Classnotes for September 29

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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
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14 Jan 9 Tue, IT83, Thu, HW7
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18 Feb 6 TE3, Tue, Thu
19 Feb 13 Tue, Thu
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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
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 29

Infinite Product Spaces

Let $\left\{X_\alpha\right\}_{\alpha\in I}$ be a collection of topological spaces.

Then we define the product $\prod_{\alpha\in I}X_\alpha=\left\{\left(x_\alpha\right)_{\alpha\in I}:x_\alpha\in X_\alpha\right\}=\left\{\left(x:I\to\bigcup_{\alpha\in I}X_\alpha\right):x(\alpha)\in X_\alpha\right\}$.

Note: The statement "If $\forall\alpha\in I\ X_\alpha\not=\varnothing$ then $\prod_{\alpha\in I}\not=\varnothing$." is equivalent to the axiom of choice in set theory, and it's nicer for things not to depend on that axiom (someone insert something more enlightening here). However, for the particular infinite product spaces we'll be using, the axiom of choice is not required to show that they're non-empty; the axiom of choice is only needed for the general case. And even if they are empty, everything we say will still be true.

Examples:

• $\{0,1\}^{\mathbb N}=\prod_{n\in\mathbb N}\{0,1\}=\{\hbox{strings of the form}\ 00110101\cdots\}$
• $\mathbb R^{\mathbb N}=\prod_{n\in \mathbb N}\mathbb R=\{a_1a_2a_3\cdots:a_i\in\mathbb R\}$
• $\mathbb R^{\mathbb R}=\prod_{\mathbb R}\mathbb R=\{f:\mathbb R\to\mathbb R,\ f\ \hbox{not necessarily continuous}\}$

The Box Topology

We use $\mathcal B=\{U\times V:U\ \hbox{open in}\ X,\ V\ \hbox{open in}\ Y\}$ as a basis for the product topology in $X\times Y$. So we'll use the basis $\mathcal B_{\mathrm{box}}=\left\{\prod_{\alpha\in I}U_\alpha:U_\alpha\ \hbox{open in}\ X_\alpha\ \hbox{for each}\ \alpha\right\}$ to generate a topology on $\prod_{\alpha\in I}X_\alpha$. We'll call this topology the box topology.

In fact, this is the wrong choice of topology, as we'll find out.

The Tychonoff Topology

Theorem: There is a unique topology (the Tychonoff topology, or the cylinder topology) on $\prod_{\alpha\in I}X_\alpha$ such that the following conditions are satisfied:

1. For each $\beta\in I$, $\pi_\beta:\prod_{\alpha\in I}\to X_\beta$ is continuous. (πβ is the projection function mapping $\left(x_\alpha\right)\mapsto x_\beta$.)
2. If $f_\alpha:Z\to X_\alpha$ is continuous for each $\alpha\in I$, then $\prod_{\alpha\in I}f_\alpha:Z\to\prod_{\alpha\in I}X_\alpha$ is continuous. ($\prod_{\alpha\in I}f_\alpha$ maps $z\mapsto\left(f_\alpha(z)\right)_{\alpha\in I}$.)

Proof:

Uniqueness: same as for $X\times Y$: see the notes on the Product Topology.

Existence:

Each πβ is continuous, so $\pi_\beta^{-1}\left(U_\beta\right)$ is open if Uβ is open in Xβ.

Finite intersections of open sets are open, so if we have $U_{\beta_1},U_{\beta_2},\ldots,U_{\beta_n}$ such that $U_{\beta_i}$ is open in $X_{\beta_i}$, then $\bigcap_{i=1}^n\pi_{\beta_i}^{-1}\left(U_{\beta_i}\right)$ must be open.

So select the basis $\mathcal B=\left\{\bigcap_{i=1}^n\pi_{\beta_i}^{-1}\left(U_{\beta_i}\right):\beta_i\in I\ \hbox{and}\ U_{\beta_i}\ \hbox{open in}\ X_{\beta_i}\ \hbox{for}\ i=1,\ldots,n\right\}$.

Motivation for the word cylinder: a cylinder is a set of points where two coordinates are constrained but the third is not. This is a bit like an open set in this basis.

Proof that this satisfies condition 2 of the theorem:

A typical basic open set is $\bigcap_{i=1}^n\pi_{\beta_i}^{-1}\left(U_{\beta_i}\right)$.

$\left(\prod_{\alpha\in I}f_\alpha\right)^{-1}\left[\bigcap_{i=1}^n\pi_{\beta_i}^{-1}\left(U_{\beta_i}\right)\right]$ $=\left\{z:\left(\prod_{\alpha\in I}f_\alpha\right)(z)\in\bigcap_{i=1}^n\pi_{\beta_i}^{-1}\left(U_{\beta_i}\right)\right\}$ $=\left\{z:f_{\beta_1}(z)\in U_{\beta_1},\ldots,f_{\beta_n}(z)\in U_{\beta_n}\right\}$ $=\left\{z:f_{\beta_1}(z)\in U_{\beta_1}\right\}\cap\cdots\cap\left\{z:f_{\beta_n}(z)\in U_{\beta_n}\right\}$ $=\bigcap_{i=1}^nf_{\beta_i}^{-1}\left(U_{\beta_i}\right)$. This is open since it is a finite intersection of open sets.

Scanned and Recorded Lecture Notes

Notes in JPG, by Annat Koren

Handwritten notes by Annat Koren
Handwritten notes by Annat Koren
Handwritten notes by Annat Koren
Handwritten notes by Annat Koren

Notes in PDF

Available here (http://katlas.math.toronto.edu/0506-Topology/index.php?title=Image:3-2-09-29.pdf).

AUDIO Of Entire Lecture

First Half of Lecture: Media:Sep2905_1stHalf.ogg.

Second Half of Lecture: Media:Sep2905_2ndHalf.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

Interesting Challenge

Let us start with a graphical challenge. Recall Σ, the Seifert surface of the trefoil from Classnotes for September 15:

Can you draw the picture on the left

on a standardly drawn punctured torus? How about, on Σ? Notice that boundaries should go to boundaries, so the entire boundary of your drawn surface should be red at the end. If you like chess better, do the same with the right-most picture. (BTW, who played this game?)

Note that these pictures can be thought of as living on $T^2-\{pt\}=({\mathbb R}^2-{\mathbb Z}^2)/{\mathbb Z}^2$, where the removed point is circled (or squared) in red below:

The chess scene above: Kasparov vs. Deep Blue, game 1. =)