# Give an example of a nonempty finite set which is neither open nor closed?

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!

• If you mean a set of real numbers, with the usual topology, i.e. the usual notion of which sets are open and closed, then every finite set is closed. If you're considering other topologies or other sets than the real numbers, it would help to say so. – Michael Hardy Nov 30 '13 at 16:28
• you have not specified the Space i.e., you have no specified the set and which topology you have on that... – user87543 Nov 30 '13 at 16:29
• @MichaelHardy Wouldn't the integers be an example of an infinite closed set seeing that it spans from negative infinity to infinity? – Billy Thorton Nov 30 '13 at 16:38

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.

• What about for usual topology? – Billy Thorton Nov 30 '13 at 16:46
• @BillyThorton: usual topology on what? Most "usual" topologies are $T_1$, if that's what you mean, even more exotic ones like Zariski topology. – tomasz Nov 30 '13 at 17:18
• $T_1$ topology I believe. (Truthfully I'm just learning topology, so my question is supposed to be introductory level) – Billy Thorton Nov 30 '13 at 17:24
• @BillyThorton: I've just written that any finite set in a $T_1$ space is closed. I don't understand the question. – tomasz Nov 30 '13 at 18:25
• @BillyThorton: and if you mean the usual topology on $\bf R$ or ${\bf R}^n$, then those two are metric spaces and therefore Hausdorff and $T_1$, so every finite set is closed. – tomasz Nov 30 '13 at 18:32

Referring to Michael Hardy's comment let $X = \{ a, b \}$ a two element set with only $\varnothing$ and $X$ open. Then $\{ a \}$ is neither open nor closed. Same with $\{ b \}$. Both sets are finite.

• I'm not sure I know what trivial topology is? – Billy Thorton Nov 30 '13 at 16:39
• @BillyThorton The trivial topology for a set $X$ is the smallest possible topology for the set. That is to say, only $\varnothing$ and $X$ are open. – Jay Nov 30 '13 at 16:44
• So then for usual topology it is impossible? – Billy Thorton Nov 30 '13 at 16:46
• Sets like $\mathbb{R}$ and spaces like metric spaces have usual topologies. Please refer to tomasz's fine answer for this situation. Using the word informally, a "random" set does not have a usual topology. – Jay Nov 30 '13 at 16:52