Skip to main content

Posts about Lean4Challenge

RSS feed (Lean4Challenge, en)

Challenge 23: The constant sequence aₙ = L converges to L

Challenges in Lean 4

Prove that the constant sequence \(a_{n} = L\) converges to \(L\).

To do this, complete the following theory:

import Mathlib.Tactic

def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop :=
  ∀ ε > 0, ∃ k, ∀ n ≥ k, |a n - L| < ε

variable {a : ℕ → ℝ}
variable {L : ℝ}

example
  (h : ∀ n, a n = L)
  : SeqLim a L :=
by sorry

How to publish your solution?

  1. Write your proof in live.lean-lang.org and copy the generated link.
  2. Post it on the Mastodon thread for this problem, with a message like:
    "My solution is at [link]."
  3. You will be able to compare your approach with other participants during the week. The official solution will be published next Monday.

Challenge 22: Subsequences with different limits imply non-convergence

Challenges in Lean 4

Prove that if a sequence has two subsequences with different limits, then the sequence is not convergent.

To do so, complete the following theory:

import Mathlib.Data.Real.Basic
import Mathlib.Tactic

-- a converges to L
def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop :=
  ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |a n - L| < ε

-- The sequence a is convergent
def SeqConv (a : ℕ → ℝ) : Prop :=
  ∃ L, SeqLim a L

-- φ is an extraction function.
def IsExtraction (φ : ℕ → ℕ) : Prop :=
  StrictMono φ

-- b is a subsequence of a.
def Subseq (b a : ℕ → ℝ) : Prop :=
  ∃ φ, IsExtraction φ ∧ b = a ∘ φ

variable {a b₁ b₂ : ℕ → ℝ}
variable {L L₁ L₂ : ℝ}

example
  (hb₁ : Subseq b₁ a)
  (hL₁ : SeqLim b₁ L₁)
  (hb₂ : Subseq b₂ a)
  (hL₂ : SeqLim b₂ L₂)
  (h : L₁ ≠ L₂)
  : ¬SeqConv a :=
by sorry

Read more…

Challenge 21: Subsequences have the same limit as the sequence

Challenges in Lean 4

A subsequence is obtained by applying an extraction function to the original sequence; that is, a strictly increasing function \(φ : ℕ \to ℕ\). For example, the subsequence \[ u_{0}, u_{2}, u_{4}, u_{6}, \dots \] is obtained with the extraction function \(\varphi\) defined by \(\varphi(n) = 2n\).

The previous definitions are formalized in Lean 4 as:

-- φ is an extraction function.
def IsExtraction (φ : ℕ → ℕ) :=
  StrictMono φ

-- v is a subsequence of u.
def Subseq (v u : ℕ → ℝ) :=
  ∃ φ, IsExtraction φ ∧ v = u ∘ φ

-- a is the limit of u.
def SeqLim (u : ℕ → ℝ) (a : ℝ) :=
  ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |u n - a| < ε

Prove that every subsequence of a convergent sequence converges to the same limit as the sequence.

To do so, complete the following Lean 4 theory:

import Mathlib.Data.Real.Basic

variable {u v : ℕ → ℝ}
variable {a : ℝ}

def IsExtraction (φ : ℕ → ℕ):=
  StrictMono φ

def Subseq (v u : ℕ → ℝ) :=
  ∃ φ, IsExtraction φ ∧ v = u ∘ φ

def SeqLim (u : ℕ → ℝ) (a : ℝ) :=
  ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |u n - a| < ε

example
  (hv : Subseq v u)
  (ha : SeqLim u a)
  : SeqLim v a :=
by sorry

Read more…

Challenge 20: If aₙ → L and M is an upper bound of aₙ, then L ≤ M

Challenges in Lean 4

Prove that if the sequence \(a_{n}\) converges to \(L\) and \(M\) is an upper bound of \(a_{n}\) (that is, \(a_{n} ≤ M\) for all \(n\)), then \(L ≤ M\).

