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2.8 Contraction Mappings |

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Definition 1

Given a set $X$, a

(1) The set $X$ itself and $\emptyset$ belong to $\tau$.

(2) Arbitrary (finite or infinite) unions $\bigcup_\alpha G_\alpha$ and finite intersections $\bigcap_\alpha G_\alpha$ belong to $\tau$.

Definition 2

A

(In this text, a topological space will usually be called $T = (X,\tau)$).

The open sets in a topology are specified. Closed sets are the complements of the open sets; $T-G$. So, $X$ and $\emptyset$ are closed (and open) sets. Closed sets are closed under arbitrary intersections and finite unions.

(1) A

(2) A point $x\in T$ is called a

(3) A point $x\in T$ is called a

(4) The set of all contact points of a set $M\subset T$ is called the

Theorem 1

The intersection $\tau = \bigcap_\alpha \tau_\alpha$ of any set of topologies on $X$ is itself a topology in $X$.

Corollary

Let $\mathscr{B}$ be any system of subsets of a set $X$. Then there exists a minimal topology containing $\mathscr{B}$, i.e. a topology $\tau(\mathscr{B})$ containing $\mathscr{B}$ and contained in every topology containing $\mathscr{B}$.

Definition 3

A family $\mathscr{G}$ of open subsets of a toplogical space $T$ is called a

Theorem 2

Given a set $T$, let $\mathscr{G}$ be a system of subsets of $G_\alpha\subset T$ with the following two properties:

(1) Every point $x\in T$ belongs to at least one $G_\alpha\in \mathscr{G}$.

(2) If $x\in G_\alpha\cap G_\beta$, then there is a $G_\gamma\in\mathscr{G}$ such that $x\in G_\gamma\subset G_\alpha\cap G_\beta$.

Suppose that $\emptyset$ and all sets representable as unions of sets $G_\alpha$ are designated as open. Then $T$ is a toplogical space, and $\mathscr{G}$ is a base for $T$.

Theorem 3

A system $\mathscr{G}$ of open sets $G_\alpha$ in a topological space $T$ is a base for $T$ if and only if, given any open set $G\subset T$ and any point $x\in G$, there is a set $G_\alpha\in\mathscr{G}$ such that $x\in G_\alpha\subset G$.

Theorem 4

If a topological space $T$ has a countable base, then $T$ contains a countable everywhere dense subset, i.e. a countable set $M\subset T$ such that $[M] = T$.

Theorem 5

If a metric space $R$ has a countable everywhere dense subset, then $R$ has a countable base.

Let $x\in X$ be some point. If there exists a countable neighborhood base at $x$, i.e. a countable system $\mathscr{O}$ of neighborhoods of $x$ with the following property: Given any open set $G$ containing $x$, there is a neighborhood $O\in\mathscr{O}$ such that $O\subset G$. Suppose every point $x$ of a topological space $T$ has a countable neighborhood base. Then $T$ is said to satisfy the

A topological space $T$ is said to satisfy the

A system $\mathscr{M}$ of sets $M_\alpha$ is called a

Theorem 6

If $T$ is a topological space with a countable base $\mathscr{G}$, then every open cover $\mathscr{O}$ has a finite or countable subcover.

Theorem 7

If a topological space $T$ satisfies the first axiom of countability, then every contact point of $x$ of a set $M\subset T$ is the limit of a convergenct sequence of points in $M$.

Definition 4

Suppose that for each pair of distinct points $x$ and $y$ in a topological space $T$, there is a neighborhood $O_x$ of $x$ and a neighborhood $O_y$ of $y$ such that $x\not\in O_y$ and $y\not\in O_x$. (Error in book!). Then $T$ is said to satisfy the

Theorem 8

Every finite subset of a $T_1$-space is closed.

Definition 5

Suppose that for each pair of distinct points $x$ and $y$ in a topological space $T$, there is a neighborhood $O_x$ of $x$ and a neighborhood $O_y$ of $y$ such that $O_x\cap O_y = \emptyset$. Then $T$ is said to satisfy the

Definition 6

A $T_1$-space $T$ is said to be

Theorem 9

Every metric space is normal.

