# Classnotes for September 22

hide timeline for printing/editing
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

## 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:

1. The map which embeds Y into X is continuous, i.e. $\iota\colon Y \rightarrow X$ is continuous.
2. 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:

1. $\iota\colon Y \rightarrow X$ is continuous
2. $f\colon Z \rightarrow Y$ 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:

1. Taking product topologies is associative, i.e. $(X \times Y) \times Z = X \times (Y \times Z)$
2. 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.
3. $X \times Y$ is homeomorphic (though not generally equal) to $Y \times X$.
4. 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.

## 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:

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

Proof:

1. $X^c = \empty$ and $\empty^c \ X$ are both open.
2. $(\cap_{(\alpha \in I)} F_\alpha)^c = (\cup_{(\alpha \in I)} (F_\alpha)^c)$ 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.
3. $(\cup_{(i\in I)} F_i)^c = (\cap_{(i \in I)} (F_i)^c)$ 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.

## 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.

1. The closure of X, denoted by Cl(A), is defined as the smallest closed set containing A, or equivalently, the intersection of all closed subsets of X containing A.
2. The interior of X, denoted by Int(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:

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

Proof:

1. If $x \in U \subset A$, then x is in the union of all open subsets of A, and is therefore in Int(A). Conversely, if $x \in Int(A)$ then Int(A) is open and $x \in Int(A) = U \subset A$.
2. Let $x \in Cl(A)$ and U be any open set containing x. If $A \cup U = \empty$ then $A \subset U^c$. Then Uc 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 Sc is an open set containing x, so it must have a non-empty intersection with A and therefore cannot be a subset of Ac, 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.

## Scanned Notes

### PDF

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

### 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