# Closed points in the finite point exclusion topology and the general notion of open sets?

I'm struggling with grasping the notion of open sets as they apply to topological spaces. For example, if we consider the finite point exclusion topology, $T$, defined by $$T = \{ A \subseteq X \mid p \notin A \ \text{or} \ A = X\}$$any singleton set of the form $\{x \mid x\in X\}$ belongs in $T$, so it is open. However, this set is obviously closed in $X$. How does one distinguish?

What got me thinking about this was a problem that asked what the closed sets in the exclude point topology are. Clearly all singleton sets and their finite unions, including $X$ itself, are closed sets in $T$. But those sets are also open? I don't understand.

• In a general topological space sets may be open and closed. This does not correspond to our intuitive understanding of open and closed. – TheGeekGreek Jun 18 '17 at 16:12
• @TheGeekGreek I apologize if my questions seem rather simple, as I've just started learning the subject. Does this mean that a set can be closed in $X$ while still being "open" in a topology on $T$? – user312437 Jun 18 '17 at 16:13
• @JoséCarlosSantos I think I understand. A set $A \subset (X,T)$ is open implies $A\in T$? – user312437 Jun 18 '17 at 16:25
• @CodyButler No. A set $A\subset X$ being open is equivalent (by definition) to $A\in T$. – José Carlos Santos Jun 18 '17 at 16:27
• @CodyButler That's right! – José Carlos Santos Jun 18 '17 at 16:32

In general a set can be both open and closed. In any topology on $X$ there are at least two examples of such sets, that is $\emptyset$ and $X$ itself. If there are more subsets of the topological space $(X,T)$ that are both open and closed, we say that the topological space $(X,T)$ is disconnected.

But this isn't the case in your question. Note that for the topology $T = \{A \subset X \mid p \not \in A \text{ or } A =X\}$, a subset $A \subset X$ is open if and only if $A$ doesn't contain $p$ or is equal to $X$. A set $B \subset X$ is closed if and only if $B^c$ is open. So $B$ is closed if and only if it contains $p$ or is equal to the empty set. Thus the only sets that are both open and closed are the empty set and $X$ itself.

• Very clear. Thank you – user312437 Jun 18 '17 at 16:33
• You're welcome. – Demophilus Jun 18 '17 at 16:34

Your intuition from the topology of $\Bbb R^n$ and metric spaces doesn't always transfer that well to general topological spaces, which can be quite strange. For example, in an arbitrary topological space, singletons need not be closed. (FYI, a space where every singleton set is closed is called a $T_1$-space, and there are certainly examples of non-$T_1$ spaces.)

Let's look at your example. Recall that by definition, a subset $A$ of a topological space $(X,T)$ is open if $A\in T$. The notion of closed-ness is also completely dependent on $T$: a subset $A$ is closed in $(X,T)$ if its complement is open (that is, $X\setminus A\in T$). Say $x\in X$. If $x\neq p$, the set $\{x\}$ is open: $p\not\in\{x\}$. In this case, $\{x\}$ is not closed: $p\not\in\{x\}$, which implies that $p\in X\setminus\{x\}$, so $X\setminus\{x\}\not\in T$. If $x = p$, $\{p\}$ is closed: $p\in\{p\}$, so that $p\not\in X\setminus\{p\}$, which means $X\setminus\{p\}$ is open. We also see that $\{p\}$ is not open, unless $X = \{p\}$: $p\in\{p\}$, so $\{p\}\not\in T$ unless $\{p\} = X$.

One can also find examples of topologies where singletons may be both open and closed (like the discrete topology on a space), either open or closed (like your example), or neither open nor closed (the generic point is neither open nor closed in the Zariski topology on $\operatorname{Spec}\Bbb Z$).

• Very detailed and clear. Thank you – user312437 Jun 18 '17 at 16:39