Classnotes for September 27

From 0506Topology

#	Week of...	Links (edit)
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20	Feb 27	Tue, Thu, HW9
21	Mar 6	Tue, Thu, HW10
22	Mar 13	Tue, Thu
23	Mar 20	Tue, Thu, HW11
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Study	Apr 17	Office Hours
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Table of contents

1 Notes in PDF, by Annat Koren

2 AUDIO Of Entire Lecture

1 Limit Points
2 Equivalent Definitions of Continuity
3 Hausdorff Spaces
4 Quotient Topology

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Notes in PDF, by Annat Koren

Available here (http://katlas.math.toronto.edu/0506-Topology/index.php?title=Image:Topology_27_September_2005.pdf)

[edit]

AUDIO Of Entire Lecture

First Quarter of Lecture: Media:Sep2705_1stQuart.ogg.

Second Quarter of Lecture: Media:Sep2705_2ndQuart.ogg.

Third Quarter of Lecture: Media:Sep2705_3rdQuart.ogg.

Fourth Quarter of Lecture: Media:Sep2705_4thQuart.ogg.

NOTE: If you are having difficulty playing the .OGG file format, here is a link to free .OGG players, for all platforms: http://wiki.xiph.org/index.php/VorbisSoftwarePlayers

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Limit Points

Definition:

A point $x$ is a limit point for a set $A$ if $x\in\overline{ A\backslash\{x\}}$ . Let $A'$ be the set of all limit points of $A$ .

Theorem:

$\overline{A}=A\cup A'$ .

Proof:

First note that $A\subset\overline{A}$ . Now, let $x\in A'$ . Then, $x\in\overline{A\backslash\{x\}}\subset\overline{A}$ . So, $A'\subset\overline{A}$ . Therefore, $A\cup A'\subset\overline{A}$ .

For the reverse inclusion, assume $x\in\overline{A}$ . If $x\in A$ we are done as $x\in A\cup A'$ . If $x\in\!\!\!\!\!/A$ , then $\overline{A\backslash\{x\}}=\overline{A}\ni x$ . So, $x\in A'$ and again $x\in A\cup A'$ .

Examples:

1. $A=\{\frac{1}{n}\}$ , $A' = {0}$ . Then, $\overline{A}=\{\frac{1}{n}\}\cup\{0\}$ . In this case, the union is disjoint.

2. $A=\mathbb{Q}$ , $A'=\mathbb{R}$ . Then, $\overline{A}=\mathbb{Q}\cup\mathbb{R}=\mathbb{R}$ . In this case, the union is not disjoint.

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Equivalent Definitions of Continuity

Definition:

Let $f:X\rightarrow Y$ and $x\in X$ . Say that $f$ is continuous at $x$ if for every neighbourhood $V$ of $f (x)$ there is a neighbourhood $U$ of $x$ such that $f(U)\subset V$ .

Theorem:

Let $f:X\rightarrow Y$ . The following are equivalent:

1. f is continuous. (If $V$ is open in $Y$ , then $f - 1 (V)$ is open in $X$ ).

2. $f(\overline{A})\subset\overline{f(A)}$

3. If $C$ is closed in $Y$ , then $f - 1 (C)$ is closed in $X$ .

4. If $B$ is a basic open set in $Y$ , then $f - 1 (B)$ is open in $X$ .

5. For all $x$ , $f$ is continuous at $x$ .

Proof:

$(1\Rightarrow 2)$ Let $x\in\overline{A}$ and $y = f (x)$ . If $y\in\overline{f(A)}^{c}$ , then $\overline{f(A)}^{c}$ is an open neighbourhood of $y$ . So, $f^{-1}(\overline{f(A)}^{c})= (f^{-1}(\overline{f(A)}))^{c}$ is an open neighbourhood of $x$ and hence it intersects $A$ .

But, $A\subset f^{-1}(f(A))\subset f^{-1}(\overline{f(A)})$ . So, $A\cap(f^{-1}(\overline{f(A)}))^{c}=\emptyset$ , a contradiction.

