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title: "Order, Product & Subpace topologies"
format: 
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            file: ../latex-header.tex
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# Order Relations

An order relation on a set $X$ is a relation < satisfying for all $a,b,c \in X$:

1. (Comparability) Either $a < b$, $a = b$, or $a > b$.
2. (Transitivity) If $a < b$ and $b < c$, then $a < c$.
3. (Non-reflexivity) $a \not< a$.

## Examples

1. The usual order on $\mathbb{R}$.
2. The (dictionary order) lexicographic order on $\mathbb{R}^2$: $(a_1, a_2) < (b_1, b_2)$ if either $a_1 < b_1$ or $a_1 = b_1$ and $a_2 < b_2$.
3. The dictionary order on words: $a < b$ if either the first letter where they differ is earlier in the alphabet in $a$ than in $b$, or $a$ is a prefix of $b$.
4. The parabolic order on $\mathbb{R}^2$ below:

![parabolic order](../order-topology.pdf)

# Order Topology

Given a set $X$ with an order relation <, the *order topology* on $X$ is generated by the basis consisting of all open intervals $(a,b) = \{x \in X : a < x < b\}$, along with intervals of the form $[a_0, b) = \{x \in X : x < b\}$ if $a_0$ is the least element of $X$, and intervals of the form $(a, b_0] = \{x \in X : a < x\}$ if $b_0$ is the greatest element of $X$.

## Examples

1. The order topology on $\mathbb{R}$ is the standard topology.
2. The (lexicographic) dictionary order topology on $\mathbb{R}^2$ has two typs of open sets:
    1. Open intervals of the form $(a \times b, c \times d) \text{ for } a < c$
    2. Vertical open intervals of the form $(a \times b, a \times d) \text{ for } b < d$

## Questions

1. Why do sets of the second type form a basis?
2. What should *rays* be in this topology? 

# Product Topology

Given topological spaces $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$, the *product topology* on $X \times Y$ is generated by the basis consisting of all products of open sets $U \times V$ where $U \in \mathcal{T}_X$ and $V \in \mathcal{T}_Y$.

## Examples
1. The product topology on $\mathbb{R} \times \mathbb{R}$ is the standard topology on $\mathbb{R}^2$.
2. The product topology on $\mathbb{R} \times \mathbb{R}$ where $\mathbb{R}$ has the standard topology and $\mathbb{R}$ has the discrete topology.
3. We will see that the torus inherits a product topology from $S^1 \times S^1$. (First, we need to define the topology on $S^1$.)
4. The product topology on $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$ is the standard topology on $\mathbb{R}^3$. 

## Question
1. Is the bases $\mathcal{B}$ actually the entire topology? (No, unions of rectangles are not rectangles.)

## Main Theorems

**Theorem 1** : A (another) basis for the product topology on $X \times Y$ is given by the collection of all products of basis elements $B_X \times B_Y$ where $B_X$ is a basis element for $X$ and $B_Y$ is a basis element for $Y$.

**Proof** :
Recall, that $\mathcal{B}$ is a basis for a topology on $X$ if for every point $x \in X$ and every open neighborhood $U$ of $x$, there is a basis element $B \in \mathcal{B}$ such that $x \in B \subseteq U$.  Let $x \in X$ and assume $U$ is an open neighborhood of $x$.  Then by definition of the product topology, there is a basis element $U_X \times U_Y$ such that $x \in U_X \times U_Y \subseteq U$.  Since $B_X$ is a basis for $X$, there is a basis element $B_X$ such that $x_X \in B_X \subseteq U_X$.  Similarly, there is a basis element $B_Y$ such that $x_Y \in B_Y \subseteq U_Y$.  Thus, we have that
$$x \in B_X \times B_Y \subseteq U_X \times U_Y \subseteq U$$
as desired.  $\square$


# Subspace Topology

Given a topological space $(X, \mathcal{T})$ and a subset $Y \subseteq X$, the *subspace topology* on $Y$ is all sets of the form $U \cap Y$ where $U \in \mathcal{T}$.  In other words, a set $V \subseteq Y$ is open in the subspace topology if there is an open set $U$ in $X$ such that $V = U \cap Y$.

Why is this a topology?

1. $Y = X \cap Y$ is open in the subspace topology since $X$ is open in $X$.

2. $\phi = \phi \cap Y$ is open in the subspace topology since $\phi$ is open in $X$.

