# Show that total orders are maximal orders

A set $$X$$ together with a binary relation $$\leq$$ such that for all $$x,y,z\in X$$,

$$O1$$. $$x\leq x$$

$$O2$$. $$x\leq y$$ and $$y\leq x$$ then $$x=y$$

$$O3$$. $$x\leq y$$ and $$y\leq z$$ then $$x\leq z$$

is called an ordered or sometimes a partially ordered.

An ordered $$(X,\leq)$$ is called a total order if for all $$x,y\in X$$, either $$x\leq y$$ or $$y\leq x$$.

Question: Show that total orders are maximal orders

My question: is there a typo in the question? What is a maximal order? How can I prove the question, can you help? Thanks...

• Maximal in the sense that there is no partial ordering of $X$ which extends a total order, except the total order itself. Feb 21 '19 at 8:29
• @AsafKaragila thanks, how should I prove the question, may you give me a hint?
– user295645
Feb 21 '19 at 8:30

General remarks. An order relation $$\leq$$ on a set $$X$$ is a subset of $$X \times X$$ (i.e. an element of the powerset $$\mathcal{P}(X \times X)$$) satisfying axioms $$O_1, O_2, O_3$$. Since the set $$\mathcal{P}(X \times X)$$ is in turn ordered by inclusion $$\subseteq$$, the question on comparing two orders on $$X$$ makes sense. In particular, an order $$\leq$$ on $$X$$ is maximal if, for every order $$\leq'$$ on $$X$$, $$\leq \,\subseteq \, \leq'$$ implies $$\leq \,=\, \leq'$$ (roughly, $$\leq$$ is maximal if there is no other order in $$X$$ putting "more order'' than $$\leq$$). Note that:

1. $$\leq \,\subseteq \, \leq'$$ means that for all $$x, y \in X$$, if $$x \leq y$$ then $$x \leq' y$$ (roughly, if $$x$$ is less than $$y$$ according to the order $$\leq$$, then they are so according to the order $$\leq'$$);

2. $$\leq \,=\, \leq'$$ means that for all $$x, y \in X$$, $$x \leq y$$ if and only if $$x \leq' y$$, i.e. the orders $$\leq$$ and $$\leq'$$ coincide.

Answer to your question. Concretely, suppose that $$\leq$$ is a total order on $$X$$. We show that $$\leq$$ is maximal, i.e. that for every order $$\leq'$$ on $$X$$, if $$\leq \,\subseteq \, \leq'$$ then $$\leq \,=\, \leq'$$. So, let $$\leq'$$ be an order on $$X$$ such that $$\leq \,\subseteq\, \leq'$$. We have to prove that $$\leq \,=\,\leq'$$, i.e. (see Point 2 above) that for all $$x, y \in X$$, $$x \leq y$$ if and only if $$x \leq' y$$. Let $$x, y \in X$$.

• Since $$\leq \,\subseteq\, \leq'$$, we already know (see Point 1 above) that if $$x \leq y$$ then $$x \leq' y$$.

• Suppose now that $$x \leq' y$$. There are two cases: either $$y \leq x$$ or $$y \not\leq x$$.

1. If $$y \leq x$$ then $$y \leq' x$$ because $$\leq \,\subseteq\, \leq'$$. From $$x \leq' y$$ and $$y \leq' x$$ it follows (by axiom $$O_2$$) that $$x = y$$; in particular, $$x \leq y$$.
2. If $$y \not \leq x$$ then $$x \leq y$$, since $$\leq$$ is a total order (i.e. either $$x \leq y$$ or $$y \leq x$$).

We have just proved that, in all cases, if $$x \leq' y$$ then $$x \leq y$$.

This complete the proof that for all $$x, y \in X$$, $$x \leq y$$ if and only if $$x \leq' y$$. Therefore, $$\leq \,=\, \leq'$$.