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?

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    Only is possible taking a coarser topology. – Martín-Blas Pérez Pinilla Feb 4 '15 at 7:51
  • extremely sorry to bother you I just wrote the wrong question@Martín-BlasPérezPinilla – Learnmore Feb 4 '15 at 9:56
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    If the space is empty or has only one point is impossible. If the space has at least two points, take simply $\tau^{'} =$ the discrete topology. – Martín-Blas Pérez Pinilla Feb 4 '15 at 10:01
  • thanks I got it @Martín-BlasPérezPinilla – Learnmore Feb 4 '15 at 10:14
up vote 5 down vote accepted

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.

If $(X,\tau)$ is disconnected, it means that $X=A\cup B$, where $A$ and $B$ are disjoint, each one open and closed. If $\tau'$ is a finer topology, then both $A$ and $B$ are still open and closed in $\tau'$, which makes $(X,\tau')$ disconnected. Thus, the above described situation is impossible.

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    Just to be clear, a finer topology has more open sets, and every set which is open in the current topology is open in the finer topology as well. In a sense, you are taking some (or all) of the open sets, and chopping them up. I believe Munkres uses the analogy of grinding gravel into smaller pieces. If you have a big rock, and you break it in half, you still have both parts to put the rock back together. – Alfred Yerger Feb 4 '15 at 7:52
  • Right, so using @AlfredYerger's metaphor, if you take two separate rocks and break them to smaller pieces, no way you will get one rock. For that you need to glue, rather than break. In topological terminology, you need to take a weaker topology, not a stronger (=finer) one. – Amitai Yuval Feb 4 '15 at 8:30
  • extremely sorry to bother I just wrote the wrong question@AmitaiYuval – Learnmore Feb 4 '15 at 9:56

Suppose we have a connected space $(X,\mathcal{T})$, where $X$ has more than 1 point. Then the topology $\mathcal{T}$ is not discrete (or else we'd have disconnected space), so there is some $D \subset X$ such that $D \notin \mathcal{T}$. Then let $\mathcal{T'}$ be the smallest topology that contains $\mathcal{T} \cup \{D, X \setminus D\}$ (i.e. the intersection of all topologies that has this as a subset). Then $\mathcal{T}'$ is finer than $\mathcal{T}$ and $X$ is disconnected in that topology, as $\{D, X \setminus D\}$ is a disconnection of $X$.

So any connected topology has lots of disconnected finer topologies.

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