3. Let $\{V_\alpha\}$ be a collection of open sets in the subspace topology.  Then for each $\alpha$, there is an open set $U_\alpha$ in $X$ such that $V_\alpha = U_\alpha \cap Y$.  Thus,
   $$\bigcup_\alpha V_\alpha = \bigcup_\alpha (U_\alpha \cap Y) = \left(\bigcup_\alpha U_\alpha\right) \cap Y$$
   which is open in the subspace topology since $\bigcup_\alpha U_\alpha$ is open in $X$.

4. Let $V_1, V_2, \ldots, V_n$ be a finite collection of open sets in the subspace topology.  Then for each $i$, there is an open set $U_i$ in $X$ such that $V_i = U_i \cap Y$.  Thus,
   $$\bigcap_{i=1}^n V_i = \bigcap_{i=1}^n (U_i \cap Y) = \left(\bigcap_{i=1}^n U_i\right) \cap Y$$
   which is open in the subspace topology since $\bigcap_{i=1}^n U_i$ is open in $X$.

It follows that the intersection of basis elements forms a basis for the subspace topology.

## Terminology

With subspaces, we need to keep track of two different types of open sets. If $Y \subseteq X$, we say $U$ is open in $Y$ if $U$ is open in the subspace topology on $Y$.  We say $U$ is open in $X$ if $U$ is open in the topology on $X$.


## Examples

1. Let $X = \mathbb{R}$ with the standard topology and let $Y = [0,1]$.  Then the open sets in the subspace topology on $Y$ are all sets of the form $(a,b) \cap [0,1]$ where $(a,b)$ is an open interval in $\mathbb{R}$.  Thus, the open sets in the subspace topology on $Y$ are:
    1. The empty set
    2. Intervals of the form $(a,b)$ where $0 < a < b < 1$
    3. Intervals of the form $[0,b)$ where $0 < b \leq 1$
    4. Intervals of the form $(a,1]$ where $0 \leq a < 1$
    5. The entire set $[0,1]$

2. Let $X = \mathbb{R}^2$ with the standard topology and let $Y$ be the $x$-axis.  Then the open sets in the subspace topology on $Y$ are all sets of the form $U \cap Y$ where $U$ is an open set in $\mathbb{R}^2$.  Since open sets in $\mathbb{R}^2$ are unions of open balls, it follows that the open sets in the subspace topology on $Y$ are unions of open intervals on the $x$-axis.  Thus, the subspace topology on $Y$ is the standard topology on $\mathbb{R}$.

3. Let $X = \mathbb{R}$ with the standard topology and let $Y = \mathbb{Q}$ be the set of rational numbers.  Then the open sets in the subspace topology on $Y$ are all sets of the form $(a,b) \cap \mathbb{Q}$ where $(a,b)$ is an open interval in $\mathbb{R}$.  Thus, the open sets in the subspace topology on $Y$ are unions of open intervals (of rationals) with rational endpoints.

4. Let $X = \mathbb{R}$ with the standard topology and let $Y = \mathbb{Z}$ be the set of integers.  Then the open sets in the subspace topology on $Y$ are all sets of the form $(a,b) \cap \mathbb{Z}$ where $(a,b)$ is an open interval in $\mathbb{R}$.  Thus, the open sets in the subspace topology on $Y$ are all subsets of $\mathbb{Z}$ since for any subset of integers, we can find an open interval in $\mathbb{R}$ that contains exactly those integers.  This is the discrete topology on $\mathbb{Z}$.

5. Let $X = \mathbb{R}$ with the standard topology and let $Y = [0,1) \cup \{2\}$.  Then the open sets in the subspace topology on $Y$ are all sets of the form $(a,b) \cap Y$ where $(a,b)$ is an open interval in $\mathbb{R}$.  Thus, the open sets in the subspace topology on $Y$ are:
    1. The empty set
    2. Intervals of the form $(a,b)$ where $0 < a < b < 1$
    3. Intervals of the form $[0,b)$ where $0 < b \leq 1$
    4. Intervals of the form $(a,1]$ where $0 \leq a < 1$
    5. The entire set $[0,1) \cup \{2\}$
    6. The singleton set $\{2\}$
    7. Unions of the above sets.

