what is total order - explanation please

sorry for the dumbest question ever, but i want to understand total order in an intuitive way, this is the defition of total order:

i) If $a ≤ b$ and $b ≤ a$ then $a = b$ (antisymmetry);
ii) If $a ≤ b$ and $b ≤ c$ then $a ≤ c$ (transitivity);
iii) $a ≤ b$ or $b ≤ a$ (totality).

totality means that any pair of the total ordered pair is mutually comparable. i dont understand what they mean under comparable, what is the defition of comparability? i can also compare the elements of partial order, where is problem? why is partial order not not mutually comparable?

can someone explain me please in simple words :(

Solution 1:

Two distinct elements are called "comparable" when one of them is greater than the other. This is the definition of "comparable". When you have a partially ordered set, some pairs of elements can be not comparable. i.e. you can have two elements $x$ and $y$ such that $x\leqslant y$ is false and $y \leqslant x$ is also false.

For example, consider the set $\mathbb{R}^2$ and a partial order defined like this: $$ (x_1,x_2) \leqslant (y_1,y_2) \quad\textrm{iff}\quad x_1\leqslant y_1 \,\textrm{and}\,x_2 \leqslant y_2. $$ With this partial order, elements $(0,0)$ and $(1,2)$ of $\mathbb{R^2}$ are comparable, because $(0,0)\leqslant(1,2)$. But elements $(0,1)$ and $(1,0)$ are not comparable, because both statements "$(0,1)\leqslant(1,0)$" and "$(1,0)\leqslant(0,1)$" are false.

Solution 2:

If $\le$ is a partial order on a set $A$, two elements $a,b$ of $A$ are comparable if either $a\le b$ or $b\le a$; otherwise they are not. For instance, in the partial order $\subseteq$ on $\wp(\Bbb N)$ the sets $\{0,2\}$ and $\{0,1\}$ are not comparable: neither is a subset of the other. On the other hand, the sets $\{0,2\}$ and $\{0,1,2\}$ are comparable: $\{0,2\}\subseteq\{0,1,2\}$.

Another name for a total order is linear order. It expresses the intuitive idea very well: you can picture a linear order $\le$ on a set $A$ as arranging the elements of $A$ in a line. The usual order $\le$ on $\Bbb R$ is a linear (or total) order: if $x,y\in\Bbb R$ are any real numbers, either $x\le y$, or $y\le x$. To put it another way, if $x$ and $y$ are any real numbers, at least one of the statements $x\le y$ and $x\ge y$ must be true. Contrast that with subsets of $\Bbb N$. If $A$ and $B$ are subsets of $\Bbb N$, it’s not the case that at least one of $A\subseteq B$ and $A\supseteq B$ must be true: we just saw that $A=\{0,2\}$ and $B=\{0,1\}$ are a counterexample. The linear order $\le$ lines up the real numbers as a single linear arrangement; the partial order $\subseteq$ does not line of the subsets of $\Bbb N$ in a linear arrangement.

Solution 3:

Suppose we have a set that is the union of members of EvilCorp and Skynet. An order on this set is " $a\leq b$ if a has equal or lower rank than b in their company ". This satisfies i) and ii) but not iii) - we can not compare a and b if they are from different companies. So this is not a total order, even though it is quite a natural order.

Solution 4:

As you know that $(D, \leq )$ is a POSET (partially ordered set) if it fulfills three properties of,

• *R*eflexivity $a\leq a$
• *A*ntisymmetry $a\leq b$ and $b\leq a$ implies $a=b$
• *T*ransitivity $a\leq b$ and $b\leq c$ implies $a\leq c$

In addition to these, if all elements in your set $D$ also are comparable with each other, for instance $D$ has $\{a,b,c,d\}$ and you can compare $a\leq b$, $a\leq c$, $a\leq d$, $b\leq c$, $c\leq d$ and so on, then it can be called as a total order. i.e. when you can have the relation $\leq$ totally on all elements.