Classnotes for September 20

From 0506Topology

#	Week of...	Links (edit)
Fall
1	Sep 12	About, Tue, Thu, Std2Disc
2	Sep 19	Tue, Thu, HW1, 14 Sets
3	Sep 26	Tue, Thu, Photo
4	Oct 3	Tue, Thu
5	Oct 10	HW2, Tue, Thu
6	Oct 17	Tue, Thu, HW3
7	Oct 24	Mon, Tue, Thu
8	Oct 31	Tue, Thu, HW4
9	Nov 7	TE1, Tue, Thu
10	Nov 14	Tue, Thu, HW5
11	Nov 21	Tue, Thu
12	Nov 28	Tue, Thu, HW6
13	Dec 5	Tue, Thu
E	Dec 12	TE2
Spring
14	Jan 9	Tue, IT83, Thu, HW7
15	Jan 16	Tue, Thu
16	Jan 23	Tue, HW8, Thu
17	Jan 30	Tue, Thu
18	Feb 6	TE3, Tue, Thu
19	Feb 13	Tue, Thu
R	Feb 20
20	Feb 27	Tue, Thu, HW9
21	Mar 6	Tue, Thu, HW10
22	Mar 13	Tue, Thu
23	Mar 20	Tue, Thu, HW11
24	Mar 27	Tue, Thu
25	Apr 3	Tue, Thu, HW12
26	Apr 10	Tue, Thu
Study	Apr 17	Office Hours
Exams	Apr 24	Final, PM

Classnotes for September 20

Table of contents

1 Announcement

1 Bases for a Topology, Ordered Sets
2 Topologies and their Bases
3 Examples
4 Product Topology
5 Scanned Notes

5.1 PDF

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Announcement

Date: Mon, 19 Sep 2005 08:57:46 -0400
From: Ida Bulat <ida@math.toronto.edu>
Subject: class announcement

Hi Andres, Dror, Joe,

Could you please announce in each of your classes this week that
it is NSERC/OGS time and that undergraduate students should come
to see me about details.  Unfortunately, I have no other way of
getting in touch with the senior undergrads.  Most of them will
be in your classes.  The deadline is fast approaching:

        TUESDAY, OCTOBER 11, 2005

Thanks very much!

                                        Ida

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Bases for a Topology, Ordered Sets

In the course of this set of notes, we will develop a way of building a topology from a collection of subsets of a set $X$ that obey certain properties and that are usually defined in a way naturally based on the structure of $X$ so that any property of the topology need only be proved to hold on this smaller collection. The sort of collection we will want to use in this way, and the properties it must have, are defined as follows:

Definition: Let $X$ be a set, and let $\mathcal{B} \subset \mathcal{P}\left(X\right)$ . Then $\mathcal{B}$ is called a basis for a topology on $X$ if it satisfies both of the following conditions:

$\forall x \in X \exists B \in \mathcal{B}$ s.t. $x \in B$ ;
$\forall B_1, B_2 \in \mathcal{B} \forall x \in B_1 \cap B_2 \exists B_3 \in \mathcal{B}$ s.t. $x \in B_3 \subset B_1 \cap B_2$ .

We will soon justify this very suggestive terminology. First, however, let us construct a non-trivial example of a basis for a topology on a set.

Definition: An ordered set is a pair $(X, < )$ , where $X$ is a set and $<$ is a binary relation that is transitive (i.e. $\forall x,y,z \in X x<y \land y<z \Rightarrow x<z$ ). A totally ordered set is an ordered set $(x, < )$ such that $\forall x,y \in X$ exactly one of the following holds:

$x < y$ ;
$y < x$ ;
$x = y$ .

Finally, if $(X, < )$ is a totally ordered set, define the relation $\le$ by $\forall x,y \in X \ x \le y \iff x < y \lor x = y$ .

Note that if $(X, < )$ is a totally ordered set, then $(X,\le)$ is an ordered set, but not a totally ordered set. The relationship between these two ordered sets corresponds to the relationship between the totally ordered set $(\mathbb{R},<)$ and the ordered (but not totally ordered) set $(\mathbb{R},\le)$ .