Let $f: X\rightarrow Y$ be a mapping between topological spaces, so $f$ associates an element $y = f(x)$ for each element $x\in X$. Then $f$ is said to be

Theorem 10

A mapping $f$ of a topological space $X$ into a topological space $Y$ is continuous if and only if the preimage $\Gamma = f^{-1}(G)$ of every open set $G\subset Y$ is open in $X$.

Theorem 10'

A mapping $f$ of a topological space $X$ into a topological space $Y$ is continuous if and only if the preimage $\Gamma = f^{-1}(F)$ of every closed set $F\subset Y$ is closed in $X$.

Theorem 11

Given topologial spaces $X$, $Y$ and $Z$, suppose $f:X\rightarrow Y$ is continuous and $\phi:Y\rightarrow Z$ is continuous. Then the mapping $\phi\circ f = \phi(f(x))$ is continuous.

If $f:X\rightarrow Y$ and $f^{-1}: Y\rightarrow X$ are continuous functions between two topological spaces $X$ and $Y$ then $f$ is called a

Urysohn's Metrization Theorem

A necessary and sufficient condition for a topological space with a countable base to metrizable is that it be normal.

$\Rightarrow$) Assume $G$ is open. Then for any point $x\in G$, it follows that $x\in G\subset G$ which is a neighborhood.

$\Leftarrow$) Assume every $x\in G$ has some neighborhood $O$ such that $x\in O\subset G$. By the property of topologies, the union of all these sets $\bigcup_\alpha O_\alpha$ is open. Our claim is that $\bigcup_\alpha O_\alpha = G$.

$\supset$) Assume $x\in G$, by our assumption, $x\in O\subset G$ so $x\in \bigcup_\alpha O_\alpha$ and so
$G\subset \bigcup_\alpha O_\alpha$.

$\subset$) Assume $x\in \bigcup_\alpha O_\alpha$, then for some $\beta$, we have $x\in O_\beta\subset G$ by our assumption. It follows that $x\in G$, which means that $\bigcup_\alpha O_\alpha\subset G$.

Hence $G = \bigcup_\alpha O_\alpha$ and it is an open set.
$\subset$) Assume $x\in \bigcup_\alpha O_\alpha$, then for some $\beta$, we have $x\in O_\beta\subset G$ by our assumption. It follows that $x\in G$, which means that $\bigcup_\alpha O_\alpha\subset G$.

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Define the closure of $M$ be the arbitrary intersection of all closed sets containing $M$. That is, if each $C_\alpha$ is a closed set such that $M\subset C_\alpha$, then $$ [M] = \bigcap_\alpha C_\alpha. $$

Given a topological space $T$, prove the following statements.

(a) $[M] = M$ if and only if $M$ is a closed set, i.e. the complement $T-G$ of an open set $G\subset T$.

$\Rightarrow$) Assume $[M] = M$. Then, by the alternative definition, $$ [M] = \bigcap_\alpha C_\alpha \;\;\Longrightarrow\;\; M = \bigcap_\alpha C_\alpha $$ which is an arbitrary intersection of closed sets, and hence $M$ is closed.

$\Leftarrow$) Assume $M$ is a closed set. We will show that $[M] = M$.

$\supset$) Let $\overline{M}$ denote the set of contact points. Then $[M] = M\cup\overline{M}$,
and so $M\subset[M]$.

$\subset$) Since $M$ is closed, then $M = C_\beta$ is a closed set containing $M$, so $$ [M] = \bigcap_\alpha C_\alpha \subset M \;\;\Longrightarrow\;\; [M] \subset M. $$ By inclusion both ways, $[M] = M$.

$\subset$) Since $M$ is closed, then $M = C_\beta$ is a closed set containing $M$, so $$ [M] = \bigcap_\alpha C_\alpha \subset M \;\;\Longrightarrow\;\; [M] \subset M. $$ By inclusion both ways, $[M] = M$.

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(b) $[M]$ is the smallest closed set containing $M$.

Let $C$ be any closed set such that $M\subset C$, then $$ [M] = \bigcap_\alpha C_\alpha \subset C $$ which shows that $[M]\subset C$. So $[M]$ is the smallest closed set containing $M$.

