A pair of F-2A Japan from the 8th Tactical Fighter Squadron take off.

A pair of F-2A Japan from the 8th Tactical Fighter Squadron take off.

2002年小松基地航空祭、飛行実験群F-2A(63-8501)

F/A-18E/F

V-22 Osprey

P-3C Orion

C-2A Greyhound

F-35C Lightning II

EA-6B Prowler

E-2C Hawkeye

Boeing E-6

Countable and Uncountable Sets

This section will deal with the concept of finite and infinite sets which were introduced in Section 3.1. This section will help attach some meaning to the size, or cardinality, of a set A when A is an infinite set.

**Definition of a Bijection, Same Cardinality**

For any nonempty sets A,B, the function f: A -> B is called a bijection, or one-to-one correspondence, if f is both one-to-one and onto.

Example

Let A = ℤ⁺ and B = 2ℤ⁺ = {2k | k ∈ ℤ⁺} = {2,4,6,…}. The function f: A -> B, defined by f(x) = 2x is a bijection.

For a₁,a₂ ∈ A, then f(a₁) = f(a₂) => 2a₁ = 2a₂ => a₁ = a₂, and so f is one-to-one.

If b ∈ B, then b = 2a for some unique a ∈ A, and f(a) = 2a = b, and so f is onto.

When f: A -> B is a bijection, then A and B are denoted as A ~ B, where A and B have the same cardinality such that |A| = |B|.

From the previous example, it was shown that ℤ⁺ and 2ℤ⁺ both have the same size, even though it seemed like 2ℤ⁺ had fewer elements than ℤ⁺. This assumption is based on the fact that 2ℤ⁺ ⊂ ℤ⁺.

Example

Instead, let g: B -> A denote another function, where A = ℤ⁺ and B = 2ℤ⁺ (only the order changed) and g is defined by g(2k) = k.

Then, g(2k₁) = g(2k₂) => k₁ = k₂ => 2k₁ = 2k₂, and so g is one-to-one.

Then, for each k ∈ A, then 2k ∈ B with g(2k) = k, and so g is also onto.

Therefore, g is a bijection and B ~ A.

Since the previous example showed the function f was a bijection, then A ~ B. Additionally, in this example, it was shown that, although B ⊂ A, it is also found that B ~ A.

The above two examples showed that, for the case of A = ℤ⁺ and B = 2ℤ⁺, then A ~ B and B ~ A. However, this is generally true for any case. This is because the function g in the previous example was actually the inverse function of f. Since a function is invertible if it is a bijection, then, if there are two nonempty sets A,B with A ~ B, then B ~ A. And it is not necessary for A = B for |A| = |B|.

Example

For B = 2ℤ⁺ = {2k | k ∈ ℤ⁺} and C = 3ℤ⁺ = {3k | k ∈ ℤ⁺}, the function h: B -> C defined by h(2k) = 3k establishes a bijection between the sets B and C. Therefore, B ~ C, C ~ B, |B| = |C|.

**Properties of Bijection**

The following are some properties of bijection for any nonempty sets A,B,C:

1. A ~ A

2. If A ~ B, then B ~ A.

3. If A ~ B and B ~ C, then A ~ C.

**Finite and Infinite Sets**

Any set A is called a finite set if A = ∅ or if A ~ {1,2,3,…,n} for some n ∈ ℤ⁺. When A = ∅, then A has no elements, and so |A| = 0. When A ~ {1,2,3,…,n}, then |A| = n.

When a set A is not a finite set, then A is called an infinite set.

From this definition, it is seen that, if A is a nonempty finite set, then there is a bijection g: {1,2,3,…,n} -> A for some n ∈ ℤ⁺. This function g provides a listing of elements of A as g(1), g(2), g(3), …, g(n), which is a listing that can be counted.

When A is an infinite set, it can be seen there is no n ∈ ℤ⁺ such that the function g: A -> {1,2,3,…,n} is a bijection. But, if A and B are infinite sets, and A ⋂ B, then |A| = |B|, then there is a bijection between A and B, and so A ~ B, B ~ A.

**Countable Sets**

A set A is a countable, or denumerable, set if A is finite or A ~ ℤ⁺.

The previous examples show that 2ℤ⁺ ~ ℤ⁺ and 3ℤ⁺ ~ ℤ⁺. Since ℤ⁺ ~ ℤ⁺, then the sets ℤ⁺, 2ℤ⁺, and 3ℤ⁺ are all countable sets. In fact, for all k ∈ ℤ, k ≠ 0, the function f: ℤ⁺ -> kℤ⁺ defined as f(x) = kx is a bijection, and so kℤ⁺ is a countable set, where |kℤ⁺| = |ℤ⁺|.

