Q:

density inequality in metric space

I am given a metric space $(X,d)$ s.t.  $Y\subseteq X$ is dense in $X$. Moreover, I have $U$ an open set in $X$. Is the statement $d(U\cap Y)\leq d(U)$ is true?  It seems so, but I cannot seem to be able to prove it.  If not, does anybody have a counter example?  Thanks for the help. 
Notation: $d(K)$ is the smallest cardinality of a dense subset of $K$.

A:

In any metric (metrizable) space, the density and weight* cardinal functions are equal.  Furthermore, the weight cardinal function is monotone with respect to subspaces (for general topological spaces).  As any subspace of a metric space is metric, we have: $$d ( U \cap Y ) = w ( U \cap Y ) \leq w ( U ) = d ( U ).$$
*The weight $w(X)$ of a topological space $X$ is the smallest cardinality of a base for $X$.