Also, it should be remembered that our definition for an ordered set is somewhat weaker than the usual definition of a partially ordered set (http://en.wikipedia.org/wiki/Partial_order), and that our definition for a totally ordered set is likewise somewhat weaker than the usual definition (http://en.wikipedia.org/wiki/Totally_ordered_set). $(\mathbb{R},<)$ is both a totally ordered set according to our definition and according to the usual definition, and $(\mathbb{R},\le)$ is both an ordered set in our definition and a partially ordered set according to the usual definition.

Actually, we can take the correspondence between ordered sets and the real line even further:

Definition: Let $X$ be a totally ordered set. Define an interval to be a subset of </math> that takes one of the following forms:

$(a,b) := \{x \in X : a < x < b\}$ , where $a,b \in X$ ;
$[a,b) := \{x \in X : a \le x < b\}$ , where $a,b \in X$ ;
$(a,b] := \{x \in X : a < x \le b\}$ , where $a,b \in X$ ;
$[a,b] := \{x \in X : a \le x \le b\}$ , where $a,b \in X$ .

Intervals of type 1. are called open, intervals of types 2. or 3. are called half-open or half-closed, and intervals of type 4. are called closed.

Now we construct our desired basis for a topology on a totally ordered set:

Proposition: Let $X$ be a totally ordered set. Define $\mathcal{B}_{(X,<)} \subset \mathcal{P}\left(X\right)$ to be the trivial topology if $X$ is empty or has one element, and otherwise definte it to be the collection consisting of:

all open intervals in $X$ ;
in the case that $\exist a_0 \in X$ s.t. $\forall x \in X a_0 \le x$ (i.e. $a 0$ is the minimal element of $X$ ), all half-open intervals $[a 0, b)$ where $b \in X$ ;
in the case that $\exist b_0 \in X$ s.t. $\forall x \in X x \le b_0$ (i.e. $b 0$ is the maximal element of $X$ ), all half-open intervals $(a, b 0]$ where $a \in X$ .

Then $\mathcal{B}_{(X,<)}$ is a basis for a topology on $X$ .

Proof: In the case that $X$ is empty or has one element, the proposition is vacuously or trivially true, so assume otherwise. It suffices to check that $\mathcal{B}_{(X,<)}$ satisfies both properties of a basis for a topology on $X$ :

If $X$ has a minimal element $a 0$ , then $\exists b \in X$ s.t. $a 0 < b$ , and hence $a_0 \in [a_0,b) \in \mathcal{B}_{(X,<)}$ . Likewise, if $X$ has a maximal element $b 0$ , then $\exists a \in X$ s.t. $a < b 0$ , and hence $b_0 \in (a,b_0] \in \mathcal{B}_{(X,<)}$ . Finally, if $x \in X$ is neither maximal nor minimal, then $\exists a,b \in X$ s.t. $a < x < b$ and hence $x \in (a,b) \in \mathcal{B}_{(X,<)}$ . Hence $\forall x \in X \exists B \in \mathcal{B}_{(X,<)}$ s.t. $x \in B$ .
Let $(a,b),(c,d) \in \mathcal{B}_{(X,<)}$ , and let $x \in (a,b) \cap (c,d)$ . For $(a,b) \cap (c,d)$ to be non-empty, we must have that $c < b$ and $a < d$ . Hence, define $e = max(a, c)$ and $f = min(b, d)$ . It is easy to see now that $(a,b) \cap (c,d) = (e,f)$ . Because both $a < x < b$ and $c < x < d$ we have that $e < x < f$ , and hence $x \in (e,f) = (a,b) \cap (c,d)$ , where $(e,f) \in \mathcal{B}_{(X,<)}$ , q.e.d.

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Topologies and their Bases

We now justify our terminology with the following result:

Theorem: Let $X$ be a set, and let $\mathcal{B}$ be a basis for a topology on $X$ . Then there exists a unique topology $\mathcal{T_B}$ such that the following equivalent conditions hold:

$\mathcal{T_B} = \{U \subset X : \forall x \in U \exists B_x \in \mathcal{B}$ s.t. $x \in B_x \subset U\}$ ;
$\mathcal{T_B}$ is the minimal topology containing $\mathcal{B}$ as a subset;
$\mathcal{T_B}$ is the collection of all possible unions of elements of $\mathcal{B}$ .

$\mathcal{T_B}$ is therefore called the topology generated by $\mathcal{B}$ , and one can say that $\mathcal{B}$ is a basis of the topology $\mathcal{T_B}$ .

