TITLE: If $(X,\tau)$ has more than $1$ point and is $T_2$ and connected, do we have $|X| =|\tau|$?
QUESTION [6 upvotes]: If $(X,\tau)$ has more than $1$ point and is $T_2$ and connected, do we necessarily have $|X| =|\tau|$?

REPLY [4 votes]: Here are:
1) a metrizable example.
Namely, consider a complete graph on $\alpha\ge\mathfrak{c}$. Its 1-skeleton, endowed with the geodesic metric with edges of length one, has cardinal $\alpha$ and has $2^{\alpha}$ open subsets (since for any subset of the set of vertices, the open ball of radius $1/2$ around this subset determines this subset.
2) a separable example. 
Namely, to fix ideas choose $S=\mathbf{R}/\mathbf{Z}$. Start from $X=S^{\mathfrak{c}}$, which is separable. Choose $D$ a countable dense subgroup (e.g., generated by any dense countable subset), and $F=S^{(\mathfrak{c})}$, the finitely supported subgroup. Then the topological group $G=FD$ is a dense subgroup, is connected because it has the dense connected subgroup $F$, and is separable because it has the dense countable subset $D$ (beware that $F$ is not separable), and $|G|=\mathfrak{c}$. Using suitable inverse images of projections, we see that $G$ has $2^{\mathfrak{c}}$ open subsets.
(Of course, a separable metrizable connected space of cardinal $\ge 2$ has cardinal $\mathfrak{c}$ and the same number of open subsets.)