# Topological space with disjoint open sets

I have to prove the following property:

Let $$(X,T)$$ be a topological space , in which all open sets are disjoint . Then it is equivalent:

$$A \subset X$$ is dense then $$A$$ is an open set

$$A \subset X$$ , non empty , then $$T _{bd(A)}=T_{dis}$$, where $$T_{dis}$$ is the discrete topology in $$bd(A)$$

I don't know where to start from , because I don't see how I can use the conditions to prove anything. I would prefer some hints as I'm trying to learn.

• $X$ itself is open and only disjoint from $\emptyset$. Hence, $T = \{X,\emptyset\}$. Oct 22 '19 at 21:48
• Moreover, since the union of two open sets is open, even if $U$ and $V$ were disjoint non-empty open sets, neither would be disjoint from $U \cup V$. So I wonder if you mean there's a disjoint basis. Oct 22 '19 at 23:21

It is clear that $$T=\{\emptyset,X\}$$. If any other non-empty open set would exist in this topology, it would not be disjoint from $$X$$, contradicting the assumption.
But then your statement that $$A$$ is dense implies $$A$$ is open is false (unless $$|X|=1$$), any proper subset of $$X$$ is dense and non-open.
If $$A$$ is a proper non-empty subset of $$X$$ then clearly $$\text{bd}(A)=X$$ and for non-singleton $$X$$ again this does not have the discrete topology.