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 $U_0$ $\cap$ $A=\phi.$

My Attempt:

(for "$\implies$")
$A$ 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....!!!
please help....!

Thank You...!!


HINT: Show that if $U_0\cap A=\varnothing$, then $U_0\cap\operatorname{cl}A=\varnothing$; do you see why this is a contradiction?

  • $\begingroup$ sir, I cannot understand how to prove: $U_0 \cap A=\phi \implies U_0 \cap cl(A)= \phi$.... $\endgroup$ – Indrajit Ghosh Jan 10 '17 at 17:30
  • $\begingroup$ @Indrajit: Every point of $U_0$ has an open nbhd (namely, $U_0$) disjoint from $A$, so no point of $U_0$ is a limit point of $A$. $\endgroup$ – Brian M. Scott Jan 10 '17 at 17:33
  • $\begingroup$ oho....yes...i got it now....thank you very much sir.. $\endgroup$ – Indrajit Ghosh Jan 10 '17 at 17:36
  • $\begingroup$ @Indrajit: You’re very welcome. $\endgroup$ – Brian M. Scott Jan 10 '17 at 17:36

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