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Give an example of a well ordered set $(X,\le)$ in which there exists an element $x_0$ such that there are infinitely many elements $x\in X$ such that $x\lt x_0$.

Let $X=${$A_i | i \in \mathbb N$}$\cup \mathbb N$ where $A_i=${$1,2,...,i$}. Let the relation be inclusion of sets. Here, $x_0=\mathbb N$ obviously. I feel intuitively that my $X$ is well ordered but I'm unsure how to prove it.

Alex
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  • To give an example of $(X,\le)$ you have to specify both $X$ and $\le.$ What is your order relation? We have to know what that is, before we can start to prove that it's a well-ordering. – bof Nov 17 '16 at 02:27
  • @bof sorry! edited! – Alex Nov 17 '16 at 02:28
  • Once you have specified your order relation, think about the definition of a well-ordered set. Show that your set and relation meet that definition. – Browning Nov 17 '16 at 02:28
  • Can you prove that $(X,\subseteq)$ is linearly ordered? Can you prove that $({A_i|i\in\mathbb N},\subseteq)$ is well-ordered? Can you prove that a linearly ordered set, which is the union of two well-ordered subsets, is well-ordered? – bof Nov 17 '16 at 02:37
  • $X={A_i | i \in \mathbb N}\cup{\mathbb N}$ – Mirko Nov 17 '16 at 04:41

2 Answers2

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Let $S\subseteq X$ be non-empty.

If $\mathbb{N}\notin S$, then $$ S=\bigcup_{i\in I}\{A_i\} $$ for a non-empty subset $I\subseteq\mathbb{N}$. Since $\mathbb{N}$ is well-ordered, it follows that there is a minimum element $i_o\in I$. Clearly, $A_{i_0}$ is the minimum of $S$.

The same applies if $$ S=\{\mathbb{N}\}\cup\bigcup_{i\in I}\{A_i\} $$ for a non-empty subset $I\subseteq\mathbb{N}$.

If $S=\{\mathbb{N}\}$ then clearly $\mathbb{N}$ is the minimum of $S$.

Guest
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I'd use cardinality for the ordering relationship. That is, $x_i \le x_j$ in your ordering when $|x_i| \le |x_j|$ using the common meaning for less-than or equal.

Effectively, your set is the natural numbers plus $\aleph_0$. The well-ordering follows directly from that.

John Fisher
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