Order Topology

The Order Basis

Let

(X, \leq)

be a total order and assume that

| X | \geq 2

then let

ℬ

be defined as follows:

All open intervals $(a, b) \in X$
All intervals of the form $[a_{0}, b)$ where $a_{0}$ is the smallest element of $X$ , if there are no smallest element then these sets are not included (as they don't exist)
All intervals of the form $(a, b_{0})$ where $b_{0}$ is the largest element of $X$ , if there are no largest element then these sets are not included (as they don't exist)

then we call

ℬ_{\leq}

the order basis for

X

The Order Basis Is a Basis

As per title.

TODO: Add the proof here.

The Order Topology

Suppose that

ℬ_{\leq}

is the order basis for

X

, then we call the topology generated by

ℬ_{\leq}

the order topology on

(X, \leq)

The Dictionary Order

Let

X_{1}, X_{2}, \dots, X_{n}

be sets with orders

\leq_{1}, \leq_{2}, \dots, \leq_{n}

, respectively. The dictionary order on the Cartesian product

X_{1} \times X_{2} \times \dots \times X_{n}

is defined as follows: For

𝐚 = (a_{1}, a_{2}, \dots, a_{n}) and 𝐛 = (b_{1}, b_{2}, \dots, b_{n})

X_{1} \times X_{2} \times \dots \times X_{n}

, we say that

𝐚 \leq 𝐛

if, for some

i \in {1, 2, \dots, n}

, we have a_j = b_j \quad \text{for all} \quad j < i \quad \text{and} \quad a_i \leq_i b_i. ParseError: Expected 'EOF', got '&' at position 40: …r all} \quad j &̲lt; i \quad \te…

Suppose we're working in $ℝ \times ℝ$ with the dictionary order, then if we consider the two points $(a, b), (c, d) \in ℝ \times ℝ$ the notationally the interval between them is $((a, b), (c, d))$ as you can see the outer parenthesis form an interval but the inner ones are simply points, this becomes incredibly confusing so the notation $x \times y$ is sometimes used to refer to points in $ℝ \times ℝ$ .

The Order Basis for the Dictionary Order on RxR

The order basis for the dictionary order on

ℝ \times ℝ

{(a \times b, c \times d) : (a < c) \lor (a = c \land b < d), a, b \in ℝ}

TODO: Add the proof here.