$ is closed, $E$ contains all it's limit points. In particular, $E$ contains all the limit points of $S$ because $E$ contains $S$ . Therefore $E$ contains $\bar{S}$ .

$◻$

Denseness

If $X$ is a topological space, then $A \subset X$ is dense if $\bar{A} = X$ .
denseness of A is a nice property, because any point $x \in X$ is either in A, or is a limit point of A.

Subspace Topology

Let $(X, T)$ be a topological space and $Y \subset X$ . Then inheriting the topology $T_{Y} = {U \cap Y : U \in T}$ , (basically taking the intersection of all open sets with $Y$ ) defines a topological space $(Y, T_{y})$ . (It is left to the reader to verify that $(Y T_{y})$ satisfies the axioms of topology)

Covers and Subcovers

Let $X$ be a topo space, and $S \subset X$ , then an open cover of $S$ ,is a collection of open sets $C = {U_{α} : α \in I}$ such that $S \subset ⋃_{α \in I} U_{a}$ .
If $C^{'} \subset C$ is also a cover of $S$ , then $C^{'}$ is a subcover of $S$ . Note: subcover simply means picking some (maybe all) elements of the cover.

Compactness

let $X$ be a topo space and $S \subset X$ . then $S$ is compact if for every open cover $C (S)$ There exists a subcover $C^{'} (S) \subset C (S)$ such that the cardinality of $C^{'}$ is finite.

Metric space

a set $X$ with a distance function $d : X \to R^{+} \cup {0}$ is a metric space if:

$d (x, y) = 0 ⟺ x = y$ (only identical points have zero distance).
$d (x, y) = d (y, x)$ (symmetry of distance).
$d (x, z) + d (z, y) \geq d (x, y)$ (triangle inequality).

Basis for a topology

let $(X, T)$ be a topological space. Then, a basis of this topology $B \subset T$ for every point $x \in X$ an open set $U$ around $x$ , there exists a "basic open set" $B \in B$ such that $x \in B \subset U$ .
Essentially, the basis is a subcollection of open sets, such that any open set is the union of some basis elements. The elements of the basis are called basic open sets.

(B1) Since $X$ is itself an open set, for every $x \in X$ , there exists a basic open set $B$ around $x$ .
(B2) For any two basic open sets $B_{1}, B_{2}$ if $x \in B_{1} \cap B_{2}$ There exists $B_{3} \subset B_{1} \cap B_{2}$ around $x$ . (this is because the (finite) intersections of open sets are open).
In fact, if there is no topology on $X$ , but we have a collection of subsets $B$ for which (B1) and (B2) holds, Then we can generate a topology from arbitrary unions of elements of $B$ . By induction on (B2) the generated topology is stable under finite intersection, and by (B1) $X$ is part of the generated topology. And by the process of generation, the generated topology is also stable under arbitrary unions.

On convergence:

Consider a set $E$ as the domain of some set of functions, targeted towards a metric space $(Y, d)$ .
The sequence of functions $f_{1}, f_{2}, \dots$ converges to $f$ pointwise, if for every $ϵ > 0$ and $x \in E$ , there exists $N (x, ϵ) \in N$ such that for $n > N (x, ϵ)$ , $d (f_{n} (x), f (x)) < ϵ$ .
The thing about pointwise convergence on $E$ is that, the tail of the sequence that is epsilon close to $f$ at an input $x$ depends not only on epsilon but $x$ as well.

On the other hand, $f_{1}, f_{2}, \dots$ uniformly converges to $f$ if for any $ϵ > 0$ , there exists $N (ϵ)$ such that no matter what choice of $x \in E$ , whenever $n > N (ϵ)$ ,
$d (f_{n} (x), f (x)) < ϵ$ .

Compact convergence

let $(X, T)$ be a toplogical space, $(Y, d)$ a metric space, and $^{X} Y$ the set of all functions from $X$ to $Y$ . A sequence of functions $f_{1}, f_{2}, \dots$ in $^{X} Y$ converges compactly to $f$ in $^{X} Y$ if for every compact subset $K$ of $X$ , the restriction of each $f_{n}$ to $K$ uniformly converges to the restriction of $f$ to $K$ . That is for any $ϵ > 0$ , there is $N (ϵ) \in N$ such that whenever $n > N (ϵ)$ , for all $x \in K$ , $d (f_{n} (x), f (x)) < ϵ$ , for any compact subset $K$ .

