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I would like to determine an ordering relation:

We are given a linear order on $\mathbb{N}$ $\leq'$ for all $m,n$ such that $ n\leq' m$ $\iff$ (n is odd and m is even) or ($n$$\leq$$m$ and $m-n$ is even) where $\leq$ corresponds to the usual order.

a) Order $6,3,9,11,12$

Here I thought, given the upper restrictions, on:$3,9,11,6,12$

b) Determine the infimum $(\mathbb{N})$:

Here I thought of $1$, whichs seems the greatest element of $K$ that is less than or equal to all elements of $(\mathbb{N})$ given the linear order.

c) Determine supremum $(\{0,1,...,2n+1\})$

Here I just dont have a clue....

d) Determine the lower bound $\{2n|n \in \mathbb{N})$

Here I thouht of $\{2n+1|n∈ℕ\}) \cup \{0\}$

Thank you all in advance....

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  • $\begingroup$ If you are given an ordering on $\mathbb{N}$, then where does $0$ come from in c)? $\endgroup$
    – Sudarsan
    Mar 31, 2015 at 18:40
  • $\begingroup$ Dont know. I havent conceptualized that. But I suppose that 0 is a part of $\mathbb{N}$. Otherwise we would see $\mathbb{N} \{\0\}$. $\endgroup$
    – Mamba
    Mar 31, 2015 at 18:43
  • $\begingroup$ Anyway, the answer posted below works. The ordering would be all odds, 0, all evens if 0 is included. $\endgroup$
    – Sudarsan
    Mar 31, 2015 at 18:49
  • $\begingroup$ @Sudarsan: For many of us $\Bbb N$ includes $0$; if we want the positive integers, we write $\Bbb Z^+$, $\Bbb Z_{>0}$, $\Bbb N^*$, $\Bbb N_{>0}$, etc. $\endgroup$ Apr 1, 2015 at 5:31
  • $\begingroup$ It’s helpful to picture the order $\le'$: $$1,3,5,7,9,\ldots,0,2,4,6,8,\ldots$$ $\endgroup$ Apr 1, 2015 at 5:33

1 Answer 1

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As for part a and b, I agree with you.

Concerning part c, finding the supremum of the set $\{0,1,2,\ldots,2n+1\}$, think about what your order does: It orders all odd numbers before any even numbers, and otherwise orders odd and even numbers as usual. This means that the supremum of the set is the largest even number in that set.

You can use a similar way of thinking for part d.

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