Library of Mathexandria

Let \( G \) be a topological space and let \( H \) be a subspace of \( G \) endowed with the subspace topology. Then \( H \times H \) endowed with the product topology is same as the subspace topology of \( H \times H \subset G \times G \) with \(G \times G \) endowed with the product topology due to: \[ (U_1 \cap H) \times (U_2 \cap H) = (U_1 \times U_2) \cap (H \times H) \] Let \( G \) be a topological group with a subgroup \( H \). We would like to show that \( H \) endowed with the subspace topology is also a topological group. Let \( p_G \) denote the product map. in \( G \) and \( p_H \) denote the product map in \( H \). We want to show that \( p_H \) is continuous. Let \( U \) be an open set in \( H \), then \(U = U_G \cap H \) for some open set \( U_G \) in \( G \). \[ \begin{array}{lcl} p_H^{-1}(U) & = & \{ (h_1, h_2) \in H \times H ~|~ h_1h_2 \in U \} \\ & = & \{ (g_1, g_2) \in G \times G ~|~ g_1g_2 \in U_G \} \cap H \times H \\ & = & p_G^{-1}(U_G) \

Our goal for this post is to review topology and prove that \( \mathbb{Z}_p \) is a topological group/ring. In order to do so, we prove properties of topological groups, inverse limits, and neighborhoods. This is my attempt to understand the following lines from Serre's A Course in Arithmetic: "The ideals \( p^n\mathbb{Z}_p \) form a basis of neighborhoods of \( 0 \); since \( x \in p^n \mathbb{Z}_p \) is equivalent to \( v_p(x) \geq n \), the topology on \( \mathbb{Z}_p \) is defined by the distance \( d(x,y) = e^{-v_p(x-y)} \). Since \( \mathbb{Z}_p \) is compact, it is complete. Finally, if \( x = (x_n) \) is an element of \( \mathbb{Z}_p \), and if \( y_n \in \mathbb{Z} \) is such that \( y_n \equiv x_n ~(\bmod{~p^n}) \), then \( \lim y_n = x \), which proves that \( \mathbb{Z} \) is dense in \( \mathbb{Z}_p \)."   Definition:  A filter \( \mathcal{F} \) on a set \( S \) is a subset of \( \mathcal{P}(S) \), the power group of \( S \) satisfying the following properti

Show that the equation \( x^2 = 2 \) has a solution in \( \mathbb{Z}_7 \). We prove a more generalized form of the problem: Let \( p \) be an odd prime and \( (a,p) = 1 \) and \( a \) be a quadratic residue modulo \( p \), then \( x^2 = a \) has a solution in \( \mathbb{Z}_p \). Lemma1 Let \( F(x) \) be a polynomial with integer coefficient, then if \( F(x) \) is solvable in \( \mathbb{Z}/p^n \mathbb{Z} \) for all \( n \), then \( F(x) \) is solvable in \( \mathbb{Z}_p \). Proof) Let \( x_n \) be the solution to \( F(x) \) in \( \mathbb{Z}/p^n \mathbb{Z} \). If \( (x_n) \in \mathbb{Z}_p \) then we are done; however, it is not true in general. We will have to somehow obtain a \( p \)-adic integer from \( (x_n) \). First observe that if \( n \geq m \), then \( x_n \equiv 0 ~\bmod{~p^m} \) because \( x_n \equiv 0 ~\bmod{~p^n} \). In particular, \( (x_n)_{n \geq m} \) has the property that \( x_n \equiv 0 ~\bmod{~p^m} \) for all \( n \). Considering \( (x_n) \) as a sequence in \( \mathb

Let \( A_n = \mathbb{Z}/p^n \mathbb{Z} \), then \( A_n \) with the canonical epimorphism \( \varphi_n: A_n \rightarrow A_{n-1} \) defines a projective system. The projective limit \( \mathbb{Z}_p \) is the subring of \[ \prod_{n \geq 1} A_n \] containing \( (x_n) \) such that \( \varphi_n(x_n) = x_{n-1} \). Suppose we give \( A_n \) the discrete topology and \( \prod A_n \) the product topology. By Tychonoff theorem, we have a compact space \( \prod A_n \).  We call \( \mathbb{Z}_p \) the \( p \)-adic integers. Theorem 1 The \( p \)-adic integers \( \mathbb{Z}_p \) is closed subset of \( \prod A_n \). Proof) We prove that the complement \( \mathbb{Z}_p^c \) is open. Let \( (x_n) \in \mathbb{Z}_p^c \), then we have some positive number \( m \) such that \( \varphi(x_m) \neq x_{m-1} \). Consider the following set \[ U = \{ x_1 \} \times \cdots \times \{ x_m \} \times \prod_{n > m} A_n \] which is open because there are only finitely many proper subsets of \( A_n \). Clearly \( x_n \

Functions and Infinity 여태까지 추상적인 이야기만 했기 때문에 예를 한번 들어보겠다. 집합 \( A = \{ a,b,c \} \) 와 집합 \( B = \{ 1, 2, 3 \} \)이 있다. 함수 \( f: A \rightarrow B \)를 다음과 같이 정의 하자. \[ f(a) = 1, f(b) = 2, f(c) = 3 \] 수학자들은 \( f \)를 반복해서 쓰는 대신 다음과 같은 표현도 쓴다. \[ a \mapsto 1, b \mapsto 2, c \mapsto 3 \] 여기서 우리는 모든 A의 원소가 유일한 B의 원소와 쌍이 되는 것을 볼 수 있다. 집합의 두 개의 크기를 비교하는 것은 이 방법을 적용시킨것이라고 생각하면 된다. 만약 \( A \)에 있는 모든 원소를 \( B \)에 있는 원소와 겹치지 않게 쌍을 지었는데 \( B \)의 원소가 남았다면 우리는 당연히 \( B \)가 더 크다고 이야기 할 것이다. 반대의 경우로는 \( A \)에 있는 모든 원소와 \( B \) 원소와 쌍을 지으려고 했지만  \( B \)의 원소가 모자라는 경우에는 \( A \)가 더 크다고 이야기 한다. 이를 수학적으로 다시 이야기를 하게 된다면, \( f \)에서 \( f(a) = f(b) \Rightarrow a = b \)가 성립할 경우 우리는 \( f \)를 단사함수(Injective Function)이라고 부른다. 모든 \( b \in B \)에 대하여 \( f(a) = b \)인 \( a \in A \)가 존재할 경우, 우리는 \( f \)를 전사함수(Surjective Function)이라고 부른다. 엄밀하지 않게 \( |A| \)를 집합 \( A \)의 크기 혹은 농도라고 정의 하자. 이 글의 맨 처음 예에서는 직관적으로 두 집합의 크기가 똑같은 것을 알 수 있다. 만약 단사 함수 \( f: A \rightarrow B \)가 존재하게 된다면 우리는 \( |A| \leq |B| \)라고 표현한다. 반대로, 전사 함수 \( f: A \r