An equivalence condition on chain and open subsets

Question

Let $U $ be an open subset of a topological space $X $, are the following equivalent:

1) $U$ is dense;

2) Every chain $A_1\subseteq A_2\subseteq ... $ of open subsets of $X $ stops if the chain $A_1 \cap U \subseteq A_2 \cap U \subseteq ... $ stops (not nessesarilly at the same step).

If is is not true what conditions can be replace by 1)?

Answer 1

It is not equivalent.
As a counterexample, let $X = \{ u_n : n \in \mathbb{N} \} \cup \{v\}$, and let the topology be generated by the family of sets (basis) $$\{u_1\}, \{u_1,u_2\}, \{u_1,u_2,u_3\}, \ldots, \{u_1,v\}.$$ Now, let $U = \{u_1,v\}$. It's clear that $U$ intersects every non-empty open set, and so it's dense.
Let $A_n = \{u_1, \ldots, a_n\}$. Then $A_n \cap U = A_1$, so that the chain $$A_1 \cap U \subseteq A_2 \cap U \subseteq \cdots$$ stops.
However, the chain $A_1 \subseteq A_2 \subseteq A_3 \subseteq \cdots$ doesn't stop, at least if the stopping of the chain is what E.Rostami claims in his/her comment.

Answer 2

A space is called Noetherian when every decreasing sequence of closed sets eventually stops. This is the same condition (take complements) as your chain condition for open sets. Your condition 2 says that $X$ is Noetherian iff $U$ is Noetherian in the subspace topology. But $X$ Noetherian already implies that all subspaces are Noetherian (it's equivalent to hereditarily compact).