We know that every Partial Order Set has to satisfy three conditions :
1- Reflexive
2- Anti-Symmetric
3- Transitive
For example : $$S= \left(\left\{1 \right\},\left\{2 \right\},\left\{3 \right\},\left\{1,2 \right\} \right)$$
And the relation is $\subseteq$ ,My question is why this Set S is a Partial Order Set Althought I Cannot prove that is Transitive which is defined as:
$$R=\{(x,y)\} ; xRy ~\text{and}~ yRz \implies xRz$$ I could not find like : if $\left\{1 \right\} \subseteq \left\{1,2 \right\}\text{ } and \text{ } \left\{1,2 \right\} \subseteq \text{ }\text{ } ?? \text{ }then\text{ } ??$
in another way how can we apply the Transitive test for this set S?
Note that statement of the form of "if $A$ then $B$" is true if $A$ is false.
Well, actually we have $\color{blue}{\{1 \}} \subseteq \color{green}{\{ 1,2\}}$ and $\color{green}{\{1,2\}} \subseteq \color{purple}{\{ 1, 2\}}$, then $\color{blue}{\{1\}} \subseteq \color{purple}{\{1,2\}}.$
Notice that for the subset operation, we always have
$$x \subseteq y \text{ and }y \subseteq z \text{ ,then }x \subseteq z$$