Generally, if A is an infinite set and A ~ ℤ⁺, then ℤ⁺ ~ A, and so there is a bijection f: ℤ⁺ -> A that provides a listing of elements of A as f(1), f(2), …, and in this way, the elements of A can be counted but never finished in counting.

Therefore, an infinite set A is a countably infinite if either the bijection f: A -> ℤ⁺ exists or the bijection g: ℤ⁺ -> A exists.

Example

Let A = {1,½,1/3,¼,…} = {1/n | n ∈ ℤ⁺}. The function f: ℤ⁺ -> A defined by f(n) = 1/n establishes a bijection between ℤ⁺ and A, and so |ℤ⁺| = |A| and A is countable.

Although all of the previous examples of countably infinite sets have been subsets of ℤ, other countably infinite sets are possible.

**Finite and Infinite Sequences**

For n ∈ ℤ⁺, a finite sequence of n terms is a function f with domain (1,2,3,…,n}. This sequence is usually written as an ordered set {x₁,x₂,x₃,…,x_{n}}, where xᵢ = f(i) for all 1 ≤ i ≤ n.

An infinite sequence is a function g with ℤ⁺ as its domain. This type of sequence is denoted by the ordered set {x₁,x₂,x₃,…}, or {xᵢ}ᵢ∈ℤ⁺, where xᵢ = g(i) for all i ∈ ℤ⁺.

Example

The set {1,½,¼,1/8,1/16} can be thought of as a finite sequence given by the function f: A -> ℚ⁺, where A = {1,2,3,4,5} and f(n) = 2^{-n+1}.

Example

The above example, where A = {1,½,1/3,¼,…} = {1/n | n ∈ ℤ⁺} can also be expressed as an infinite sequence {1/n}_{n∈ℤ⁺} given by the function g: ℤ⁺ -> ℚ⁺, where g(n) = 1/n for each n ∈ ℤ⁺.

Example

For sequences, the terms do not have to be distinct. For instance, let f: ℤ⁺ -> ℤ, where x_{n} = f(n) = (-1)^{n + 1}, for each positive integer n. Then the infinite sequence is {x_{n}}_{n∈ℤ⁺} = {x₁,x₂,x₃,x₄,…} = {1,-1,1,-1,1,…}, but the range of f is only the two element set {1,-1}.

If A is a nonempty countable set, then A can be written as a sequence of distinct elements.

If A is finite, then A ~ {1,2,3,…,n} and {1,2,3,…,n} ~ A for some n ∈ ℤ⁺. Therefore, f: {1,2,3,…,n} -> A is a bijection. Then let aᵢ = f(i) for each 1 ≤ i ≤ n. Since f is a bijection, then {a₁,a₂,a₃,…,a_{n}} is a sequence of n distinct elements of A.

If A is a countably infinite set, then either f: ℤ⁺ -> A or g: A -> ℤ⁺ is a bijection. Let aᵢ = g(i) for all i ∈ ℤ⁺. Since g is a bijection, the elements {a₁,a₂,a₃,…} are distinct elements and A = {a₁,a₂,a₃,…}.

**Subsequences**

The infinite sequence {a₁,a₂,a₃,…} = {aᵢ}ᵢ∈ℤ⁺ is a subsequence of ℤ⁺ if for all i ∈ ℤ⁺, aᵢ ∈ ℤ⁺ and aᵢ < a_{i+1}.

Let {x_{n}}_{n∈ℤ⁺} and {y_{n}}_{n∈ℤ⁺} be two infinite sequences. Then {y_{n}}_{n∈ℤ⁺} is a subsequence of {x_{n}}_{n∈ℤ⁺} if there exists a subsequence {a_{k}}_{k∈ℤ⁺} of ℤ⁺, when for each k ∈ ℤ⁺, y_{k} = x_{ak}.

Example

{1,3,5,7,…} is a subsequence of ℤ⁺, where it can be given by the function f: ℤ⁺ -> ℤ⁺ where a_{n} = f(n) = 2n – 1.

Example

For n ∈ ℤ⁺, let x_{n} = 1/n and let y_{n} = 1/(3n). Then {x_{n}}_{n∈ℤ⁺} = {1,½,1/3,¼,…} and {y_{n}}_{n∈ℤ⁺} = {1/3,1/6,1/9,…}. Consider the subsequence {a_{k}}_{k∈ℤ⁺} of ℤ⁺ where a_{k} = 3k for each k ∈ ℤ⁺. Then for all n ∈ ℤ⁺, y_{n} = 1/(3n) = x₃_{n} = x_{an}, and so {y_{n}}_{n∈ℤ⁺} is a subsequence of {x_{n}}_{n∈ℤ⁺}.

