A question on a dense subspace Suppose $Z$ is a topological space; and $X$ is dense in $Z$. Then do we have  $W(X)= W(Z)$?
Where $W(X)$, $W(Z)$ denote the weight of the $X$ and $Z$ respectively. 
What I've tried: On one hand, $W(X)\le W(Z)$, clearly; On the other hand, for any open set $U$ of $Z$, we have $U\cap X$, an open set in $X$, because $X$ is dense in $Z$. So $W(X)= W(Z)$. Am I right?
Thanks ahead.
 A: The weight of a space does not necessarily equal the weight of a dense subspace.  
As an example, note that $\mathbb N$ is clearly second-countable ($w(\mathbb{N}) = \aleph_0$), but its Stone–Čech compactification $\beta \mathbb{N}$ has weight $2^{\aleph_0}$.  This can be generalised for the Stone–Čech compactification of any infinite discrete space  (see, e.g., Engelking, General Topology, Theorem 3.6.11, pp.174-5).
A: Arthur’s example shows that $w(Z)$ can be as large as $2^{w(X)}$. If $Z$ is regular, this is the biggest possible value for $w(Z)$. In that case $Z$ has a base $\mathscr{R}$ of $w(Z)$ regular open sets, and if $U\subseteq Z$ is regular open, then $U=\operatorname{int}_Z\operatorname{cl}_Z(U\cap X)$, so each $U\in\mathscr{R}$ is uniquely determined by $U\cap X$. If $\mathscr{B}$ is a base for $X$ of cardinality $w(X)$, $U\cap X=\bigcup\mathscr{B}_U$ for some $\mathscr{B}_U\subseteq\mathscr{B}$, and $U$ is uniquely determined by $\mathscr{B}_U$. Thus, $w(Z)=|\mathscr{R}|\le 2^{|\mathscr{B}|}=2^{w(X)}$.
If $Z$ is not regular, however, $w(Z)$ can be much larger. Let $\kappa$ be any infinite cardinal, let $Z$ be a set of cardinality $\kappa$, and give $Z$ the cofinite topology; clearly $w(Z)=|Z|=\kappa$. Now let $X\subseteq Z$ with $|X|=\omega$. Then $X$ is dense in $Z$, and $w(X)=\omega$.