To do so, complete the following Lean 4 theory:

import Mathlib.Data.Real.Basic
import Mathlib.Tactic

def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop :=
  ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |a n - L| < ε

def IsUB (a : ℕ → ℝ) (M : ℝ) : Prop :=
  ∀ n, a n ≤ M

variable {a : ℕ → ℝ}
variable {L M : ℝ}

example
  (ha : SeqLim a L)
  (hM : IsUB a M)
  : L ≤ M :=
by sorry

Read more…

Challenge 19: Convergent sequences are Cauchy sequences

Challenges in Lean 4

Prove that convergent sequences are Cauchy sequences

To do so, complete the following Lean 4 theory:

import Mathlib.Data.Real.Basic
import Mathlib.Tactic

variable {u : ℕ → ℝ}

def SeqLim (u : ℕ → ℝ) (a : ℝ) : Prop :=
  ∀ ε > 0, ∃ k, ∀ n ≥ k, |u n - a| < ε

def SeqConv (u : ℕ → ℝ) :=
  ∃ a, SeqLim u a

def IsCauchy (u : ℕ → ℝ) :=
  ∀ ε > 0, ∃ k, ∀ p ≥ k, ∀ q ≥ k, |u p - u q| < ε

example
  (h : SeqConv u)
  : IsCauchy u :=
by sorry

Read more…

Challenge 18: Convergence of the product of convergent sequences

Challenges in Lean 4

This week's challenge is to prove in Lean 4 that if the sequence \(aₙ\) converges to \(L\) and \(bₙ\) converges to \(M\), then \(aₙbₙ\) converges to \(LM\).

To do so, complete the following Lean 4 theory:

import Mathlib.Data.Real.Basic
import Mathlib.Tactic

def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop :=
  ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |a n - L| < ε

variable {a b c : ℕ → ℝ}
variable {L M : ℝ}

example
  (ha : SeqLim a L)
  (hb : SeqLim b M)
  (hc : ∀ n, c n = a n * b n)
  : SeqLim c (L * M) :=
by sorry

Read more…

Challenge 17: Convergent sequences are bounded

Challenges in Lean 4

This week's challenge is to prove in Lean 4 that if a sequence \(a_n\) is convergent, then \(a_n\) is bounded; that is, there exists an \(M ∈ ℝ\) such that for every \(n ∈ ℕ\), \(|a_n| ≤ M\).

To do so, complete the following Lean 4 theory:

import Mathlib.Data.Real.Basic
import Mathlib.Tactic

open Finset

def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop :=
  ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |a n - L| < ε

def SeqConv (a : ℕ → ℝ) : Prop :=
  ∃ L, SeqLim a L

def SeqBdd (a : ℕ → ℝ) : Prop :=
  ∃ M, ∀ n, |a n| ≤ M

variable (a : ℕ → ℝ)

example
  (ha : SeqConv a)
  : SeqBdd a :=
by sorry

Read more…

Challenge 1: The sequence 1/n converges to 0

Challenges in Lean 4

In Lean, a sequence \(a₀, a₁, a₂,...\) can be represented by a function \(a : ℕ → ℝ\) so that \(a(n)\) is \(aₙ\).

The statement that \(L\) is the limit of the sequence \(a\) is defined by

def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop :=
  ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |a n - L| < ε

Prove that if \(aₙ = 1/n\) for all \(n\), then the sequence \(a\) converges to 0.

To do so, complete the following Lean 4 theory:

import Mathlib.Data.Real.Basic
import Mathlib.Tactic

variable (a : ℕ → ℝ)

def SeqLim (a : ℕ → ℝ) (L : ℝ) : Prop :=
  ∀ ε > 0, ∃ k : ℕ, ∀ n ≥ k, |a n - L| < ε

example
  (ha : ∀ n, a n = 1 / n)
  : SeqLim a 0 :=
by sorry

Read more…