I am trying to solve the following exercise:
Show that a subset $A$ of a topological space $(X,\tau)$ is nowhere dense if and only if for each non empty open set $U$ in $X$, there exists a non empty open set $U_0$ such that $U_0$ $\subset U$ and $U_0$ $\cap$ $A=\phi.$
$A$ is nowhere dense $\implies$ $int\ cl(A)=\phi$.
Let $U\in \tau$ and $U \neq \phi$.We have to construct a nonempty open set $U_0$ which will serve the condition.
Now clearly, $X\setminus cl(A) \in \tau$ and $X\setminus cl(A) \neq \phi$. (Since $int\ cl(A)=\phi$)
Now we choose $U_0=U \cap X\setminus cl(A)\ (\neq \phi)$ and $U_0 \in \tau.$
clearly, $U_0$ $\subset U$ and $U_0$ $\cap$ $A=\phi.$ Hence we have done.
now , the problem occurs (to me..!!) in proving the converse part.
In this case I was trying:
Let the given condition holds.
To show, $int\ cl(A)=\phi$.
If possible let $int\ cl(A) \neq \phi$.
Then $U= int\ cl(A) \neq \phi$ Then there exists a $U_0 (\neq \phi) \in \tau$ such that $U_0 \cap A = \phi$. but here i stopped ...and couldn't show any contradiction....!!!