Since the inequality holds for all $x \in K$ , we can say that $sup_{x \in K} d (f_{n} (x), f (x)) > ϵ$ for all $n > N (ϵ)$ .
Stating the theorem compactly, for all compact $K \subset X$ , $lim_{n \to \infty} sup_{x \in K} d (f_{n} (x), f (x)) = 0$

In some sense, We want to put a topology on ${#X} Y$ such that if $f_{n} ⟶ f$ (that is the sequence $f_{n}$ converges to $f$ , both in $^{X} Y$ as a topological space) if and only if $f_{n}$ compactly converges to $f$ .

Take any compact set $C \subset X$ , any $ϵ > 0$ and $f \in \ {#XY}$ . Define the open ball \ Basis element: $$B_C(f, \epsilon) = { g \in

{ #XY}
: \sup_{x \in C} \ d(f(x),g(x)) < \epsilon}$$

Then, using $κ X$ to denote all compact subsets of $X$ , $$\mathscr B = \bigcup_{C \in \kappa X} \ \bigcup_{\epsilon > 0} \ \bigcup_{f \in
{ #XY}
} \ B_{C}(f,\epsilon)$$

We now have to verify that $B$ is a basis, by (B1) and (B2).
So (B1) basically states (in this context) that for each $f \in \ {#XY}$ there exists $B \in B$ such that $f \in B$ .
well, pick any compact $C$ and $ϵ > 0$ .
Clearly $B_{C} (f, ϵ) \in B$ , and $f \in B_{C} (f, ϵ)$ as $d (f (x), f (x)) = 0 < ϵ$ .

Lemma

If $g \in B_{C} (f, ϵ)$ and $δ_{c} = ϵ - sup_{x \in C} d (f (x), g (x))$ then $B_{C} (g, δ_{c}) \subset B_{c} (f, ϵ)$ .
proof: let $h \in B_{c} (g, δ_{c})$ , then $sup_{x \in C} d (h (x), g (x)) < ϵ - sup_{x \in C} d (g (x), f (x))$ . Rearrange and use triangle inequality to get $sup_{x \in C} d (h (x), g (x)) < ϵ$ .

let $h \in B_{C} (f, ϵ) \cap B_{D} (t, υ)$ as defined in the lemma, $B_{C} (h, δ_{C}) \subset B_{C} (f, ϵ)$ and $B_{D} (h, δ_{D}) \subset B_{D} (g, υ)$ . It follows that $B_{C \cup D} (h, min (δ_{C}, δ_{D})) \subset B_{C} (f, ϵ) \cap B_{D} (t, υ)$ and contains $h$ . ( $C, D$ are compact so $C \cup D$ is compact). Finally showing (B2).

Okay back to fucking reality. We have put a topology on $^{X} Y$ . But what does convergence in a topological space mean?

Convergence of sequences in topological spaces

A sequence $x_{n}$ converges to $x$ (written $x_{n} ⟶ x)$ in a topological space $X$ if for every nbhd $V$ around $x$ , there exists $N (V) \in N$ such that for all $n > N (V)$ $x_{n} \in V$ .
Equivalently $x_{n} ⟶ x$ if and only if for each basis element $B$ around $x$ there exists $N (B) \in N$ such that for all $n > N (B)$ $x_{n} \in B$ .
proof: if $V$ is a nbhd of $x$ , then there is an open set $U$ where $x \in U \subset V$ . But $U$ is the union of some basis elements, one of them has to contain $x$ , call it $B *$ . So it's sufficient to show that there is a tail of the sequence contained in $B *$ .
But all Basis elements $B$ around $x$ are themselves nbhd's around $x$ . so it is necessary to show that there is a tail of the sequence contained in any such B.

