Classnotes for September 22

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 22

Table of contents

1 The Induced Subset Topology

2 Closed Sets

3 Interior and Closure

4 Scanned Notes

4.1 PDF
4.2 AUDIO Of Entire Lecture

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The Induced Subset Topology

Given a space $X$ with a topology $\mathcal{T}$ , there is a natural way to induce a topology $\mathcal{T}^\prime$ on a subset $Y \subset X$ that is in many ways compatible with $\mathcal{T}$ . We would expect of such a natural topology certain nice properties such as:

The map which embeds $Y$ into $X$ is continuous, i.e. $\iota\colon Y \rightarrow X$ is continuous.
A function $f$ from another space $Z$ to $Y$ continuous whenever $g\colon Z \rightarrow X$ defined as $g = \iota \circ f$ .

This first condition puts a lower bound, so to speak, on the fineness of $\mathcal{T}^\prime$ : it must have at least the open sets that $\mathcal{T}$ grants Y. The second condition gives an upper limit on the fineness in a similar way. The following theorem demonstrates that these 'bounds' agree and yield the sought-after natural subset topology.

Theorem: If $X$ is a topological space and $Y \subset X$ , there exists a unique topology $\mathcal{T}_Y$ on $Y$ such that:

$\iota\colon Y \rightarrow X$ is continuous
$f\colon Z \rightarrow Y$ is continuous whenever $g\colon Z \rightarrow X$ defined as $g = \iota \circ f$ is continuous

and it is explicitly described as: $\mathcal{T}_Y = \{U \cap Y: \forall\;open\; U \subset X\}$

The proof of this theorem is left as an easy exercise. The induced subset topology can be thought of as restricting a topology to a subset.

If $X$ and $Y$ are topological spaces and $U \subset X$ and $V \subset Y$ , then the set $U \times V \subset X \times Y$ could be endowed with two potentially distinct natural topologies. One being as a product of subsets (i.e. endow $U$ and $V$ with subset topologies, as described above, from $X$ and $Y$ respectively and then give $U \times Y$ the product topology) or as a subset of a product (i.e. endow $X \times Y$ with the product topology and give $U \times V \subset X \times Y$ the induced subset topology). However, as one might expect, these two processes yield identical topologies.

There are some further properties of induced product and subset topologies worth noting:

Theorem:

Taking product topologies is associative, i.e. $(X \times Y) \times Z = X \times (Y \times Z)$
If $Z \subset Y \subset X$ with $X$ being a topological space, then the subset topology induced on $Z$ by considering it as a subset of $Y$ (with a subset topology induced from $X$ ) is equal to the subset topology induced on $Z$ by considering it directly as a subset of $X$ .
$X \times Y$ is homeomorphic (though not generally equal) to $Y \times X$ .
If $(a,b) \subset X$ is an interval within an ordered set $X$ , then the interval topology induced on $(a, b)$ by its ordering is the same as the subset topology induced from $X$ with its interval topology.

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Closed Sets

Defintion: A subset F of a topological space X is called closed if $F^c = {X \setminus F} = \{x \in X : x \notin F\}$ is open.

Closed sets have properties that are strikingly similar to those of open sets:

Theorem: If $X$ is a topological space then:

$X, \empty$ are both closed.
An arbitrary intersection of closed sets is a closed set.
The finite union of closed sets is a closed set.

Proof:

$X^c = \empty$ and $\empty^c \ X$ are both open.
$(\cap_{(\alpha \in I)} F_\alpha)^c = (\cup_{(\alpha \in I)} (F_\alpha)^c)$ by de Morgan's Law. This set is the union of open sets and is therefore open. So, its complement $(\cap_{(\alpha \in I)} F_\alpha)$ is closed.
$(\cup_{(i\in I)} F_i)^c = (\cap_{(i \in I)} (F_i)^c)$ by de Morgan's Law. This set is the finite intersection of open sets and is therefore open. So its complement $(\cup_{(i\in I)} F_i)$ is closed.

Note that a set can be both closed and open; the empty set and the entire space, for example, are always both closed and open. Also, note that the definition of a topology could just as well have used closed sets rather than open sets. The choice of open sets as the basic elements of topology was arbitrary.

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Interior and Closure

We will introduce two very useful operators on sets that are topological in nature:

Definition: Let $A \subset X$ where $X$ is a topological space.

The closure of $X$ , denoted by $C l (A)$ , is defined as the smallest closed set containing $A$ , or equivalently, the intersection of all closed subsets of $X$ containing $A$ .
The interior of $X$ , denoted by $I n t (A)$ , is defined as the largest open set contained in $A$ , or equivalently, the union of all open subsets of $X$ contained in $A$ .

As an exercise, one many show that arbitrary applications of the interior, closure, and complement operators to a subset $A$ of a topological space can yield, at most, 14 distinct sets. Also, one can find a subset of $\mathbb{R}$ which yields 14 distinct sets by applying these operators in different sequences. Curious or lazy students can find the proof and example here.

The interior and closure of a set have very useful characterisations:

Theorem:

$x \in Int(A)$ if and only if there exists an open set $U$ such that $x \in U \subset A$ .
$x \in Cl(A)$ if and only if for every open set $U$ such that $x \in U$ , $\;U \cap A \ne \empty$ .

Proof:

If $x \in U \subset A$ , then $x$ is in the union of all open subsets of $A$ , and is therefore in $I n t (A)$ . Conversely, if $x \in Int(A)$ then $I n t (A)$ is open and $x \in Int(A) = U \subset A$ .
Let $x \in Cl(A)$ and $U$ be any open set containing $x$ . If $A \cup U = \empty$ then $A \subset U^c$ . Then $U c$ is a closed set containing $A$ , so $Cl(A) \subset U^c$ . This implies that $Cl(A)^c \subset U$ but $x \in U$ , so $x \notin Cl(A)$ , giving a contradiction. To prove the converse, suppose every open set $U$ containing $x$ intersects $A$ but $x \notin Cl(A)$ . That is, there is a closed set $S$ containing $A$ which does not contain $x$ . Then, $x \in S^c \subset A^c$ . But then $S c$ is an open set containing $x$ , so it must have a non-empty intersection with $A$ and therefore cannot be a subset of $A c$ , giving a contradiction.

The power of this theorem is made greater by noting that if the topological space in question has a basis, the condition on the sets $U$ can be strengthened to not just open but also basic.

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Scanned Notes

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PDF

Available here (http://katlas.math.toronto.edu/0506-Topology/index.php?title=Image:2-1-09-22.pdf).

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AUDIO Of Entire Lecture

First Half of Lecture: Media:Sep2205_1stHalf.ogg.

Second Half of Lecture: Media:Sep2205_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

Retrieved from "http://katlas.math.toronto.edu/0506-Topology/index.php?title=Classnotes_for_September_22"

Classnotes for September 22

From 0506Topology

The Induced Subset Topology

Closed Sets

Interior and Closure

Scanned Notes

PDF

AUDIO Of Entire Lecture

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