# Is the Opposite of the Open Closed in Topology? - On

Following is the topological proof of "infinitude of primes"

If you see above proof, it first defines its own toplogy and comments

" This topology has two notable properties... 1. the complement of a finite set cannot be a closed set."

I understood the complement of a finite set on $\Bbb Z$ results in infinite set, thus it could be possibly become open if it meets its topology-definition. However, why it suddenly says "cannot be a closed set"?

It looks like I am confusing something. Help me to figure out where I wrongly misunderstanding.

• The wikipedia page is wrong. In this sentence "finite set" should be further qualified as "nonempty finite set". – Lord Shark the Unknown Mar 10 '18 at 8:00

What it says is that the complement of a finite (non-empty) set cannot be closed. This is so because, for this topology, a non-empty finite set cannot be open, since every non-empty open set contains an infinte sequence.

• but cannot be open doesn't mean that it's closed. isn't it? – delinco Mar 10 '18 at 8:01
• @delinco No. Sets are not doors. A set can be both closed and open, it can be both non-closed and non-open, it can be closed and non-open and it can be non-closed and open. – José Carlos Santos Mar 10 '18 at 8:04
• Still confusing how the fact "nonempty finite set cannot be open" derives it cannot be closed. – delinco Mar 10 '18 at 8:15
• @delinco What is derived is that the complement is not closed. That follows from the definition of closed set: a set is closed if and only if its complement is open. – José Carlos Santos Mar 10 '18 at 8:28

Start by a "a finite set $F$ cannot be open". This is true if $F \neq \emptyset$ (the empty set is finite and is explicitly mentioned as being open, as it should be in any topology), because if we have $x \in F$ we have a whole infinite sequence $x \in S(a,b) \subseteq F$, which cannot happen if $F$ is finite.

A set $C \subseteq \mathbb{Z}$ is closed by definition if $\mathbb{Z}\setminus C$ is open. So if $\mathbb{Z}\setminus C$ is finite and $\mathbb{Z} \neq C$ (i.e. the complement of $C$ is non-empty and finite), then the complement of $C$ cannot be open, and so $C$ cannot be closed.

• I don't know that definition of closed constricted via complentary. Is that closed definition unique in this problem only or universal to the general topology? – delinco Mar 10 '18 at 8:28
• @delinco It's a definition in general topology: a set is closed iff its complement is open. It's a very basic and essential fact. – Henno Brandsma Mar 10 '18 at 8:37
• @delinco that same fact is also used in the proof of 2. as they written a set as the complement of an open set to show it is closed. – Henno Brandsma Mar 10 '18 at 22:12