Compact convergence topology

Let $(X, T)$ be a topological space and $(Y, d)$ be a metric space. The compact convergence topology on function from $X$ to $Y$ , $^{X} Y$ is the one generated by the following basis set:

\mathscr B = \bigcup_{C \in \kappa X} \ \bigcup_{\epsilon > 0} \ \bigcup_{f \in \ {#XY} } \ B_{C}(f,\epsilon)

Where

B_C(f, \epsilon) = \{ g \in \ {#XY} : \sup_{x \in C} \ d(f(x),g(x)) < \epsilon\}

And in this topology on $^{X} Y$ $f_{n} ⟶ f$ if and only if $f_{n}$ converges compactly to $f$ . This follows directly by considering any basis $B^{*} = B_{C} (g, δ)$ that contains $f$ .
We must have $N (B^{*})$ such that for all $n > N (B^{*})$
$f_{n} \in B^{*}$ , equivalently, $sup_{x \in C} d (f_{n} (x), g (x)) < δ$ . Hence, $sup_{x \in C} d (f_{n} (x), g (x)) + sup_{x \in C} d (g (x), f (x)) < 2 δ$ . using the triangle inequality and properties of $sup$ , we get $sup_{x \in C} d (f_{n} (x), f (x)) < 2 δ = ϵ$ . Notice here that $C$ is any compact set and $ϵ > 0$ is also arbitrary.

Alternate definition for closed sets

Let $X$ be a topological space. Then $E \subset X$ is closed if and only if $X - E$ is open.
Well suppose $E$ is closed, then any limit point of $E$ is in $E$ . Therefore, no point $p \in X - E$ is a limit point of $E$ . Hence, for each $p \in X - E$ There exists a NBHD $V_{p}$ around $p$ such that $V_{p} \cap E = \emptyset$ . Equivalently, There exists an open set $U_{p}$ around each $p$ such that $U_{p} \cap E = \emptyset$ . Hence $⋃_{p \in X - E} U_{p} = X - E$ . Hence $X - E$ is open (it is a union of open sets).
Now suppose $X - E$ is open, and (for the sake of contradiction) suppose $q^{*} \in X - E$ is a limit point of $E$ . Write $X - E = ⋃_{α \in I} B_{α}$ as a union of some collection of basis elements. By the axiom of choice, $q^{*}$ is in some $B_{α^{*}}$ But since $q^{*}$ is a limit point of $E$ , $B_{α^{*}}$ contains some point $e \neq q^{*} \in E$ . This means that $e \in ⋃_{α \in I} B_{α}$ , which means that $e \in X - E$ , which is a contradiction. Hence no point $q^{*} \in X - E$ is a limit point of $E$ , therefore $E$ contains all its limit points ( $E$ is closed)

Note: whenever we have a definition of something, A := B iff ("def 1"), and then we also have A := B iff ("some condition") we can alternatively take that condition itself as a definition, and recover the older definition as a theorem instead. That is to say, if we define closed sets as ones that contain all their limit points, we can show the theorem: E is closed iff its complement in X is open. Dually, we can define E is closed iff its complement in X is open, and recover E is closed iff it contains all its limit points as a theorem

Continuous maps on topological spaces

let $X, Θ_{X}$ and $Y, Θ_{Y}$ be topological spaces. A map $f : X \to Y$ is a homomorphism (continuous) if for any open subset $V \in Θ_{Y}$ of $Y$ , the pre image of $V$ , $f^{- 1} (V)$ is open in $X$ , that is $f^{- 1} (V) \in Θ_{X}$ (note that since $\emptyset$ is always open in any topology, there is no need for our map to be surjective, if $V$ is not hit by $f$ at all, then its pre-image is empty and so we are okay).
If $f : X \to Y$ is bijective, and both $f, f^{- 1}$ are continuous, then $f$ is a topological isomorphism (called "Homeomorphism")

Given three topological spaces $X, Y, Z$ , if $f : X \to Y, g : Y \to Z$ are both continuous, then $g \circ f : X \to Z$ is continuous. For all $V_{Z} \in Θ_{Z}$ , $g^{- 1} (V_{Z}) \in Θ_{Y}$ . Hence $f^{- 1} (g^{- 1} (Θ_{Z})) \in Θ_{X}$ .
if $f : X \to Y$ is a continuous map, let $S \subset X$ if we put the subspace topology (inheriting from $Θ_{X}$ ) on $S$ , then the restriction $f |_{S} : S \to Y$ is continuous. let $V$ be any open set of $Y$ . $f^{- 1} |_{S} (V) = f^{- 1} (V) \cap S$ . Since $f^{- 1} (V)$ is open in $X$ , by the definition of the subspace topology, $f^{- 1} (V) \cap S$ is open in $S$ .