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.

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).