$(2\Rightarrow 3)$ Assume $A\subset Y$ is closed. Let $x\in\overline{f^{-1}(A)}$ . We want to show $x\in f^{-1}(A)$ , i.e., $f(x)\in A$ .

We have $f(x)\in f(\overline{f^{-1}(A)})\subset \overline{f(f^{-1}(A))}\subset\overline{A}=A$ as required.

$(3\Rightarrow 1)$ Let $U$ open in $Y$ . $U c$ is closed. So, $f - 1 (U c) = (f - 1 (U)) c$ is closed. Therefore, $f - 1 (U)$ is open.

$(4\Leftrightarrow 1)$ Since every basic set is open then $1\Rightarrow 4$ . Since every open set is the union of basic sets and since the inverse image is well-behaved with respect to unions we have $4\Rightarrow 1$ .

$(1\Rightarrow 5)$ Let $x\in X$ and let $V$ be an open set containing $f (x)$ . Then, $U = f - 1 (V)$ is an open set containing $x$ such that $f(U)\subset V$ .

$(5\Rightarrow 1)$ Let $V$ open in $Y$ and $x\in f^{-1}(V)$ . So, $f(x)\in V$ . Then, there is an open neighbourhood $U$ of $x$ such that $f(U)\subset V$ . So, $U\subset f^{-1}(V)$ . Therefore, $f - 1 (V)$ is open.

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Hausdorff Spaces

Definition:

$X$ is called Hausdorff or $T 2$ if whenever $x,y\in X$ , $x\neq y$ , then there exist open sets $U, V$ such that $x\in U$ , $y\in V$ and $U\cap V=\emptyset$ .

Examples:

1. $\mathbb{R}^{n}_{\textrm{std.}}$ For points $x$ and $y$ , the balls of radius $\frac{d(x,y)}{3}$ about each point will be disjoint. So, this space is Hausdorff.

2. $(\mathbb{R},\mathcal{T})$ , $\mathcal{T}=\{(a,\infty)\}\cup\{\emptyset,\mathbb{R}\}$ If $x,y\in\mathbb{R}$ , $x < y$ , then any open set of $X$ will also contain $y$ . So, this space is not Hausdorff.

Theorem:

Singletons are closed in $T 2$ spaces. If $x\in X$ and $X$ is $T 2$ , then ${x}$ is closed.

Proof:

For any $y\neq x$ let $V y$ be the $V$ from the definition of $T 2$ . Then, $X\backslash\{x\}=\bigcup_{y\neq x}V_{y}$ is open. So, ${x}$ is closed.

Theorem:

Let $X$ be a $T 2$ space. Let $A\subset X$ and $x\in A'$ . Then, every neighbourhood of $x$ intersects $A$ in an infinite set.

Proof:

Let $U$ be an open set containing $x$ such that $x\in\overline{A\backslash\{x\}}$ . Assume $U\cap A\backslash\{x\}=\{x_{i}\}_{i=1}^{n}$ . Then, $V=U\backslash\{x_{1},\ldots,x_{n}\}= U\bigcap(\cap_{i=1}^{n}\{x_{i}\}^{c}$ is open in and $x\in V$ . Hence, $V\cap A\backslash\{x\}$ is non-empty, but

$V\cap(A\backslash\{x\})\subset U\cap(A\backslash\{x\})= \{x_{1},\ldots,x_{n}\}$

Also, $V\cap(A\backslash\{x\})\subset(\cap\{x_{i}\})^{c}= \{x_{1},\ldots,x_{n}\}^{c}$ , a contradiction.

Theorem:

1. If $X$ and $Y$ are $T 2$ , then so is $X\times Y$ .

2. If $X$ is $T 2$ and $Y\subset X$ , then $Y$ is $T 2$ .

Proof:

1. Let $(a,b),(a',b')\in X\times Y$ such that $(a,b)\neq(a',b')$ . W.l.o.g. assume $b\neq b'$ . Since $Y$ is $T 2$ we can find $U$ , $U'$ open in $Y$ such that $b\in U$ , $b'\in U'$ and $U\cap U'=\emptyset$ . Then, $X\times U$ , $X\times U'$ are open in $X\times Y$ , $(a,b)\in X\times U$ , $(a',b')\in X\times U'$ and $(X\times U)\cap(X\times U')=(X\cap X)\times(U\cap U')= X\times\emptyset=\emptyset$ .

