Number of Primes is Infinite/Proof 2

Proof
Define a topology on the integers $\Z$ by declaring a subset $U \subseteq \Z$ to be an open set it is either:
 * the empty set $\O$

or:
 * a union of sequences $\map S {a, b}$, where:
 * $\map S {a, b} = \set {a n + b: n \in \Z} = a \Z + b$

In other words, $U$ is open every $x \in U$ admits some non-zero integer $a$ such that $\map S {a, x} \subseteq U$.

The open set axioms are verified as follows:

$(O1): \quad$ All unions of open sets are open:

For any set of open sets $U_i$ and $x$ in their union $U$, any of the numbers $a_i$ for which $\map S {a_i, x} \subseteq U_i$ also shows that $\map S {a_i, x} \subseteq U$.

$(O2): \quad$ The intersection of two (and hence finitely many) open sets is open:

Let $U_1$ and $U_2$ be open sets.

Let $x \in U_1 \cap U_2$ (with integers $a_1$ and $a_2$ establishing membership).

Set $a$ to be the lowest common multiple of $a_1$ and $a_2$.

Then:
 * $\map S {a, x} \subseteq \map S {a_i, x} \subseteq U_1 \cap U_2$

$(O3): \quad$ By definition, $\O$ is open: $\Z$ is just the sequence $\map S {1, 0}$, and so is open as well.

The topology is quite different from the usual Euclidean one, and has two notable properties:


 * $(1): \quad$ Since any non-empty open set contains an infinite sequence, no finite set can be open. Put another way, the complement of a finite set cannot be a closed set.
 * $(2): \quad$ The basis sets $\map S {a, b}$ are both open and closed: they are open by definition, and we can write $\map S {a, b}$ as the complement of an open set as follows:


 * $\displaystyle \map S {a, b} = \Z \setminus \bigcup_{j \mathop = 1}^{a - 1} \map S {a, b + j}$

The only integers that are not integer multiples of prime numbers are $-1$ and $+1$, that is:


 * $\displaystyle \Z \setminus \set {-1, + 1} = \bigcup_{\text {$p$ prime} } \map S {p, 0}$

By the first property, the set on the cannot be closed.

On the other hand, by the second property, the sets $\map S {p, 0}$ are closed.

So, if there were only finitely many prime numbers, then the set on the would be a finite union of closed sets, and hence closed.

Therefore by Proof by Contradiction, there must be infinitely many prime numbers.

Also see

 * Euclid's Theorem and its corollary