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(c) The closure operator, i.e. the mapping of $T$ into $T$ carrying $M$ into $[M]$ satisfies:

1) If $M\subset N$ then $[M]\subset[N]$

2) $[[M]] = [M]$

3) $[M\cup N] = [M]\cup[N]$

4) $[\emptyset] = \emptyset$

2) $[[M]] = [M]$

3) $[M\cup N] = [M]\cup[N]$

4) $[\emptyset] = \emptyset$

(1) Assume $M\subset N$, and we will show that $[M]\subset[N]$. By definition of closed sets, let $C_\alpha$ be the closed sets containing $N$. Then, $$ [N] = \bigcap_\alpha C_\alpha. $$ Since $M\subset N$, then $M\subset C_\alpha$ for each $\alpha$, so: $$ M \subset \bigcap_\alpha C_\alpha = D_\gamma. $$ So, $D_\gamma$ is a closed set containing $M$. Again, let $D_\beta$ be the closed sets containing $M$. By definition of the closure, $$ [M] = \bigcap_\beta D_\beta = \bigcap_{\beta\not=\gamma}D_\beta\cap D_\gamma \subset D_\gamma = [N]. $$ Therefore, $[M]\subset[N]$.

(2) Since $[M]$ is a closed set, then by (a), $[[M]] = [M]$.

(3)

$\subset$) We have that $M\cup N\subset [M]\cup[N]$, which is a closed set containing $M\cup N$ since it is a finite union
of closed sets. So,
$$
[M\cup N] = \bigcap_\alpha C_\alpha \subset [M]\cup[N].
$$
$\supset$) Since $M\subset M\cup N$, it follows from (1) that $[M]\subset [M\cup N]$, and similarly, $[N]\subset[M\cup N]$.
Since both of these are subsets, we can conclude that $[M]\cup[N]\subset[M\cup N]$.

By inclusion both ways, $[M\cup N] = [M]\cup[N]$.

(4) Since $\emptyset$ is a closed set, then by (a), $[\emptyset] = \emptyset$.

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Verifying the three properties of a partial order.

(Reflexive). For some topology $\tau$, then obviously, $\tau \subset \tau$, so $\tau\leq\tau$ is true.

(Transitivity). If $\tau_1\leq \tau_2$ and $\tau_2\leq \tau_3$, we have $\tau_1\subset\tau_2\subset\tau_3$ from which it clearly follows that $\tau_1\subset\tau_3$, and $\tau_1\leq\tau_3$.

(Antisymmetry). If $\tau_A\leq\tau_B$ and $\tau_B\leq\tau_A$, then they are included in each other, so $\tau_A = \tau_B$ which shows that antisymmetry is satisfied.

All properties are met, so we can conclude that $\leq$ is a partial ordering on $\mathscr{T}$.

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There are minimal and maximum elements. The mimium element is the trivial topology, $\{\emptyset, X\}$ which must be included
in all other topologies. The maximum element is the power set of all possible subsets of $X$, denoted by $\mathcal{P}(X)$ which
is a topology, and which contains all other topologies.
Yes, this is possible. Consider the set $X = \{a, b, c\}$ with the following topologies. $$ \tau_1 = \{\emptyset, \{b\}, X\}, $$ $$ \tau_2 = \{\emptyset, \{c\}, X\}. $$ These are distinct topologies. If we set $A = \{a\}$, then the relative topologies become the trivial topology on $A$: $$ \tau_1\cap A = \{\emptyset, \{a\}\} = \tau_2\cap A. $$

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Note that $A\cap B = \{b\}$ and that there is no open set containing $b$ that is a subset of $A\cap B$. This breaks with the second property of Theorem 2, so $\mathscr{G}$ is NOT a valid base for a topology.

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Assume that $M$ doesn't have a limit point. Then for any $x\in M$ there exists some $G_x\in\mathscr{G}$ such that $G_x\cap M = \{x\}$. For another point $y\in M$ we can similarly find $G_y\cap M = \{y\}$. This means that $x\not= y\Rightarrow G_x\not= G_y$ which defines an injection from $M$ to $\mathscr{G}$ so $|M|\leq|\mathscr{G}|$ which suggests that $M$ is countable, but this is a contradiction. Therefore there must be a limit point in $M$.

