Countability of a partially ordered set

Question

First of all, thank you in advance to everyone who will look at my question, I saw a question yesterday and I thought of such a solution to the question, but I was not sure if it was my mistake or not, I would be very grateful if you could help me. Here is the question $$(A,\leq) \text{ is an poset and for every } x\in A \text { the set }R_x=\{y \in A : y\leq x\} \text { or the set } \\ L_x=\{y \in A : x \leq y\} \text{ is finite set then A is a countable set }$$ I am adding my solution below

Answer 1

The set $A$ is not finite, so it will not be countable

Let $x_0 \in A$ and let $R_{x_0}=\{y \in A : y \leq x_0 \}$ be finite and $L_{x_0}=\{y \in A : x_0 \leq y \}$ for this element. Then $R_{x_0}=\{x_1,x_2,\ldots x_n\}$ such that $x_1,x_2,\ldots x_n \in A$ and $x_0 \leq x_i$ for every $i=1,2,\ldots n$. If $ x_i \leq y $ for $y \in A $ then $x_0 \leq y$ will be satisfied if $\{y \in A : x_i \leq y \} \subseteq \{y \in A : x_0 \leq y \}=L_{x_0}$ (for $i=1,2,\ldots , n$)

Hence it follows that $\bigcup_{i=1}^n \{y \in A : x_i \leq y \} = L_{x_0}$. For every $i=1,2,\ldots n $ $\{y \in A : x_i \leq y \} \subseteq L_{x_0}$, so $\bigcup_{i=1}^n \{y \in A : x_i \leq y \} \subseteq L_{x_0}$.

Let see $ L_{x_0} \subseteq \bigcup_{i=1}^n \{y \in A : x_i \leq y \}$. Conversely, let us assume the converse, in which case $\exists y_1 \in L_{x_0} $ such that $y_1 \notin \bigcup_{i=1}^n \{y \in A : x_i \leq y \} $.

If $y_1 \in L_{x_0}$ then $x_0\leq y_1$ and $y_1 \notin L_{x_0}$ then $x_0\leq y_1 \bigcup_{i=1}^n \{y \in A : x_i \leq y \}$ $x_i \leq y_1$ for every i. In this case $y_1 \in R_{x_0}$ but in this case a new element appears in our finite set, which is a contradiction. Similarly, the assumption that $L_{x_0}$ is finite will also lead to a contradiction. Then A is finite and therefore countable.