Definition:Ordered Sum

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Definition

Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.

Let:

the order type of $\struct {S, \preceq_1}$ be $\theta_1$
the order type of $\struct {T, \preceq_2}$ be $\theta_2$.


Let $S \cup T$ be the union of $S$ and $T$.

We define the ordering $\preceq$ on $S$ and $T$ as:

$\forall s \in S, t \in T: a \preceq b \iff \begin{cases}

a \preceq_1 b & : a \in S \land b \in S \\ a \preceq_2 b & : \neg \paren {a \in S \land b \in S} \land \paren {a \in T \land b \in T} \\ & : a \in S, b \in T \end{cases}$

That is:

If $a$ and $b$ are both in $S$, they are ordered as they are in $S$.
If $a$ and $b$ are not both in $S$, but they are both in $T$, they are ordered as they are in $T$.
Otherwise, that is if $a$ and $b$ are in both sets, their ordering in $S$ takes priority over that in $T$.


The ordered set $\struct {S \cup T, \preceq}$ is called the ordered sum of $S$ and $T$, and is denoted $S + T$.

The order type of $S + T$ is denoted $\theta_1 + \theta_2$.


General Definition

The ordered sum of any finite number of ordered sets is defined as follows:

Let $S_1, S_2, \ldots, S_n$ be tosets.

Then we define $T_n$ as the ordered sum of $S_1, S_2, \ldots, S_n$ as:

$\forall n \in \N_{>0}: T_n = \begin {cases}

S_1 & : n = 1 \\ T_{n - 1} + S_n & : n > 1 \end {cases}$


Informal Interpretation

We can consider the ordered sum $\struct {S \cup T, \preceq}$ as:

First the whole of $S$, ordered by $\preceq_1$
After that, the set $T \setminus S$, ordered by $\preceq_2$, where $T \setminus S$ denotes set difference.


Caution

Note the way the definition of an ordered sum has been worded.

Suppose $a, b \in S \cap T$.

Suppose:

$a \preceq_1 b$ (through dint of $a, b \in S$
$b \preceq_2 a$ (through dint of $a, b \in T$.

Then because $a, b \in S$, we have that $a \prec b$.

But, also because $a, b \in S$, we do not consider the fact that $a, b \in T$ and so the relation $b \preceq_2 a$ is ignored.


Also see

  • Results about ordered sums can be found here.


Sources