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I am not understanding this def. of Nowhere dense set in a metric space. Though I know another form of def. which states that a subset A of Metric Space M is nowhere dense iff, Int(Clo A)= Null set.

Shivam Kumar
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  • $\operatorname{int}(\operatorname{cl}(A)) = \varnothing$ means $\operatorname{cl}(A)$ contains no nonempty open set, and $A$ is a subset of $\operatorname{cl}(A)$... – arkeet Jun 21 '18 at 02:12
  • Since Int(Clo A) = Null Set, implies Clo(A) contains no interior points,i.e. there is no point x in Clo(A) such that there exists an open sphere S(x,r) centred at x and contained in Clo(A) . Sir, I am not getting how to proceed further. – Shivam Kumar Jun 21 '18 at 02:17
  • You almost have a proof by contradiction in your comment. Assume A contains an open set of M and see where that leads you. – Steve B Jun 21 '18 at 03:29
  • Ok! sorry I have edited my question I have used Clo(A) instead of A – Shivam Kumar Jun 21 '18 at 03:50

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Argue by contradiction. Assume that $A$ contains a non-empty open set of $M$, call it $U \subset A$. Let $x \in U$ be any point. Clearly, $x$ is an interior point of $U$, so there exists an open sphere $S(x,r)$ centered at $x$ and contained in $U$.

Since $U$ is contained in $A$, $S(x,r)$ is contained in $A$.

Since $A$ is contained in $\mbox{cl } A$, $S(x,r)$ is contained in $\mbox{cl } A$. We get that there is a point $x \in U \subset \mbox{cl } A$ such that $S(x,r)$ is contained in $\mbox{cl } A$.

Then, $x$ is an interior point of $\mbox{cl } A$, a contradiction as the interior of the closure is given to be empty.

Sarvesh Ravichandran Iyer
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