23 Jan 2020 - Abhishek Nandekar

The Topological Approach:

The infinitude of the set of Prime numbers can be proved using basic topological constructs. In this section, we look at the motivations behind the notions of open and closed sets, then, we’ll discuss the basics of topology and finally we’ll prove the infinitude of Primes.

The Open Set:

Consider the set $\mathbb R$. Consider a real number $y \in \mathbb R$. This element $y$ will still be closer to zero than any other element $y_1 \notin \mathbb R$. In this manner, one may speak of whether two subsets of a topological space are “near” without defining the metric.

In general, a family of sets containing zero, used to approximate zero is referred to as the neighbourhood basis of zero. Any member of this neighbourhood basis is the open set.

Consider an arbitrary set $X$. Given $x \in X$ we can define a collection of sets “around” $x$, including $x$ itself, used to approximate $x$. This collection of sets contains all open sets.

Definitions:

$U \subseteq \mathbb R^n$ is open if,

\(\forall x \in U, \exists \text{ } \epsilon > 0\) such that,

$$\{|y-x| < \epsilon \implies y \in U | y \in \mathbb R^n\}$$

For any collection of open sets,

• Union any arbitrary number of open sets is open,
• Any finite intersection of open sets is open.

Axioms of Topology

$X$ is a set, $\tau$ is a collection of subsets of $X$. If,

1. $\phi, X \in \tau$
2. $\tau_i’ \subseteq \tau$, and $\cup_{i=1}^N \tau_i’ \in \tau$, $N$ can be infinite.
3. $U_1 \in \tau$, and $U_2 \in \tau$, then $U_1 \cap U_2 \in \tau$

Then, $\tau$ is a topology on $X$ and the elements of $\tau$ are open sets.

The ordered pair $(X, \tau)$ is called a topological space.

Examples:

Let $X = {1, 2, 3}$.

• If $\tau = {\phi, X}$. In this case, all three axioms are satisfied. Hence $\tau$ is a genuine topology on $X$.
• If $\tau = {\phi, {1}, X}$, axiom 3 is not satisfied. Hence, $\tau$ is not a topology $X$.
• If $\tau = {\phi, {1}, {2}, {1, 2}, X}$, union of any subsets of $\tau$ is an element of $\tau$, and intersection of any elements of $\tau$ is an element of $\tau$. Hence all three axioms of topology are satisfied and $\tau$ is a genuine topology on $X$.

Closed Sets

Closed sets are the compliments of open sets.

Theorem 1

Any finite union of closed sets is closed.

Proof:

Consider the above given topological space $T = (X, \tau)$. Let $\bigcup_{i=1}^n V_i$ be a finite union of closed sets of $T$.

Then from De Morgan’s laws:

$$X \setminus \bigcup_{i=1}^n V_i = \bigcap_{i=1}^n (X \setminus V_i)$$

By definition, $X \setminus V_i$ is an open set in $T$.

Using Axiom 3, we can say,

$$\bigcap_{i=1}^n(S \setminus V_i) = S \setminus \bigcup_{i=1}^n V_i$$

LHS is a finite intersection of open sets, and hence is open.

$\implies$RHS is also an open set. Hence $S \setminus \bigcup_{i=1}^n V_i$ is open, implying that $\bigcup_{i=1}^n V_i$ is closed. $\square$

Proof for the infinitude of Primes

Let $O \subseteq \mathbb Z$. $O$ is an open set if it is either empty or, $\forall a \in O$, $\exists b>0$ such that

\(N_{a, b} = \{a + nb\big|n\in \mathbb Z\}\) and $N_{a,b} \subseteq O$.

Clearly, the union of open sets is open again. Let $N_{a, b_1} \subseteq O_1$ and $N_{a, b_2} \subseteq O_2$, and $a \in O_1 \cap O_2$. Then, $N_{a, b_1b_2} \subseteq O_1 \cap O_2$, which is also an open set. So, this family of sets is a bona fide topology in $\mathbb Z$

By this definition, we can observe two facts:

• Any nonempty open set is infinite.
• Any $N_{a, b}$ is closed as well.

To prove the second fact, we can say that

$$N_{a, b} = \mathbb Z \setminus \bigcup_{i=1}^{b-1} N_{a+i, b}$$

Observe the fact that $\forall n \in \mathbb Z \setminus {-1, 1}$, there exists a prime divisor $p \in \mathbb P$. Hence we can say,

$$\bigcup_{p \in \mathbb P} N_{0, p} = \mathbb Z \setminus \{-1, 1\}$$

Now, if $\mathbb P$ is a finite set, then LHS is a finite union of closed sets and hence, is closed ( proved in Theorem 1 ).

$\implies$ ${-1, 1}$ should be an open set. But, as stated earlier, any nonempty open set is infinite. Hence, this is a contradiction. $\mathbb P$ should be an infinite set. $\square$