What happens if singleton set is not closed

Question

I'm reading Munkres and now is learning the separation axioms.

When he starts to discuss regularity and normality, he says "Suppose one-point sets are closed in $X$". Our Prof. also didn't explain much in the class.

So I'm quite curious what happens if singleton set is not closed. Will it be open in some cases or neither open nor closed?

Also I can't come up with an example of such space.

For $ x\in X$, $\overline{\lbrace x\rbrace}=\lbrace x\rbrace$ does not seem to depend on the topology or the set.

Any explanation will be appreciated.

Answer 1

Consider $X=\mathbb N$ and call $U\subseteq X$ open iff $U=\emptyset$ or $U=[n,\infty)=\{\,x\in \mathbb N\mid x\ge n\,\}$ for some $n$. In this topology, the singleton $\{1\}=X \setminus [2,\infty)$ is closed, whereas $\{2\}$ is not closed (and also not open); the closure of $\{2\}$ is $\{1,2\}$.

Answer 2

If a singleton $\{x\}$ is not closed, what it really means is that the constant net $\langle x\rangle$ has more than one limit point. Regardless as to $\{x\}$ being open or not.

It could, for example, be dense, in which case every point will be a limit point. For example if you consider $(X,\tau)$ where $\tau=\{X,\varnothing\}$. If $X$ is not a singleton itself, then no single point is closed, since $X\setminus\{x\}$ is never open. But this means that if $x,y\in X$ are any two points, then for every open neighborhood of $y$, we can find $x$ as well. So $\overline{\{x\}}=X$.

You can note that in the topology $\{X,\varnothing,\{x\}\}$, the singleton $\{x\}$ is open and dense; and if $y\neq x$ then $\overline{\{y\}}=X\setminus\{x\}$. So singletons which are open and "just not-closed" can be "quite dense".

So if singletons are not closed, $(X,\tau)$ is not a Hausdorff space, so it is not a metric space, and so on. But we can even say more, it's not a $T_1$ space, since being $T_1$ is the same as the singletons being closed.

Answer 3

A nice counterexample is the Sierpiński space $S=\{0,1\}$ with the topology $\big\{\varnothing,\{1\},\{0,1\}\big\}$: in this space the set $\{0\}$ is closed, but the set $\{1\}$ is not: the only open nbhd of $0$ is $\{0,1\}$, which contains $1$, so $0\in\operatorname{cl}\{1\}$.

A space $X$ with the property that $\{x\}$ is closed for every $x\in X$ is said to be a $T_1$-space, though the $T_1$ property is usually not defined this way.

Definition. A space $X$ is $T_1$ if whenever $x$ and $y$ are distinct points of $X$, there are open sets $U$ and $V$ such that $x\in U$ and $y\notin U$, and $y\in V$ and $x\notin V$.

A good (and quite easy) exercise is to prove that this is equivalent to the statement that $\{x\}$ is closed for each $x\in X$.

Answer 4

Consider $X=\left\{a,b,c\right\}$. We are going to define a topology on $X$ by defining all open sets. Put $\tau=\left\{\emptyset, X, \left\{a,b\right\},\left\{c\right\}\right\}$. Check that $\tau$ indeed determines a topology. Notice that the point $\left\{a\right\}$ is not open nor closed.

Answer 5

One example is the indiscrete space, where only the whole space and the empty set are open (and hence closed) - so, if there is more than one point, singletons would not be closed. This space has the property that all pairs of points $x,y$ are topologically indistinguishable - meaning if $S$ is an open set then $x\in S$ if and only if $y\in S$ - that is, the topology gives no information to tell them apart, which makes it somewhat trivial - however, one can find spaces where the points are topologically distinguishable, but where singletons are not closed (and plenty of the other answers do this, so I won't provide additional examples).

I doubt you'd have many familiar examples of such topologies if you haven't been introduced to more "pathological" topologies that like to be counterexamples to every statement you want. One should note that the condition that singletons are closed is (equivalent to) a separation axiom making a space a $T_1$ space, which is a fairly weak separation axiom.

Answer 6

In many fields of mathematics, most of the spaces in consideration are such that singletons are indeed closed subsets.

However, there are also theories for which the previous statement is false. For example, in algebraic geometry one of the basic concepts is a scheme, and the topology of a general scheme is such that singletons are not necessarily closed.

Answer 7

First a bit of terminology: a T1 space is a space where points are closed. Now, the simplest example of a space that is not T1 is just a set with more than one element with the trivial topology (the only open sets are the whole set and the empty set). In this case, the closure of a singleton space is the whole space. You can find more elaborate examples in the wiki page: http://en.wikipedia.org/wiki/T1_space

Though not precisely related to your question, note that the interesting distinction is between T1 spaces and Hausdorff spaces, not T1 vs not T1. T1 but not Hausdorff show up often enough, algebraic varieties with the Zarisky topology being prime examples.

Answer 8

According to the usual definitions of the concept of separation in topology. All points are closed if and only if your topological space is T1 separated or a T1 space.

It might happen that you have a T0 space which is not T1. The following example comes from the Steen-Seebach's counter examples in topology (by the way, if you are interested by pathological spaces, you want to read this book) :

Take $X:=[-1,1]$. Its open sets are : $[-1,b)$ for $b>0$ and $(a,1]$ for $a<0$ and $(a,b)$. This is a topology.

You can see that the point $0$ won't be closed so is not T1. Nevertheless it is easily seen to be T0.

Answer 9

If a space is not $T_1$, points needn't be closed. But of course such a space cannot be a regular or normal one.

Answer 10

If your topological space comes from a metric space then all points are closed (proof: fix a point $P$. Now given another point that is distance $d>0$ from your point, consider the open ball centre that other point and radius $d$; it's an open set that doesn't contain $P$ but does contain the other one; now take the union).

But in a general topological space, points may not be closed, or open. Here is a silly example: let $X$ be any set (of size bigger than 1) and then just declare that the empty set and $X$ are open, and nothing else is. That's a topological space and no points are open and no points are closed.

Whether or not you think such examples are "weird" depends very much on what kind of mathematics you do. For example 99 percent of the topological spaces studied by people thinking about Lie groups might well have the property that all points are closed. However algebraic geometers work every day with topological spaces which have non-closed points -- for example on an algebraic variety like the affine plane, if you're using Grothendieck's foundations, then there will be a non-closed point corresponding to every reduced irreducible algebraic curve in that plane, and also a generic point whose closure is the entire plane.

Here is a simple example with an open point: take a set $X$ with two points, $s$ and $\eta$ (this is the traditional notation) and let the open sets be the empty set, the whole of $X$, and $\{\eta\}$. That is, $\{s\}$ is the only subset of $X$ that isn't open. Then $s$ is closed, $\eta$ is open, and the closure of $\eta$ is all of $X$ so it contains $s$. This is how an algebraic geometer models an infinitesimal neighbourhood of a point: you can think of $X$ as a small open disc, then $s$ is a point in $X$ and $\eta$ is all the other points put together.