Before offering a proof, we will find it useful to prove the following lemma:

Lemma: Let $X$ be a set, let $I$ be an index set, and suppose that $\forall \alpha \in I \ \mathcal{T}_\alpha$ is a topology on $X$ . Then $\bigcap_{\alpha \in I} \mathcal{T}_\alpha$ is a topology on $X$ .

Proof of Lemma: It suffices, as usual, to check the three properties of a topology:

By the definition of a topology, $\forall \alpha \in I \ \emptyset, X \in \mathcal{T}_\alpha$ and hence $\emptyset, X \in \bigcap_{\alpha \in I} \mathcal{T}_\alpha$ .
If $K$ is an index set, and if $\forall t \in K \ U_t \in \bigcap_{\alpha \in I} \mathcal{T}_\alpha$ , then $\forall \alpha \in I$ we have that $\forall t \in K \ U_t \in \mathcal{T}_\alpha$ and hence that $\bigcup_{t \in I} U_t \in \mathcal{T}_\alpha$ Hence $\bigcup_{t \in I} U_t \in \bigcap_{\alpha \in I} \mathcal{T}_\alpha$ .
If $K$ is a finite index set, and if $\forall i \in K \ U_i \in \bigcap_{\alpha \in I} \mathcal{T}_\alpha$ , then $\forall \alpha \in I$ we have that $\forall i \in K \ U_i \in \mathcal{T}_\alpha$ and hence that $\bigcap_{i \in K} U_i \in \bigcap_{\alpha \in I} \mathcal{T}_\alpha$ , q.e.d.

Proof of the Theorem: The three conditions listed in the statement of the theorem define three different collections of subsets of $X$ . We do not know a priori whether or not these collections are actually topologies, so we now prove that this is indeed the case, one by one.

First, let $\mathcal{T}_1 = \{U \subset X : \forall x \in U \exists B_x \in \mathcal{B}$ s.t. $x \in B_x \subset U\}$ . Let us now check that the three properties of a topology hold:

It is vacuously true that $\emptyset \in \mathcal{T}_1$ , and it follows from the definition of a basis for a topology that $X \in \mathcal{T}_1$ .
If $I$ is an index set, and if $\forall \alpha \in I \ U_\alpha \in \mathcal{T}_1$ , then $x \in \bigcup_{\alpha\in I} U_\alpha$ implies that $x \in U_\beta$ for some $\beta \in I$ , and hence $\exists B_x \in \mathcal{B}$ s.t. $x \in B_x \subset U_\alpha \subset \bigcup_{\alpha\in I} U_\alpha$ . Thus $\bigcup_{\alpha\in I} U_\alpha \in \mathcal{T}_1$ .
If $I$ is a finite index set, and if $\forall i \in I \ U_i \in \mathcal{T}_1$ , then $x \in \bigcap_{i\in I} U_i$ implies that $\forall i \in I x \in U_i$ , and hence $\forall i \in I \exists B_x^i \in \mathcal{B}$ s.t. $x \in B_x^i \subset U_i$ . Thus $x \in \bigcap_{i \in I} B_x^i \subset \bigcap_{i \in I} U_i$ , where $\bigcap_{i \in I} B_x^i \in \mathcal{B}$ by mathematical induction. Hence, $\bigcap_{i \in I} U_i \in \mathcal{T}_1$ .

Thus $\mathcal{T}_1$ is a topology on $X$ which satisfies condition 1. in the theorem, and it is easy to see that $\mathcal{T}_1$ contains $\mathcal{B}$ .

Next, let $\mathcal{T}_2$ be the intersection of all topologies on $X$ that contain $\mathcal{B}$ . This intersection exists because, at the very least, the discrete topology and $\mathcal{T}_1$ are topologies on $X$ that contain $\mathcal{B}$ , and is a topology itself by the lemma. That it is indeed the minimal topology containing $\mathcal{B}$ (and thereby being the topology defined by condition 2.) follows from its construction - any other topology containing $\mathcal{B}$ is guaranteed to contain it, and thus can at most be equal to it. That $\mathcal{T}_2$ contains $\mathcal{B}$ follows immediately from the definition of $\mathcal{T}_2$ .

