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  1. Describe all the partial orders on $\{1,2,3,4\}$ where the set of minimal elements are $\{2,4\}$ and the set of maximal elements is $\{1,3\}$

  2. Describe all the partial orders on $\{1,2,3,4\}$ where 2 is the minimum and 1 is the maximum.

  3. Describe the total orders over $\mathbb N$ with a maximum and no minimum.

  4. Describe the total orders over $\mathbb N$ with no maximum and no minimum.

No need to prove and we can describe it with a graph.

I'm not sure I understand the question, well, for 1. I think the set is $A=\{\text{x is even}: x> 2 , x \in \mathbb Q \}\cup \{\text{x is odd}: x<3, x \in \mathbb Q\}$

For 2, I think it's $\{1\}$ on it's own and $\{x \in \mathbb Q: x\le 2 $}

3: it's like $\omega+0$ but there should be no minimum so I'm not sure how to cut it off. Probably 4 has the same principle.

Community
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shinzou
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1 Answers1

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The question asks you to describe all the partial orders with certain properties.

For example, all the partial orders of the set $\{1,2,3\}$ whose minimum element is $1$ are:

2   3     2    3
 \ /      |    |
  1       3    2
          |    |
          1    1

The first two parts of the question give similar description of orders of $\{1,2,3,4\}$.

The last two are more difficult, since there are continuum many linear orders of $\Bbb N$, with or without a maximum. But recall that every linear order can be realized as a subset of $\Bbb Q$, so by removing a particular set we can get linear orders.

Asaf Karagila
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  • Oh I understand now about the first two. I'm not sure I understand about the other two, by 'realized' you mean there's a 1-1/onto function between a linear order and the rationals ? By particular set you mean maybe the set of single elements ? ${{i} } i\in\mathbb N$ – shinzou May 24 '14 at 22:11
  • So there can't be a minimum, the next smallest subsets have two elements: 1<2, 1<3, ..., 2<3, 2<4, ..., 3<4, ... those can't be a minimum. – shinzou May 24 '14 at 22:15
  • No, I meant that every countable linear order $(A,\leq_A)$ is isomorphic to some $S\subseteq\Bbb Q$ when ordered with the natural order of the rational numbers. For example, if you add one more element on top of the natural numbers, then this is isomorphic to the set ${1-\frac1n\mid n\in\Bbb N\setminus{0}}\cup{1}$. The fact that we are asked to order $\Bbb N$ doesn't mean that it has to agree with the natural order of $\Bbb N$. – Asaf Karagila May 24 '14 at 22:21
  • @kuhaku See for example here on the fact, that every countable linearly ordered set can be realized as a subset of $\mathbb Q$. – Martin Sleziak Jun 05 '14 at 16:41