0
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ $X$ itself is open and only disjoint from $\emptyset$. Hence, $T = \{X,\emptyset\}$. $\endgroup$
    – amsmath
    Commented Oct 22, 2019 at 21:48
  • $\begingroup$ 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. $\endgroup$ Commented Oct 22, 2019 at 23:21

1 Answer 1

2
$\begingroup$

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.

So your question is hopelessly flawed.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .