How to find an example of a topological space $(X,\tau)$ such that it is connected but when we consider a finer topology $\tau^{'}$on the same set $X$,then we get $(X,\tau ^{'})$ to be disconnected?
[Math] A finer topology making a connected set disconnected
connectednessgeneral-topology
Related Solutions
This is too long to be a comment and is currently only a partial answer.
Following the idea that the existence of an open, totally disconnected set should relate to the connectedness of the space, I read about ways a space can be connected. A locally connected space is one where every point in a open set has a connected, open neighborhood and felt most relevant. I wasn't able to show that being locally connected was a sufficient condition for the lack of an open, totally disconnected set. So, I ended up adding connected back as a requirement (and T1 ended up being helpful during the proof) which led to,
Let $(X,\tau)$ be a locally connected, connected, T1 space that is not the topology on a one point set. Then, $(X, \tau)$ is not weakly disconnected.
Proof:
The proof strategy will be to use proof by contrapositive. Let $(X, \tau)$ be a weakly disconnected space. Let $U$ be an open, non-empty totally disconnected set. Let $(U, \tau_{U})$ be the subspace topology of $U$. As $U$ is totally disconnected and non-empty, its connected components are singletons. Here, we'll split the proof into two cases. Either all of its connected components are open or they there exists a connected component that is not open.
If there exists a connected component that is not open then its connected components are not all open which means it is not locally connected. As locally connected is hereditary for open subspaces, $(X,\tau)$, can not be locally connected.
If there exists a connected component that is open, call that component $A$. As $U$ is totally disconnected, $A$ must be a singleton. Since $U$ is open, any open set in $U$ is also open in $X$. This means $A$ is an open singleton in $X$. As $X$ is a T1 space, singletons are closed making $A$ a clopen set. Since $X$ is not a one point set, $A$ is non-trivial clopen set. This means $(X,\tau)$ is not connected.
In either case, $(X, \tau)$ can not have the three properties of being locally connected, connected, and T1. This completes the proof.
The issue is I'm not sure if those three properties are a necessary condition, so I still lack a nice way of describing all spaces that lack an open, totally disconnected set. While my current proof strategy uses the fact the space is T1, I'm skeptical that there isn't an alternate way to drop that condition.
Hints:
- An image of a connected set via continuous mapping is connected.
- A nonempty subset of a discrete space is connected if and only if it is a singleton (i.e. contains exactly one point).
Best Answer
As Martín-Blas Pérez Pinilla points out in a comment, there is a very easy example. Suppose $\langle X, \tau\rangle$ is connected. Then $\tau$ is a subset of the power set of $X$ by definition, and the power set of $X$ is a topology for $X$, called the discrete topology. It is easy to show that if $X$ is given the discrete topology it is disconnected unless it contains fewer than two points.
I would like to add that there is an important pattern of reasoning here. The question asks you to show that $X$ can become disconnected if we make its topology finer. It should be clear that while making the topology finer can turn a connected space into a disconnected space, the opposite never happens: you cannot turn a disconnected space into a connected space by making the topology finer. This is because $X$ is disconnected if we can find a partition of $X$ into open sets, and if these sets are open in one topology, they are open in a finer topology because this is the definition of “finer”.
So the thing to try is the finest possible topology, and if that topology does not make $X$ disconnected, then nothing can. The finest possible topology is the discrete topology, and that does solve the problem.