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

Question

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

Answer 1

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.

Answer 2

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 half-open 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 $\kappa\geq 2^\omega$.

Answer 3

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), 116-121) 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 $\mu(X \cap (a, b)) = 0.5 (b-a)$, 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 $h(b) < 0.1$.

Answer 4

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 well-known 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 Golomb-like space, or vice versa?

Cardinality of a set of countable connected Hausdorff spaces

Continuous self-maps in the Golomb space that are neither increasing nor decreasing

Answer 5

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/1994-122-03/S0002-9939-1994-1287102-2/S0002-9939-1994-1287102-2.pdf. It's fairly clear from the construction that $\omega_1$ of the $U_{\alpha i}$'s (the sub-basic open sets) must be different.

Answer 6

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.)