Let $X$ be a set with more than two elements. Define a relation $R$ on $P (X)$, the power set of X, by $(A, B) \in R$ if and only if $A \subseteq B$. Show that $R$ is a partial order on $P (X)$. Is it a well ordering? Is it a total ordering?
So there is a set $X$, that has more than 2 elements. The power set of $X$ has $(A,B)\in R$ if and only if $A \subseteq B$. How would I show if it is partial order on $P(X)$? Or if it is well and/or total ordering?
If I am correct, partial ordering is when reflexive, anti symmetric, and transitive.
I am not to sure about total ordering.
Well ordering = a set having a least element, if I am not mistaken.
(Discrete Mathematics)
Best Answer
If you really understand the definitions this should be easy: (and it is. There moral to this story is going to be "trust yourself"...)
Reflexive: Is $A \subset A$ for all $A \subset X$? Well, ... obviously.
Antisymmetric: Does $A \subset B; B \subset A \implies A = B$? Well, .... obviously.
Transitivity: If $A \subset B; B \subset C \implies A \subset C$? Well....
So is it a partial order? Well....
How about total order?
Must any two sets be such that either $A \subset B$ or $B \subset A$? In other words, is it impossible for $A \not \subset B$ and $B \not \subset A$? What about $\{0,1\}$ and$\{1,2\}$? Must one be a subset of the other?
Is it well-ordered? Let $X = \mathbb N$. Does {even numbers, odd numbers, prime numbers, square numbers} have a least element? As I listed above $\{0,1\}$ and $\{1,2\}$ which of those two are smaller?
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Note by definition, to be well-ordered an ordering must be a total order. Otherwise if $a \not \le b$ and $b \not \le a$ then $\{a,b\}$ can't have a smallest element because $a$ and $b$ can't even be compared.
So if $\subset$ is not a total ordering, it can't be a well ordering.