Now, let $\mathcal{T}_3$ be the collection of all possible unions of elements of $\mathcal{B}$ . We once more check that the three properties of a topology hold:

It is vacuously true that $\emptyset \in \mathcal{T}_3$ , and on the other hand, by definition, $\forall x \in X \exists B_x \in \mathcal{B}$ s.t. $x \in B_x$ , and hence $X = \bigcup_{x \in X} B_x \in \mathcal{T}_3$ .
If $I$ is an index set, and if $\forall \alpha \in I \ U_\alpha \in \mathcal{T}_3$ , then $x \in \bigcup_{\alpha\in I} U_\alpha$ implies that $x \in U_\beta$ for some $\beta \in I$ , and hence $\exists B_x^\beta \in \mathcal{B}$ s.t. $x \in B_x^\beta \subset U_\beta \subset \bigcup_{\alpha\in I} U_\alpha$ , since $U β$ is a union of elements of $\mathcal{B}$ . Hence $U_\beta = \bigcup_{x \in U_\beta} B_x^\beta$ , and thus $\bigcup_{\alpha \in I}U_\alpha = \bigcup_{\alpha \in I}\bigcup_{x \in U_\alpha}B_x^\alpha \in \mathcal{T}_3$ .
If $I$ is a finite index set, and if $\forall i \in I \ U_i \in \mathcal{T}_3$ , then $x \in \bigcap_{i\in I} U_i$ implies that $\forall i \in I x \in U_i$ , and hence $\exists B_x^i \in \mathcal{B}$ s.t. $x \in B_x^i \subset U_i$ , since $U_i \in \mathcal{T}_3$ and is hence the union of elements of $\mathcal{B}$ , one of which will contain $x$ . Thus $x \in \bigcap_{i \in I} B_x^i \subset \bigcap_{i \in I} U_i$ , where $\bigcap_{i \in I} B_x^i \in \mathcal{B}$ by mathematical induction. Hence, $\bigcap_{i \in I} U_i = \bigcup_{x \in \bigcap_{i \in I} U_i} \left( \bigcap_{j \in I} B_x^j \right) \in \mathcal{T}_3$ .

Hence, $\mathcal{T}_3$ is indeed a topology on $X$ which clearly contains $\mathcal{B}$ .

Thus, each of the three conditions in the statement of the theorem are actually definitions of topologies on $X$ . To prove the existence and uniqueness of the desired topology $\mathcal{T_B}$ and the equivalence of the three conditions listed in the theorem, it suffices entirely to prove that $\mathcal{T}_1=\mathcal{T}_2=\mathcal{T}_3$ , in which case we define $\mathcal{T_B} := \mathcal{T}_1 = \mathcal{T}_2 = \mathcal{T}_3$ . Uniqueness of $\mathcal{T_B}$ follows immediately from the uniqueness of $\mathcal{T}_3$ as the minimal topology on $X$ containing $\mathcal{B}$ . First, it is easy to see that $\mathcal{T}_2 = \mathcal{T}_3$ , because conditions 2. and 3. are readily seen to be equivalent. Next, it follows from the definition of $\mathcal{T}_2$ that $\mathcal{T}_2 \subset \mathcal{T}_3$ . So, consider some $U \in \mathcal{T}_3$ . By definition, $U$ is some union of elements of $\mathcal{B}$ , and we know that $\mathcal{B} \subset \mathcal{T}_2$ . Hence, $U \in \mathcal{T}_2$ . Thus $\mathcal{T}_3 \subset \mathcal{T}_2$ and therefore $\mathcal{T}_3 = \mathcal{T}_2$ . Hence, $\mathcal{T}_1 = \mathcal{T}_2 = \mathcal{T}_3$ , q.e.d.

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Examples

Given the material we have covered thus far this time, the following definition is a natural and expected one:

Definition: Let $(X, < )$ be an ordered set, and let $\mathcal{B}_{(X,<)}$ be the basis for a topology on $X$ defined above. The unique topology $\mathcal{T}_{ord}$ generated by the basis $\mathcal{B}_{(X,<)}$ is called the order topology on $X$ .

As a concrete example of an order topology, we have the following:

Example 1: Consider $\mathbb{R}\times\mathbb{R}$ with the ordering $<$ given by $(a, b) < (c, d)$ if $a < c$ , or $a = c$ and $b < d$ . This ordering is called the dictionary ordering on $\mathbb{R}\times\mathbb{R}$ , and the order topology corresponding to the dictionary ordering is called the dictionary topology. A basic open set in the dictionary topology is given by $\{(x,y)\in\mathbb{R}\times\mathbb{R}\,|\,(a,b)<(x,y)<(c,d)\}$ , for some $(a,b),(c,d)\in\mathbb{R}\times\mathbb{R}$ .

