If $(X,\tau)$ has more than $1$ point and is $T_2$ and connected, do we necessarily have $X =\tau$?

$\begingroup$ Thanks for all these interesting examples, I am looking forward to studying them all in detail! $\endgroup$ – Dominic van der Zypen May 21 '18 at 17:13

1$\begingroup$ A metrizable one: (if I'm correct it seems to me that all the examples given so far are not). If $I$ is any set of cardinal $\alpha\ge c$ fix any connected graph structure on $I$ (so it has $\alpha$ edges), and $X$ the corresponding 1skeleton, with its geodesic metric. Then $X$ has cardinal $\alpha$ and has a discrete subset of cardinal $\alpha$ whose points have pairwise distance $\ge 1$, so the open $1/2$ball around each such subset makes a different open subset and there are $2^\alpha$ open subsets. (Note: clearly a metrizable connected space with $\ge 2$ points has cardinal $\ge c$.) $\endgroup$ – YCor May 8 at 7:01

$\begingroup$ Thank you for this example  it would definitely merit a full answer (which I am of course happy to upvote) $\endgroup$ – Dominic van der Zypen May 9 at 9:22
Consider the topology on $\mathbb{R}^2$ generated by subsets that are open in some line from the origin. This topological space is connected, has the cardinality of continuum, and has $2^c$ open subsets.

$\begingroup$ (any $\mathbb{R}$ linear space in place of $\mathbb{R}^2$ would also work, for examples of higher cardinality) $\endgroup$ – Pietro Majer May 21 '18 at 14:32

3$\begingroup$ This example is essentially like the star graph, with $\kappa$ many edges joined at a single point. There is a similarity with the long line example: with the long line, you have $\kappa$ many unit intervals joined endtoend, but with the star graph, you join them only on one side, letting them otherwise float freely from one another. $\endgroup$ – Joel David Hamkins May 21 '18 at 17:24

1

The $\frak{c}$long line is $T_2$, connected and size continuum $\frak{c}$, but has $2^{\frak{c}}$ many open sets, since there is a size continuum discrete subset.
The more familiar $\omega_1$long line is $T_2$, connected (even path connected, also locally connected), and size $2^\omega$. There are at least $2^{\omega_1}$ many open sets, since there is a size $\omega_1$ discrete family (such as the centers of the halfopen intervals used to construct the long line), and so you can place intervals around each of them as you like, making $2^{\omega_1}$ many distinct open sets.
If $2^\omega<2^{\omega_1}$, for example, if CH holds (but that hypothesis is weaker than CH), then the ordinary long line itself is an example. But in any case, the $\kappa$long line is an example for every cardinal $\kappa\geq 2^\omega$.

1$\begingroup$ The completed long line, with the point at $\omega_1$, would be compact Hausdorrf and connected, size $2^\omega$, but with $2^{\omega_1}$ many open sets. $\endgroup$ – Joel David Hamkins May 21 '18 at 14:15
The density topology $\tau$ on $\mathbb{R}$ is also a natural counterexample. It consists of all (Lebesgue) measurable $X \subseteq \mathbb{R}$ such that every point $x \in X$ is a density one point of $X$ which means: Whenever $\{I_n: n \geq 1\}$ is a sequence of open intervals containing $x$ whose lengths decrease to $0$, $$\lim_{n \to \infty} \frac{\mu(X \cap I_n)}{\mu(I_n)} = 1$$
It is clear that $\tau$ is a ccc topology on $\mathbb{R}$ that extends the usual topology. It is also clear that every conull set is in $\tau$ so that $\tau = 2^{\mathfrak{c}}$.
Claim: (Theorem 3 in C. Goffman, D. Waterman, Approximately continuous transformations, Proc. Amer. Math. Soc. 12 (1961), 116121) Every interval $I \subseteq \mathbb{R}$ is $\tau$connected.
Proof: Towards a contradiction, suppose $X, Y$ are non empty members of $\tau$ that partition $I$. It suffices to construct a nested sequence $\{ (a_n, b_n): n \geq 1\}$ of intervals $(a_n, b_n) \subseteq I$ such that $a_n < a_{n+1} < b_{n+1} < b_{n}$, $b_n  a_n \to 0$ and $\mu(X \cap (a_n, b_n)) = 0.5 (b_n  a_n)$. Since then $a = \lim a_n \in I$ cannot be a density one point of either one of the sets $X, Y$.
To construct such a sequence of intervals, use the following facts.
(1) If $a < b$ are in $I$ and $\mu(X \cap (a, b)) = 0.5 (ba)$, then both $X, Y$ meet $(a, b)$.
(2) If $a < b$, $a \in X$ and $b \in Y$, then for all sufficiently small $r > 0$, there exists $a < c < d < b$ such that $d  c = r$ and $\mu(X \cap (c, d)) = 0.5 (d  c)$.
(1) is trivial and (2) holds because for all sufficiently small $r > 0$, the functions $h:[a, b] \to \mathbb{R}$ defined by $$h(x) = \frac{\mu((x  r, x + r) \cap X)}{2r}$$ is continuous and satisfies $h(a) > 0.9$ and $h(b) < 0.1$.

$\begingroup$ Very nice explanation  thanks Ashutosh! $\endgroup$ – Dominic van der Zypen May 21 '18 at 17:12

$\begingroup$ No problem. The following may be interesting: (1) Does the density topology on $\mathbb{R}^n$ (for $n \geq 2$) preserve old connected sets? (2) If not, is there another topoogy $\tau$ on $\mathbb{R}^n$ ($n \geq 2$) that refines the usual topology, has size $2^{\mathfrak{c}}$ and preserves old connected sets. $\endgroup$ – Ashutosh May 21 '18 at 17:23
The Golomb space $(\mathbb N,\tau)$ (a "universal" counterexample to many questions), also gives a counterexample with $\mathbb N=\aleph_0$ and $\tau=\mathfrak c$.
I recall that the Golomb space is the set $\mathbb N$ of natural numbers endowed with the topology $\tau$ generated by the base consisting of the arithmetic progressions $a+\mathbb N_0b=\{a+nb:n\ge 0\}$ where $a,b$ are relatively prime.
It is wellknown that the Golomb space is connected and Hausdorff. Since it contains a countable disjoint family of open sets (like any infinite Hausdorff space), its topology has cardinality $\mathfrak c\le\tau\le\mathcal P(\mathbb N)=\mathfrak c$.
In place of the Golomb space one can take any other countable Hausdorff connected space.
Such spaces have appeared in other questions of Dominic van der Zypen:
Is there a connected $T_2$topology on $\mathbb{Q}$ that is coarser than the Euclidean one?
Is $\mathbb{Q}$ the continous image of a Golomblike space, or vice versa?
Cardinality of a set of countable connected Hausdorff spaces
Continuous selfmaps in the Golomb space that are neither increasing nor decreasing
This is probably overkill, but assuming the Continuum Hypothesis there is a connected Hausdorff space $X$ such that
 $X=\omega$; and
 $\tau=\omega_1$.
It is Example 2.1 in this paper. http://www.ams.org/journals/proc/199412203/S00029939199412871022/S00029939199412871022.pdf. It's fairly clear from the construction that $\omega_1$ of the $U_{\alpha i}$'s (the subbasic open sets) must be different.
Here are:
1) a metrizable example.
Namely, consider a complete graph on $\alpha\ge\mathfrak{c}$. Its 1skeleton, 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.)