I had this question that I was struggling to come up with an example for (there may not be an example, in which case why?):
Give an example of a nonempty finite set which is neither open nor closed?
I can come up with examples for infinite sets, but I wasn't sure about finite sets. I know that a closed set can be infinite or finite (e.g. the integers), but what about open sets? Must they be always infinite?
Thanks in advance for all your help! I really appreciate it!
Best Answer
In most applications, topological spaces are assumed to be Hausdorff ($T_2$), i.e. that any two distinct points have disjoint neighbourhoods. This is true among others for metric spaces, like the real numbers, and it immediately implies the $T_1$ condition which says that for any pair of distinct points, each has a neighbourhood not containing the other, which is actually equivalent to saying that any singleton set is closed.
Now, finite unions of closed sets are closed, so if a topological space $X$ is $T_1$, then any finite set is closed. On the other hand, if a space is not $T_1$, then if you take the pair of points witnessing that, i.e. $x,y\in X$ such that for any open $U\ni y$ we have $x\in U$, then the set $\{x\}$ will not be closed, an in fact any set containing $x$ but not containing $y$, finite or not, will not be closed. In particular, any finite set which is not open and does not contain $y$ is neither open nor closed.
For example, any nonempty finite subset of an infinite space with trivial topology is neither open nor closed.
Note that even if a space is not $T_1$, it can still be the case that every finite set is either open or closed, for example if you take $X=\{x,y\}$ with open sets $\emptyset,X,\{x\}$, then $\{y\}$ is closed and all other subsets of $X$ are open.