6. Let $X = \mathbb{S}^1$ be the unit circle in $\mathbb{R}^2$ with the subspace topology inherited from the standard topology on $\mathbb{R}^2$.  Then the open sets in the subspace topology on $X$ are all sets of the form $U \cap \mathbb{S}^1$ where $U$ is an open set in $\mathbb{R}^2$.  Since open sets in $\mathbb{R}^2$ are unions of open balls, it follows that the open sets in the subspace topology on $X$ are unions of open arcs on the circle.

7. Let $X = \mathbb{S}^1 \times \mathbb{S}^1$ be the torus with the product topology inherited from the standard topology on $\mathbb{R}^2 \times \mathbb{R}^2 \cong \mathbb{R}^4$.  Then the open sets in the product topology on $X$ are all sets of the form $U_1 \times U_2$ where $U_1$ and $U_2$ are open sets in $\mathbb{S}^1$.  Since open sets in $\mathbb{S}^1$ are unions of open arcs on the circle, it follows that the open sets in the product topology on $X$ are unions of products of open arcs on the two circles.

8. Let $X$ be the torus with the subspace topology inherited from the standard topology on $\mathbb{R}^3$ via the usual embedding of the torus in $\mathbb{R}^3$.  Then the open sets in the subspace topology on $X$ are all sets of the form $U \cap X$ where $U$ is an open set in $\mathbb{R}^3$.  Since open sets in $\mathbb{R}^3$ are unions of open balls, it follows that the open sets in the subspace topology on $X$ are unions of open "patches" on the torus.  It follows from Theorem 3 below that this topology is the same as the product topology on $X$.

## Some basic Theorems

**Lemma 2** : If $U \subseteq Y \subseteq X$, and $U$ is open in $Y$ and $Y$ is open in $X$, then $U$ is open in $X$. 

**Proof** : Since $U$ is open in $Y$, there is an open set $V$ in $X$ such that $U = V \cap Y$.  Since $Y$ is open in $X$, we have that $U = V \cap Y$ is open in $X$ as the intersection of two open sets.  $\square$

**Theorem 3** : If $A \subset X$ and $B \subset Y$, then the subspace topology on $A \times B$ as a subset of $X \times Y$ is the same as the product topology on $A \times B$ where $A$ and $B$ have the subspace topologies inherited from $X$ and $Y$, respectively.

**Proof** : Let $\mathcal{T}_1$ be the subspace topology on $A \times B$ as a subset of $X \times Y$ and let $\mathcal{T}_2$ be the product topology on $A \times B$ where $A$ and $B$ have the subspace topologies inherited from $X$ and $Y$, respectively.  We will show that $\mathcal{T}_1 \subseteq \mathcal{T}_2$ and $\mathcal{T}_2 \subseteq \mathcal{T}_1$.

($\mathcal{T}_1 \subseteq \mathcal{T}_2$) Let $U \in \mathcal{T}_1$.  Then there is an open set $V$ in $X \times Y$ such that $U = V \cap (A \times B)$.  Since $V$ is open in the product topology on $X \times Y$, there is a basis element $U_X \times U_Y$ such that $x \in U_X \times U_Y \subseteq V$.  Thus,
$$U = V \cap (A \times B) \supseteq (U_X \times U_Y) \cap (A \times B) = (U_X \cap A) \times (U_Y \cap B)$$
Since $U_X$ is open in $X$, we have that $U_X \cap A$ is open in the subspace topology on $A$.  Similarly, $U_Y \cap B$ is open in the subspace topology on $B$.  Thus, $(U_X \cap A) \times (U_Y \cap B)$ is a basis element for the product topology on $A \times B$, and so $U$ is open in $\mathcal{T}_2$.

($\mathcal{T}_2 \subseteq \mathcal{T}_1$) Let $U \in \mathcal{T}_2$.  Then there are open sets $U_A$ in $A$ and $U_B$ in $B$ such that $U = U_A \times U_B$.  Since $U_A$ is open in the subspace topology on $A$, there is an open set $V_A$ in $X$ such that $U_A = V_A \cap A$.  Similarly, there is an open set $V_B$ in $Y$ such that $U_B = V_B \cap B$.  Thus,
$$U = U_A \times U_B = (V_A \cap A) \times (V_B \cap B) = (V_A \times V_B) \cap (A \times B)$$
Since $V_A$ is open in $X$ and $V_B$ is open in $Y$, we have that $V_A \times V_B$ is open in the product topology on $X \times Y$.  Thus, $U$ is open in $\mathcal{T}_1$.  $\square$



