Prove/disprove: there's a max element on the order $\leq \ $,as $ \ A\leq B; \ $ if $ \ \forall b_1,b_2\in B\to(b_1<b_2 \to|[b_1,b_2]\cap A|\geq2)$

Question

Let $P_{\aleph_{0}}(\mathbb{N}) = \{x\in P(\mathbb{N}):|x|=\aleph_{0}\} $

we'll define the following relation on $P_{\aleph_{0}}(\mathbb{N})$ for $A, B \in P_{\aleph_{0}}(\mathbb{N})$:

$A\preccurlyeq B$ if for every $b_1, b_2 \in B \longrightarrow\left(b_1 < b_2 \longrightarrow\vert [b_1, b_2] \cap A\vert\geq2\right)$

$ \left\langle \preccurlyeq,P_{\aleph_{0}}(\mathbb{N})\right\rangle $ is a Partially ordered set.

prove/ disprove: there exist a maximum element on the order $\preccurlyeq$

my attempt:

I'd like to disprove that argument by contradiction.

suppose there exist a maximum element on the order $\preccurlyeq$, Let it be $A$.

by a definitio; $|A| = \aleph _0$ . therefore by the Well -ordering principle,I'll define $A$ as following $ A = \{ a_i \in A \vert i \in \mathbb{N}\}:$ $$a_1 < a_2 < a_3 < \cdots < a_n < \cdots$$

let $B = \{ a_{2i} \in A \vert i\in \mathbb{N} \}$, hence $ B $ is the even elements of $A$. that means that for every $${b_i} \in B;$$ $$ b_1 < b_2 < \cdots < b_n < \cdots $$ therefore, for every $i$, $a_i < b_i = a_{2i}$ let $b_k, b_j \in B \ $ WLOG $ \ b_k < b_j \ $ and also $ \ b_k = a_{2k}, \ b_j = a_{2j} \ $ for some $ \ k, j \in \mathbb{N}$ $ |b_j - b_k| = |a_{2j}- a_{2k}| > 2$ and of course, $|b_j - b_k| = |a_{2j}- a_{2k}| > |a_j - a_k| $

this is a contradiction to the assumption that $A$ is the maximum element, as we found a bigger element then $A$. therefore. there exist no maximum element.

is that correct?

Answer 1

I'd prove by contradiction that there exist no maximum element in the order $\preccurlyeq$.

Suppose there exist an object in $P_{\aleph_{0}}(\mathbb{N})$ on the order $\preccurlyeq$, Let it be $X$.

$X \in P_{\aleph_{0}}(\mathbb{N}) \Longrightarrow |X| = \aleph_0$

I'll define a function $$f:\mathbb{N} \to X$$ such that $f$ is well ordered function (on the order < over $\mathbb{N}$) and onto.

let the set $$B = \{ f(n) \vert n\in \mathbb{N}_{even} \}$$ clearly, $B \subseteq X$, therefore $|B| \leq \aleph_0$.

I'll define a function $$g:\mathbb{N}_{even} \to B$$ such that $g(n) = f(n)$ for every $n \in \mathbb{N}_{even}$. $f$ is one-to-one therefore $g$ is also one-to-one, $\Longrightarrow \aleph_o = |\mathbb{N}_{even}| \leq |B| $, thus $|B| = \aleph_o$.

$B \subseteq X \subseteq \mathbb{N}$, hence $B \in P_{\aleph_{0}}(\mathbb{N}$.

Let $n_1, n_2 \in B$ such that $n_1 < n_2 $.

because $B \subseteq X$ then $n_1 , n_2 \in X $ so $$|[n_1 , n_2 ] \cap X \geq 2$$ consequently, $B \preccurlyeq X$, which contradicts the assumption that $X$ is the maximum element in the order.

lemma I'll prove that the function $f$, which is defined to be a well ordered and invertible function exist.

I will define $f$ in recursion: $$f(0) = Min (X) $$

$$f(n+1) =_{ \ <} \ \left( X \ / \bigcup_{0 \leq i \leq n} f(i)\right) $$

Answer 2

Let $A\leq B$.

suppose there exist a maximum element, $B$.

$$B = \{ b_{1},b_{2},b_{3},\cdots\}$$

such that: $i<j\ \Longrightarrow\ b_{i}<b_{j}$.

Let $A = \{ n\in\mathbb{N}\vert n<b_{3}\vee n<b_{4}\}$

hence, for a given $b_3 , b_4 $:

$$A\cap[b_{3},b_{4}]=\emptyset$$

Therefore $$ 0 = \vert\emptyset\vert=\vert A\cap[b_{3},b_{4}]\vert<2$$

which contradicts the assumption that $A\leq B$. that means that for every $B$ there exist some $A$ which doesn't maintain the order $ A\leq B$, thus there exist no maximum element $B$.