Find in the ordered set $(\mathcal P(\Bbb N),\subseteq)$ a chain in which there is neither maximum nor minimum.

Question

Find in the ordered set $(\mathcal P(\Bbb N),\subseteq)$ a chain in which there is neither maximum nor minimum.

I've found a chain in which there is no minimum: $\Bbb N \supseteq \Bbb N / \{0\} \supseteq \Bbb N / \{0,1\} \supseteq \Bbb N / \{0,1,2\} \supseteq \ldots$.

Could you please help me?

Answer 1

Here is one: $$ \cdots \subseteq \{4, 6, 8, \ldots\}\subseteq \{2, 4, 6, \ldots\}\subseteq\{0,2,4,\ldots\}\\ \subseteq \{0, 1, 2, 4, 6, \ldots\}\subseteq\{0,1,2,3,4,6,8,\ldots\}\subseteq\cdots $$ As the chain continues leftwards removes one even number at a time, and as it continues rightwards it adds one odd number at a time.

Answer 2

$\cdots\supsetneq 2^1\Bbb N\cup\{1,3,5\}\supsetneq 2^1\Bbb N\cup\{1,3\}\supsetneq 2^1\Bbb N\cup\{1\}\supsetneq 2^1\Bbb N\supsetneq 2^2\Bbb N\supsetneq 2^3\Bbb N\supsetneq \cdots,$

where for a natural number $k$, $k\Bbb N$ is the set of all multiples of $k$.

Answer 3

Your chain as no minimum, but a maximum. Likewise, $\emptyset\subset\{1\}\subset\{1,2\}\subset\{1,2,3\}\subset\ldots$ has no maximum, but a minimum. We can combine these by using each trick for its own copy of $\Bbb N$ (i.e., one for odd and one for even numbers). So let $$\mathcal C_1=\{\,A\in \Bbb N\mid \exists n\colon A=\{x\in\Bbb N\mid x\ge n\,\} \,\}$$ be your chain without minimum $$\mathcal C_2=\{\,A\in \Bbb N\mid \exists n\colon A=\{x\in\Bbb N\mid x< n\,\} \,\}$$ my chain without maximum and $$ \mathcal C=\{\,2A\mid A\in\mathcal C_1\}\cup \{(2A+1)\cup2\Bbb N\mid A\in\mathcal C_2\,\}.$$