Supremum and ordinals

Question

Question: Show that sup{$ \xi \dotplus 1 : \xi \in A $} is the least ordinal that is greater than each element of $A$.

I tried to get a better feel for this by letting $A=3=${$0,1,2$}. Then I know that the least ordinal greater than each element of $3$ is $3$ itself. So, sup{$ \xi \dotplus 1 : \xi \in 3 $}=$\bigcup_{\xi \in 3}${$\xi \dotplus 1$}={$0 \dotplus 1$}$\cup${$1 \dotplus 1$}$\cup${$2 \dotplus 1$}$=${$1,2,3$}.

I take it that {$1,2,3$} and {$0,1,2$} are the same because there's an order-preserving isomorphism from one to the other.

Now I'm having trouble generalizing.

Answer 1

Suppose $\alpha=\sup\{\xi+1\colon\xi\in A\}$ then $\alpha$ is bigger than all $\xi\in A$ for obvious reasons.

Suppose $\beta<\alpha$ is also greater than all $\xi\in A$. Since $\alpha$ is the supremum of $\xi+1$ we have that it is the least ordinal greater or equal than $\xi+1$ for $\xi\in A$; in particular we have that $\beta\le\xi_0+1\le\alpha$ for some $\xi_0\in A$.

Since $\beta<\alpha$ we have that if $\xi_0+1=\alpha$ then $\beta=\xi_0$, in contradiction to the fact $\beta$ was strictly greater than all $\xi\in A$, and if $\beta<\xi_0+1$ then $\beta=\xi_0$, once again in contradiction to the aforementioned fact.

Answer 2

You haven't formed the union correctly (you inadvertantly added another layer of braces). What you actually get in your example is $\cup\{1,2,3\}=1\cup 2\cup 3$, and this is just $3$ itself, since $1=\{0\}\subset\{0,1\}= 2\subset\{0,1,2\}= 3$.

In general, it is clear that the supremum of $\{\xi+1:\xi\in A\}$ is at least as large as $\xi+1$ for every $\xi\in A$, and so you just have to argue that nothing smaller will also be larger than each of these.

Answer 3

Let $\alpha = \sup \{\xi+1:\xi\in A\}$. For each $\eta \in A$, $\eta \in \eta+1 \subseteq \bigcup\limits_{\xi \in A}(\xi+1) = \sup \{\xi+1:\xi\in A\} = \alpha$, so $\eta < \alpha$. That is, $\alpha$ is greater than each member of $A$.

Now suppose that some ordinal $\beta < \alpha$. Then $\beta \in \alpha = \bigcup\limits_{\xi \in A}(\xi+1)$, so there must be some $\xi \in A$ such that $\beta \in \xi+1$. But then $\beta < \xi+1$, and hence $\beta \le \xi$. Thus, $\beta$ is not greater than every member of $A$.

Putting the pieces together, we see that $\alpha$ must be the least ordinal greater than all members of $A$.

Answer 4

The sets $\{ 1, 2, 3 \} $ and $\{ 0, 1, 2 \} $ are different sets. They have different elements.

Back to the question. For any set $X$ of ordinals there is an ordinal that is greater than or equal to each and every ordinal in $X$. Since the ordinals are well-ordered there is a least ordinal that is greater than or equal to each and every ordinal in $X$. This ordinal is the supremum of $X$. If $X$ happens to have a largest element then that largest element is the supremum of $X$. (Compare this with the real numbers.)

For your problem there are two cases: either $A$ has a largest element or $A$ does not have a largest element.

Suppose the first case holds. Let $\xi^*$ be the largest element of $A$. The element $\xi^* + 1$ is larger than every element in $A$ and is the supremum of the elements of $\{ \xi + 1 \colon \xi \in A \} $. Any ordinal that is strictly smaller than $\xi^*$ will be no larger than at least one element of $A$, namely $\xi ^*$. In this case we have the desired result.

Suppose the second case holds. In this case $\sup A$ is the least ordinal greater than every ordinal in $A$. To see this choose $\xi_{0} \in A$. By our assumption that the second case holds there is a $\xi_{1} \in A$ satisfying $\xi_{0} < \xi_{1} \leq \sup A$. Since $\xi_{0}$ was an arbitrarily chosen element of $A$ we get $\sup A$ is the least ordinal greater than every element of $A$. Now we show that $\{ \xi + 1 \colon \xi \in A \} = \sup A$. Select $\xi_{0} \in A$. Not that $\xi_{0} < \xi_{0} + 1 \leq \xi_{1}$, where $\xi_{1}$ is some ordinal in $A$ that is greater than $\xi_{0}$. Since for every element $\alpha \in B = \{ \xi + 1 \colon \xi \in A \} $ there is another element of $A$ which is greater than $\alpha$ the supremum of $A$ is no smaller than the supremum of $B$. But then the two supremum are equal.