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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 R. Consider a real number yR. This element y will still be closer to zero than any other element y1R. 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 xX 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:

URn is open if,

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

{|yx|<ϵyU|yRn}

For any collection of open sets,

Axioms of Topology

X is a set, τ is a collection of subsets of X. If,

  1. ϕ,Xτ
  2. τiτ, and Ni=1τiτ, N can be infinite.
  3. U1τ, and U2τ, then U1U2τ

Then, τ is a topology on X and the elements of τ are open sets.

The ordered pair (X,τ) is called a topological space.

Examples:

Let X=1,2,3.

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,τ). Let ni=1Vi be a finite union of closed sets of T.

Then from De Morgan’s laws:

Xni=1Vi=ni=1(XVi)

By definition, XVi is an open set in T.

Using Axiom 3, we can say,

ni=1(SVi)=Sni=1Vi

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

RHS is also an open set. Hence Sni=1Vi is open, implying that ni=1Vi is closed.

Proof for the infinitude of Primes

Let OZ. O is an open set if it is either empty or, aO, b>0 such that

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

Clearly, the union of open sets is open again. Let Na,b1O1 and Na,b2O2, and aO1O2. Then, Na,b1b2O1O2, which is also an open set. So, this family of sets is a bona fide topology in Z

By this definition, we can observe two facts:

To prove the second fact, we can say that

Na,b=Zb1i=1Na+i,b

Observe the fact that nZ1,1, there exists a prime divisor pP. Hence we can say,

pPN0,p=Z{1,1}

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

1,1 should be an open set. But, as stated earlier, any nonempty open set is infinite. Hence, this is a contradiction. P should be an infinite set.