If A is nowhere dense in M, and if G is a nonempty open set in M, prove that A is nowhere dense in G.
I tried by contradiction but could not figure it out. I found that we can use following result Given a subset S of a topological space Z, the following are equivalent:
S is nowhere dense in Z. For every non-empty open V⊆Z, there is a non-empty open W⊆V such that S∩W=∅. how to prove this result and apply in our context.
Use the equivalence at the end of your question and the fact that if $V\subseteq G$ is open in $G$, then $V$ is also open in $M$, since $G$ is open in $M$.
Let $V$ be a non-empty open subset of $G$. $G$ is open in $M$, so $V$ is open in $M$. $A$ is nowhere dense in $M$, so there is a non-empty open $W\subseteq V$ such that $A\cap W=\varnothing$. Thus, every non-empty open subset of $G$ has a non-empty open subset disjoint from $A$, so $A$ is nowhere dense in $G$.