Definition. Let $X$ denote a set. Whenever $\tau$ is a topology on $X$, write $\tilde{\tau}$ for the collection of subsets of $X$ that are connected from the viewpoint of the space $(X,\tau)$.

In general, we cannot recover $\tau$ from $\tilde{\tau}$. For instance, let $\tau$ denote the usual topology on $\mathbb{Q}$, and $\sigma$ denote the discrete topology on $\mathbb{Q}$. Then $\tilde{\tau} = \tilde{\sigma}$, but $\tau \not = \sigma$.

I'm wondering whether we can use local connectedness to get rid of this problem.

Question. Let $X$ denote a set. Suppose $\tau$ and $\sigma$ are topologies on $X.$ Suppose $(X,\tau)$ and $(X,\sigma)$ are locally connected spaces. From $\tilde{\tau} = \tilde{\sigma}$, can we deduce $\tau = \sigma$?

If not, can we replace local connectedness with something else, so that from $\tilde{\tau} = \tilde{\sigma}$, we can deduce $\tau = \sigma$?

  • $\begingroup$ At the very least, you should assume that both $\tau$ and $\sigma$ are Hausdorff, otherwise there are easy examples. $\endgroup$ – Moishe Kohan Apr 30 '16 at 13:47
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    $\begingroup$ @studiosus Perhaps I'm a bit dense today but I don't see them. Which examples do you have in mind? $\endgroup$ – Marc Paul Apr 30 '16 at 14:28
  • $\begingroup$ This already fails for $X=\{ a,b\}$: the topology $\{a\}, \emptyset, X$ makes all sets connected, as does the trivial topology $\emptyset, X$. $\endgroup$ – user138530 Apr 30 '16 at 19:42
  • $\begingroup$ @MarcPaul: See my answer. $\endgroup$ – Moishe Kohan May 1 '16 at 4:36

The answer to your first question is NO.

$T_0$ and $T_1$ (see my comment below @studiosus's answer) examples have been given.

$T_2+regular$ example: Let $X=[0,1]\times[0,1)$. We define topologies $\sigma$ and $\tau$ on $X$ by specifying a basis for each. $\\$

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$\sigma$: Neighborhood of a point with positive $y$-coordinate is a vertical open interval. Neighborhood of a point with $y$-coordinate $0$ is a Euclidean ball.

$\tau$: Neighborhood of a point with positive $y$-coordinate is same as in $\sigma$. Neighborhood of a point $(x,0)$ is a vertical open tube around the point, minus $\{x\}\times [r,1)$ for some $r>0$.

$(X,\sigma)$ and $(X,\tau)$ are easily seen to be Hausdorff (and more) and locally connected (all of the basic open sets are connected).

$\tau\neq \sigma$ obviously.

$\tilde \tau=\tilde \sigma$: $\tilde \sigma \subseteq \tilde \tau$ because $\tau\subseteq \sigma$. Now let $C\in \tilde \tau$. If $C$ is contained in a vertical line then $C\in \tilde \sigma$. Otherwise, $C$ must be closed downward, i.e., if $(x,y)\in C$ then $(x,z)\in C$ for all $z<y$, and also $C\cap [0,1]\times \{0\}$ must be connected. Again $C\in \tilde \sigma$.

Note: These spaces are neither second countable nor separable, but they are probably metrizable, or very close anyway.

Can we replace local connectedness with something else, so that from $\tilde{\tau} = \tilde{\sigma}$, we can deduce $\tau = \sigma$?

I suggest replacing locally connected with locally connected+metric+compact Hausdorff (if you have fewer open sets, then a connected basis means more).

  • $\begingroup$ ..... Very nice! $\endgroup$ – Moishe Kohan May 2 '16 at 7:05
  • $\begingroup$ @studiosus thank you! I started thinking about strange locally connected spaces, and the Alexandroff square came to mind. These examples are sort of based on that. $\endgroup$ – Forever Mozart May 2 '16 at 7:18
  • $\begingroup$ @studiosus sorry I meant the "wheel without the hub". The Alexandroff square is not locally connected. $\endgroup$ – Forever Mozart May 2 '16 at 7:22

Here is the explanation of my earlier comment.

Given a topological space $(X,\tau)$ and a point $x'\notin X$, we define the set $X'=X\cup \{x'\}$ and the topology $\tau'$ on $X'$ as follows:

Elements of $\tau'$ are the empty set and all the subsets of the form $U\cup \{x'\}$, $U\in \tau$. It is then easy to see that $(X',\tau')$ is locally connected: If $V$ is any nonempty element of $\tau'$, it contains the point $x'$ and, hence, is connected.

Now, let us take your example: $X={\mathbb Q}$, $\tau_1$ is the discrete topology on $X$ and $\tau_2$ is the standard topology on $X$. As above, we define the topological spaces $(X', \tau_i'), i=1, 2$, $X'=X\cup \{x'\}$. Both are locally connected. The connected nonempty subsets of both $(X', \tau_1'), (X',\tau_2')$ are exactly the singletons in $X$ as well as all the subsets of the form $Z\cup \{x'\}$, $Z\subset X$ arbitrary.

Thus, you obtain two locally connected topologies on the same set which share the same collection of connected subsets.

I do not particularly like this example since it is not Hausdorff.

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    $\begingroup$ Your topologies are $T_0$ only. But note that you can easily get them to be $T_1$ (and still locally connected) by adding $\omega$ with the cofinite topology instead of just one point $x'$. So your open sets would be of the form $U\cup (\omega\setminus F)$. Connected sets would be the same for both topologies generated like this. $\endgroup$ – Forever Mozart May 1 '16 at 4:46
  • $\begingroup$ @ForeverMozart: Yes, very true, but I would prefer Hausdorff examples. $\endgroup$ – Moishe Kohan May 1 '16 at 4:49

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