2. Let $a,b\in Y\subset X$ . Since $X$ is $T 2$ , there exist $U,\,V\in\mathcal{T}_{X}$ such that $a\in U$ , $b\in V$ and $U\cap V=\emptyset$ . Then, $U\cap Y, V\cap Y \in \mathcal{T}_{Y}$ , $a\in U\cap Y$ , $b\in V\cap Y$ and $(U\cap Y)\cap(V\cap Y)=U\cap V\cap Y=\emptyset$ .

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Quotient Topology

Definition:

Let $X$ be a set. An equivalence relation on $X$ is a relation $\sim$ that is

1. Reflexive. $x\sim x$ for all $x\in X$ .

2. Symmetric. If $x\sim y$ , then $y\sim x$ .

3. Transitive. If $x\sim y$ and $y\sim z$ , then $x\sim z$ .

Theorem:

Given $p: X\rightarrow X/\!\sim$ where $X$ is a topological space. There exists a unique topology on $X/\!\sim$ such that

1. $p$ is continuous.

2. If $f: X/\!\sim\,\rightarrow Y$ is given and $g=f\circ p$ is continuous, so is $f$ .

Proof:

Uniqueness: Let $\mathcal{T}$ and $\mathcal{T}'$ be two topologies on $X/\!\sim$ satisfying 1 and 2. Let $X/\!\sim=(X/\!\sim\,,\,\mathcal{T})$ and $Y=(X/\!\sim\,,\,\mathcal{T}')$ . Then, by 2, $f=\textrm{Id}:(X/\!\sim\,,\,\mathcal{T})\rightarrow (X/\!\sim\,,\,\mathcal{T}')$ is continuous. This gives $\mathcal{T}'\subset\mathcal{T}$ . Reversing the roles of $\mathcal{T}$ and $\mathcal{T}'$ gives the opposite inclusion. Therefore, $\mathcal{T}=\mathcal{T}'$ .

Existence: Define $\mathcal{T}=\{V\subset X/\!\sim|p^{-1}(V)\,\textrm{open}\}$ . $\mathcal{T}$ is clearly a topology. Condition 1 is satisfied by the construction of $\mathcal{T}$ . Let $f:X/\!\sim\rightarrow Y$ be given and suppose $g=f\circ p$ is continuous. Let $U$ open in $Y$ . Then $g - 1 (U) = p - 1 (f - 1 (U))$ is open in $X$ . Therefore, $f^{-1}(U)\in\mathcal{T}$ and so $f$ is continuous.

Examples:

1. $X = [0,1]$ , $0\sim1$ but no other equivalences other than $x\sim x$ . The map $p:[0,1]\rightarrow[0,1]/0\sim1$ is given by $p (x) = x$ if $x\in(0,1)$ , $p (0) = p (1) = φ$ . A neighbourhood of $φ$ is $[0,\epsilon)\cup(1-\delta,1]$ . $[0,1]/\!\sim$ is homeomorphic to $S 1$ .

2. $\mathbb{R}^{2}/\mathbb{Z}^{2}$ . $(x,y)\sim(x',y')$ if $(x-x',y-y')\in\mathbb{Z}^{2}$ . $\mathbb{R}^{2}/\mathbb{Z}^{2}$ is homeomorphic to the torus.

One might hope that the following statement would be true: If $X$ is $T 2$ , then so is $X/\!\sim$ . However, this is actually false. An example is the orbits of a hyperbolic flow, $\mathbb{R}^{2}/\sim$ where $(x,y)\sim(\lambda x,\lambda^{-1}y)$ for $λ > 0$ . In $X/\!\sim$ any open neighbourhood of the orbit $(0,0)$ will contain the orbit represented by represented by $(1,0)$ . Therefore, these two points of the quotient space cannot be separated by disjoint open sets and so the space is not Hausdorff.