Subset Relation: Is the subset relation a partial order? I read in a Wikipedia entry (subset in german http://de.wikipedia.org/wiki/Teilmenge):
"Every set is a subset of itself"
But for example, if A is a set of all sets, with maximum 5 Elements, than A isn`t a Subset of itself.
so the subset-relation isn`t a partial order and the wikipedia entry is wrong, isnt it?
 A: The set $A$ is a subset of the set $B$ is every element of $A$ is an element of $B$. Since every element of $A$ is an element of $A$ the set $A$ is a subset of $A$. This shows the subset releation if reflexive.
If $A \subseteq B$ and $B \subseteq C$ then every element of $A$ is an element of $B$ and every element of $B$ is an element of $C$. Pick $a \in A$. Since $A \subseteq B$ we have $a \in B$. Since $a \in B$ and $B \subseteq C$ we have $a \in C$. Recalling that $a$ was an arbitrarily chosen element of $A$ we see that $A \subseteq C$. Thus the subset relation is transitive.
I will leave anti-symmetry as an exercise.
A: Here subset notation $\subseteq$ is the "inclusive or" statement i.e A may be equal to A. 
A: I was relatively confused by the wikipedia portion of your question, but yes, the subset/inclusion relation is a partial order(ing). In order to prove this, we have to show that it is reflexive, anti-symmetric, and transitive. 
Reflexive: The set $A$ will always be a subset of itself because it contains all the elements within itself. Effectively, $A = A$ because $A \subseteq A$ and vice versa.
Anti-Symmetric: If set A includes set B ($A \subseteq B$) and ($B \subseteq A$), then $B = A$. This would prove it is anti-symmetric.
Transitive: If $A \subseteq B$ and $B \subseteq C$, then $A \subseteq C$.
Now we have proved that the subset/inclusion relation is a partial ordering. 
