Chapter 3 - Topological Spaces

3.9 Basic Concepts

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Results

Given a set $X$, a topology in $X$ is a system $\tau$ of subsets $G\subset X$, called open sets (relative to $\tau$), with the following two properties.
(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 $\tau$.




A topological space is a pair $(X,\tau)$ consisting of a set $X$ and a topology $\tau$ defined in $X$.
(In this text, a topological space will usually be called $T = (X,\tau)$).



Closed Sets
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.



Concepts from Metric Spaces
(1) A neighborhood of a point $x$ in a topological space $T$ is any open set $G\subset T$ containing $x$.
(2) A point $x\in T$ is called a contact point of a set $M\subset T$ if every neighborhood of $x$ contains at least one point of $M$.
(3) A point $x\in T$ is called a limit point of a set $M\subset T$ if every neighborhood of $x$ contains infinitely many points of $M$.
(4) The set of all contact points of a set $M\subset T$ is called the closure of $M$, and is denoted by $[M]$.





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




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 $\mathscr{B}$.




A family $\mathscr{G}$ of open subsets of a toplogical space $T$ is called a base for $T$ if every open set in $T$ can be represented as a union of sets in $\mathscr{G}$.




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 $T$.



Useful theorem for determining whether a given system of open sets is a base.


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 $x\in G_\alpha\subset G$.




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 $[M] = T$.




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



First Axiom of Countability
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 first axiom of countability (or that it is a first-countable set).



Second Axiom of Countability
A topological space $T$ is said to satisfy the second axiom of countability if it has at least one countable base.



Covering of Topological Spaces
A system $\mathscr{M}$ of sets $M_\alpha$ is called a cover or covering of a topological space $T$, and $\mathscr{M}$ is said to cover $T$ if, $$ T = \bigcup_\alpha M_\alpha. $$ An open (closed) cover consists of open (closed) sets. A subcover is a subset of the covering that also covers $T$.




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




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 $M$.



General topological spaces are too different from metric spaces. Some additional assumptions makes the resemblance clearer.


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 first axiom of separability, and is called a T₁-space.




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



A stronger seperation axiom; each pair of distinct points has a pair of disjoint neighborhoods.


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 second axiom of separability, and is called a T₂-space or a Hausdorff space.



Spaces more general than Hausdorff are rarely studied. Instead, an even stronger axiom of seperability is used. All pairs of disjoint closed, sets has a pair of disjoint neighborhoods.


A $T_1$-space $T$ is said to be normal if for each pair of disjoint closed sets $F_1$ and $F_2$ in $T$, there are open sets $O_1\supset F_1$ and $O_2\supset F_2$ such that $O_1\cap O_2 = \emptyset$.




Every metric space is normal.



Continuous Mappings
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 continuous at the point $x_0\in X$ if, given any neighborhood $V_{y_0}$ of the point $y_0 = f(x_0)$, there is a neighborhood $U_{x_0}$ around $x_0$ such that $f(U_{x_0})\subset V_{y_0}$. The mapping $f$ is said to be continuous on $X$ if it is continuous at every point of $X$.




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 $X$.




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 $X$.




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.



Homeomorphic Mappings
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 homeomorphic mapping or just a homeomorphism. The spaces will have the same topological properties, and are just two different 'representatives' of the same space.



Topological spaces can be introduced by specifying the open sets, specifying a base, or introducing a notion of convergence. One of the most important ways is by specifying a metric space in $T$.


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

Problem 1

Given a topological space $T$, prove that a set $G\subset T$ is open if and only if every point of $x\in G$ has a neighborhood contained in $G$.

Proof.
$\Rightarrow$) 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 $\bigcup_\alpha O_\alpha = G$. Hence $G = \bigcup_\alpha O_\alpha$ and it is an open set.

Problem 2

Closure - Alternative Definition
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 $$ [M] = \bigcap_\alpha C_\alpha. $$



Problem
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 $G\subset T$.

Proof.
$\Rightarrow$) 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 $[M] = M$.



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

Proof.
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 $M$.



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

Proof.
(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)

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


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

Problem 3

Consider the set $\mathscr{T}$ of all possible topologies defined in a set $X$, where $\tau_1\leq \tau_2$ means that $\tau_1$ is weaker than $\tau_2$ ($\tau_1\subset\tau_2$). Verify that $\leq$ is a partial ordering of $\mathscr{T}$. Is there a maximum and minimum element? If so, what are they?

Proof.
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 $\mathscr{T}$. 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.

Problem 4

Can two distinct topologies, $\tau_1$ and $\tau_2$ in $X$ generate the same relative topology in a subset $A\subset X$?

Proof.
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 $A$: $$ \tau_1\cap A = \{\emptyset, \{a\}\} = \tau_2\cap A. $$

Problem 5

Let $$ X = \{a,b,c\},\quad A = \{a, b\},\quad B = \{b, c\} $$ and let $\mathscr{G} = \{\emptyset, X, A, B\}$. Is $\mathscr{G}$ a base for a topology in $X$?

Proof.
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.

Problem 6

Prove that if $M$ is an uncountable subset of a topological space with a countable base, then some point of $M$ is a limit point of $M$.

Proof.
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 $M$.

Problem 7

Prove that the topological space $T = \{\emptyset, \{b\}, \{a, b\}\}$ is connected. (A space is connected if the only open and closed sets are the empty set and the entire space $X$).

Proof.
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.

Problem 8

Prove that a topology satisfying the second axiom of countability satisfies the first axiom of countability.

Proof.
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.

Problem 8 - Additional

Prove that a topology satisfying the $T_2$/Hausdorff property also satisfies the $T_1$ property.

Proof.
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 $T_1$.

Problem 9

Give an example of a topological space satisfying the first axiom of countability but not the second axiom of countability.

Proof.
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.

Problem 10

Let $\tau$ be the system of sets consisting of the empty set and every subset of the closed interval $[0,1]$ obtained by deleting a finite or countable number of points from $X$. Verify that $T = (X,\tau)$ is a topological space. Prove that $T$ satisfies neither the second nor the first axiom of countability. Prove that $T$ is a $T_1$-space, but not a Hausdorff ($T_2$) space.

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

Showing that this is a topology

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.

Countability Axioms

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.

This topology is T₁

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.

This topology is NOT Hausdorff (T₂)

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.

Problem 11

Skipped...

Problem 12

Prove the converse of Theorem 8. The statement of Theorem 8 can be expressed as follows:

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 $T_1$.

Proof.
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.

Problem 13

Skipped...

Problem 14

Skipped...