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The empty set and $\{a, b\}$ are open and closed. The set $\{b\}$ is only open since $\{a,b\}-\{b\} = \{a\}$ is closed as it is not specified in the topology. Hence, the topology $T$ is connected.

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Assume that $T$ satisfies the second axiom of countability, i.e. $T$ has a countable base $\mathscr{G}$. By Theorem 3, for any $O_x\subset T$ containing $x$, there is a set $G_\alpha\in\mathscr{G}$ such that $x\in G_\alpha\subset O_x$, which shows that the topology satisfies the first axiom of countability.

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Assume that the topological space $T$ satisfies the Hausdorff property. Then for any two distinct points $x$ and $y$, there are open sets containing them $O_x$ and $O_y$ such that $O_x\cap O_y = \emptyset$. Since $x\in O_x$ and $x\not\in O_x\cap O_y$, then $x\not\in O_y$, and similarly, $y\not\in O_x$. We can conclude that $T$ is $T_1$.

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A discrete, uncountable topology satisfies the first axiom of countability, but not the second axiom. In the discrete topology, every distinct point is an open set, so it cannot be covered by a countable base.

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Set $X = [0,1]$, let $G_\alpha$ be open sets in $\tau$ and $F_\alpha$ be closed sets.

As specified, $\emptyset\in\tau$. If we remove 0 points from $X$, we are left with $X$, so $X\in\tau$. Next we need to show that open sets are closed under arbitrary unions and finite intersections.

Let $G_\alpha\in\tau$ for all $\alpha\in I$ for some arbitrary index set $I$. Let $F_\alpha = X - G_\alpha$. Then: $$ \bigcup_\alpha G_\alpha = \bigcup_\alpha (X - F_\alpha) = X - \bigcap_\alpha F_\alpha. $$ Since closed sets are closed under arbitrary intersections, and $\bigcap_\alpha F_\alpha \subset F_\beta$ for some closed set that is at most countable, the arbitrary union is $X$ with a countable number of points removed. So, $\bigcup_\alpha G_\alpha\in\tau$ and $\tau$ is closed under arbitrary unions.

Let $G_k\in\tau$ for $k = 1,2,\ldots, N$. Then: $$ \bigcap_k G_k = \bigcap_k(X - F_k) = X - \bigcup_k F_k $$ A finite union of closed sets is closed, and a finite union of countable sets is countable, which means $\bigcap_k G_k \in\tau$ and $\tau$ is closed under finite intersections.

All properties are verified, so $\tau$ is topology.

Since $A_2\Rightarrow A_1$ as shown in Problem 8, the converse states that $\neg A_1 \Rightarrow\neg A_2$. We only need to show that this topology does not satisfy the first axiom of countability and the second result being false follows. Let $G$ be an open set containing $x$ which is $X$ where all rational points - except $x$ if it is rational - have been removed (which is countable). Then there are not a countable number of open sets $O\ni x$ such that $O\subset G$. There is only the one set, $G$! So the first axiom of countability is not satisfied. And then, neither is the second axiom.

Let $x,y\in X$ be two distinct points. Set $O_x = [0, y)\cup(y, 1]$ and $O_y = [0, x)\cup(x, 1]$ which are open sets since we only remove a single point. Then $x\in O_x$ and $y\in O_y$, but $x\not\in O_y$ and $y\not\in O_x$, which shows that this is a $T_1$ space.

Let $x$ and $y$ be two distinct points in $X$. Then any open set containing $x$ will have at most a countable set of $X$ removed and similarly for open sets containing $y$. Then the intersection $O_x\cap O_y \not= \emptyset$. In fact, it will be an uncountable set, so this topology is NOT Hausdorff.

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If the topological space $T$ is a $T_1$ space, then every finite subset is closed.

The converse is therefore:

If every finite subset of a topological space $T$ is closed, then $T$ is $T_1$.

Assume that $T$ is some topological space and that every finite subset is closed. In particular, this means that the point $\{x\}$ is a closed set, and $X - \{x\}$ is open. Take two distinct points $x,y$. Then $x\in O_x = X-\{y\}$ and $y\in O_y = X-\{x\}$, but by construction, $x\not\in O_y$ and $y\not\in O_x$. Hence, $T$ satisfies the first axiom of separability and is a $T_1$-space.

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