Example of Partial Order that’s not a Total Order and why?

Think about the subsets of $\{0,1\}$. They are: $\emptyset, \{0\}, \{1\}$, and $\{0,1\}$. Now, we can make these subsets into a partial order with $\subset$. For instance, $\emptyset \subset \{0\}$ and $\{1\} \subset \{0,1\}$. You can show this satisfies the axioms for a partial order:

$$(A \subset A \text{ and } A \subset B, \text{ and } B \subset C) \Rightarrow A \subset C \\ \\$$ $$A \subset B, B \subset A \Rightarrow A = B$$

But a total order $<$ drops the first axiom above and replaces it with the following:

$x < y$ or $y < x$ for all $x,y$

And we see that our example of subsets of $\{0,1\}$ does not satisfy this. For instance, neither $\{0\} \subset \{1\}$ nor $\{1\} \subset \{0\}$ are true. In a total order, we want to be able to compare any two elements. In a partial order, we don't.