Example 2: Consider $\mathbb{R}$ , and let $\mathcal{T}_l$ be the topology generated by the basis ${[a, b): a < b}$ .

Now, which topology on $\mathbb{R}$ is finer, $\mathcal{T}_l$ or $\mathcal{T}_{std}$ ? The answer is indeed $\mathcal{T}_l$ : it is easy to check that $\mathcal{T}_{std} \subset \mathcal{T}_l$ , but on the other hand $[0,1)$ is an element of $\mathcal{T}_l$ but not of $\mathcal{T}_{std}$ , so that the inclusion is a strict one.

We can also ask whether we get more or fewer continuous functions when we have the domain and codomain be either $(\mathbb{R},\mathcal{T}_{std})$ or $(\mathbb{R},\mathcal{T}_l)$ .

	Domain
		$\left ( \mathbb{R},\mathcal{T}_{l} \right )$	$\left ( \mathbb{R},\mathcal{T}_{std} \right )$
Codomain	$\left ( \mathbb{R},\mathcal{T}_{l} \right )$	?;	constant functions;
	$\left( \mathbb{R},\mathcal{T}_{std} \right )$	more continuous functions;	continuous (old definition) functions;

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Product Topology

Theorem Let $X$ and $Y$ be topological spaces. There exists a unique topology, the product topology, on $X\times Y$ such that

1. $\Pi_{X}:\,X\times Y \rightarrow X$ is continuous. $\Pi_{Y}:\,X\times Y \rightarrow Y$ is continuous.

2. If $f:\, Z \rightarrow X$ and $G:\, Z \rightarrow Y$ are continuous, then $(f\times g):\, Z \rightarrow X\times Y$ is continuous.

Proof:

Let $\mathcal{B}=\{U\times V\,|\,U\, \textrm{open\,in}\, X,\, V \textrm{open\,in }\, Y\}$ . This is a basis for a topology. So, take $\mathcal{T}_{X\times Y}=\mathcal{T_{B}}$ . Then, $Π X$ is continuous since if $U$ is open in $X$ , then $\Pi^{-1}_{X}(U)= U\times Y\in\mathcal{T_{B}}$ . Similarly, for $Π Y$ .

To prove 2, assume $f:Z\rightarrow X$ and $g:Z\rightarrow Y$ are continuous. Let $W\in X\times Y$ . We want to show the continuity of $(f\times g)$ . (In general, it is sufficient to verify continuity on basic open sets $B$ . If $U$ is open then $U=\bigcup B_{\alpha}$ and $f^{-1}(U)=f^{-1}(\bigcup B_{\alpha})=\bigcup f^{-1}(B_{\alpha})$ . So, without loss of generality we may assume $W=U\times V$ . Then,

$(f\times g)^{-1}(U\times V)$	$=\{z\in Z\,\|\,f(z)\in U\,\textrm{and}\,g(z)\in V\}$
	$=\{z\in Z\,\|\,(f(z),g(z))\in U\times V\}$
	$= \{z\in Z\,\|\,f(z)\in U\,\textrm{and}\,g(z)\in V\}$
	$=f^{-1}(U)\cap g^{-1}(V)$
	but the latter is open in Z as an intersection of two open sets

To prove uniqueness, assume $\mathcal{T}_{a}$ and $\mathcal{T}_{b}$ are topologies on $X\times Y$ such that 1 and 2 hold for both. To show the topologies are equal it will be sufficient to show that $\textrm{Id}: (X\times Y,\,\mathcal{T}_{a}) \rightarrow (X \times Y,\,\mathcal{T}_{b})$ is continuous. Note that $Id(x, y) = (x, y) = (Π X (x, y),Π Y (x, y))$ . So, $\textrm{Id}=\Pi_{X} \times \Pi_{Y}$ . Then, $Π X$ and $Π Y$ are continuous by 1 for $\mathcal{T}_{a}$ . So, $\Pi_{X} \times \Pi_{Y}$ is continuous by 2 for $\mathcal{T}_{b}$ .