Fundemental Properties of Well-Ordering (Checking My Answers)

Question

1. Give two different examples of well-ordered sets one of which is infinite.

My Answer: $\mathbb{N}$ and $\left\{0\right\}$.

2. Show that every subset of a well ordered set is a well ordered set with respect to the induced order.

My Proof: Let $X$ be well-ordered. Let $A\subseteq X.$ Since $X$ is well-ordered , there is an $a\in A$ such that $a\leq x$ for any $x\in X$. Consider $A\subseteq A$. $A$ is well-ordered because there is an $a\in A$ such that $a\leq y$ for any $y\in A$.

3. Show that minimal element $m$ of a nonempty subset $A$ of a totally ordered set $(X,<)$ is unique.

My Proof trying: Let $(X,<)$ be well-ordered. Let $A\subseteq X$ which $m_1,m_2$ are minimal elements of $A$. We need to show $m_1=m_2$, right? If yes,can you help? If no, can you give a hint?

4. If $X$ is well-ordered then $X\cup\left\{X\right\}$ is also well-ordered.

My Proof: Let $\emptyset\ne A\subseteq X\cup\left\{X\right\}$. If $A\subseteq X$ then $A$ does have a minimal element because $X$ is well-ordered. If $A\subseteq \left\{X\right\}$, then $A=X$ so clear.

Can you check my answers? Can you help if my answer is false? Thanks...

Answer 1

"Give two different examples of well-ordered sets one of which is infinite. My Answer: N and {0}." These are sets. Since you were asked to give examples of well-ordered sets, you must also state what ordering you have for each set.

"Show that every subset of a well ordered set is a well ordered set with respect to the induced order. My Proof: Let X be well-ordered. Let A⊆X. Since X is well-ordered , there is an a∈A such that a≤x for any x∈X."

Since X is well ordered there is a∈ X such that …. How do you argue that there is such an "a" in A? Suppose X is the set of all positive integers and A is the set of all even positive integers with the usual order. They are both well ordered: X has "1" as minimum, A has "2" as minimum. But if your x= 1, there is no a in A such that a≤ 1!

" Consider A⊆A. A is well-ordered because there is an a∈A such that a≤y for any y∈A."

I don't understand this at all!

"Show that minimal element m of a nonempty subset A of a totally ordered set (X,<) is unique. My Proof trying: Let (X,<)be well-ordered. Let A⊆X of which m1,m are minimal elements of A. We need to show m1=m2, right?"

Yes. that is right.

"If yes,can you help? If no, can you give a hint?"

That's a strange phrasing! Wouldn't a hint help you? Saying that a thing is "unique" means that there are not two of them. And typically the simplest way to prove a negative is by "contradiction". Suppose there exist two "minimum" members, m1 and m2. Since m1 is minimum, m1 is less than or equal to every member of the set. In particular, m1≤ m2. Since m2 is minimum, m2 is less than or equal to every member of the set. In particular, m2≤ m1.

"If X is well-ordered then X∪{X} is also well-ordered. My Proof: Let ∅≠A⊆X∪{X}. If A⊆X then A does have a minimal element because X is well-ordered If A⊆{X}, then A=X so clear."

What if A is neither a subset of X not {X}? If A is a subset of PUQ, it does not follow that A is a subset of P or a subset of Q!

"Can you check my answers? Can you help if my answer is false? Thanks.."

Answer 2

You correctly applied the definition of being well-ordered only at the first part of your proof for 4.

You should use the same definition for 2. (The part ' Consider $A\subseteq A$. ... ' starts to get unclear. You'd rather want to consider an arbitrary $B\subseteq A$.)

For 3., total ordering means that either $m_1\le m_2$ or $m_2\le m_1$ holds, but they both are assumed to be minimal.

For 4., your idea is basically right, though the proof is not written correctly.. Note that we consider $x<X$ for each $x\in X$.
Let $A\subseteq X\cup\{X\}$ nonempty. If $A\cap X\ne\emptyset$, then the minimal element of $A\cap X$ is minimal in $A$. Else, $A=\{X\}$ which trivially has minimal element: $X$.

Answer 3

For the first one, you've got the right idea, but you should probably specify that $\Bbb N$ is in the usual order, since we can define many orders on $\Bbb N$ that aren't well-orders.

For the second one, you must instead choose an arbitrary subset of $A,$ say $C.$ Then since $C$ is in turn a subset of $X,$ we can find the desired least element, which proves that $A$ is well-ordered by the induced order.

For the third, note that $m_1\le m_2$ and $m_2\le m_1.$ What can we conclude?

For the fourth, you haven't yet covered all the options, as $X\cup\{X\}$ has subsets that are neither subsets of $X$ nor equal to $\{X\}$--for example, $X\cup\{X\},$ itself! Rather, you should consider the cases:

• $A\cap X\ne\emptyset,$ in which case the least element of $A\cap X$ is the least element of $A,$ since $X$ is the greatest element of $X\cup\{X\}.$
• $A=\{X\},$ in which case it's clear.