Proposition 2

Let $Y$ be a subspace of a topological space $X$, and let $A$ be a subset of $Y$. If $A$ is nowhere dense relative to $Y$, then $A$ is nowhere dense relative to $X$. Conversely, if $Y$ is open in $X$ and $A$ is nowhere dense relative to $X$, then $A$ is nowhere dense relative to $Y$.

Suppose that $A$ is nowhere dense relative to $Y$. If the closure $\overline{A}$ of $A$ in $X$ contains a non-empty open set $U$, then $U \cap A$ is not empty (by the definition of closure); hence $U \cap Y$ is a non-empty open set relative to $Y$, and is contained in the closure $\overline{A} \cap Y$ of $A$ with respect to $Y$, which is contrary to the hypothesis.

Now suppose that $Y$ is open in $X$ and that $A \subset Y$ is nowhere dense relative to $X$. If $U$ is open in $Y$ and is not empty, then $U$ is open in $X$ and therefore contains a non-empty set $V$ which is open in $X$ (and a fortiori in $Y$) and does not meet $A$; hence $A$ is nowhere dense relative to $Y$.

The second part of Proposition 2 is clearly not valid if $Y$ is not open in $X$; consider, e.g., the situation where $Y \neq \emptyset$ is nowhere dense in $X$, and $A = Y$.