**Infinite Countable Sets and Subsets**

If S is an infinite countable set, and A ⊆ S, then A is a countable set no matter the set A.

**Uncountable Sets**

Up to this point, it seemed as though every infinite set has turned out to be countable.

However, dealing with the infinite set of real numbers ℝ, not only is ℝ uncountable, but even (0,1] in ℝ is also uncountable.

Example

To prove that the set (0,1] = {x | x ∈ ℝ ᴧ 0 < x ≤ 1} is not a countable set, assume the set (0,1] is countable, such that f: ℤ⁺ -> (0,1] is a bijection. Then the set can be written in an infinite sequence of distinct terms, where for every i ∈ ℤ⁺, rᵢ is a real number: (0,1] = {r₁,r₂,r₃,…} = {r_{n}}_{n∈ℤ⁺}. To avoid repeats, it is agreed that every real number rᵢ is written in decimal form such that 0.5 becomes 0.499…, so that no decimal expansion ends. Writing such decimal expansions for real numbers r₁, r₂, r₃, …, yields the following:

r₁ = 0.a₁₁a₁₂a₁₃a₁₄…

r₂ = 0.a₂₁a₂₂a₂₃a₂₄…

r₃ = 0.a₃₁a₃₂a₃₃a₃₄…

⋮

r_{n} = 0.a_{n1}a_{n2}a_{n3}a_{n4}…

⋮

where aᵢ_{j} ∈ {0,1,2,3,…,8,9} for all i,j ∈ ℤ⁺.

Consider the real number r = 0.b₁b₂b₃…, where, for each k ∈ ℤ⁺, r is constructed as follows:

Note that 2 and 3 are randomly chosen integers.

Then r ∈ (0,1). However, for every k ∈ ℤ⁺, r ≠ r_{k}, because their nth digits differ, and so r ∉ {r₁,r₂,r₃,…}, which results in a contradiction, because it was assumed that the interval (0,1] was inside the countable infinite sequence {r₁,r₂,r₃,…}. Therefore, the interval (0,1] is uncountable.

Therefore, when a set is not uncountable, it is called uncountable. When a set A is uncountable, then ℤ⁺ and A do not have the same cardinality, and so A ≁ ℤ⁺. The cardinality of A is then greater than the cardinality of ℤ⁺, where |ℤ⁺| < |A|, despite both A and ℤ⁺ being infinite sets.

Additionally, it can happen where the subset of a uncountable set is countable, such as ℤ ⊆ ℝ. However, if the subset A is uncountable, and A ⊆ B, then B is uncountable as well, such as (0,1] ⊆ ℝ. Therefore, ℝ is uncountable.

Recall that an infinite set is countable if there either exists the bijection f: A -> ℤ⁺ or g: ℤ⁺ -> A. However, for uncountable sets, the roles of A and ℤ⁺ of f cannot be reversed. If there is a one-to-one function f: ℤ⁺ -> A, the set A is uncountable, such as f: ℤ⁺ -> ℝ. Therefore, |ℤ| < |ℝ|.

Example

To prove that the set of positive rational numbers ℚ⁺ is a countable infinite set, where ℚ⁺ ~ ℤ⁺ and |ℚ⁺| = |ℤ⁺|, and f: ℤ⁺ -> ℚ⁺ or g: ℚ⁺ -> ℤ⁺ is a bijection, then a bijection of ℚ⁺ and ℤ⁺ needs to be created, where all of the elements in ℚ must be considered.

Begin by considering all of the positive rational numbers of the form x/y for some x,y ∈ ℤ⁺. Then the following table displays how every single positive rational number is included in a sequence using the arrows as direction of ordering:

The fractions that are circled and crossed out are skipped and not included in the sequence set, because they have already been included in the list.

Therefore, the set of positive rational numbers can have each of its distinct numbers listed as an infinite sequence {1,2,½,1/3,3,4,3/2,2/3,…}, which creates a countable set. Therefore, ℚ⁺ is countable.

Additionally, to show that the set ℚ is countable, the infinite sequence created from the set of positive rational number ℚ⁺ is taken and every negative number is placed beside its corresponding positive number to account for every negative of every positive rational number. Therefore, ℚ is countable, and so f: ℤ⁺ -> ℚ is a bijection.

The following proof is to show that, if A is any set, then |A| < |℘(A)|.

Example

To prove that if A is any set, |A| < |℘(A)|, consider two cases of the set A, where A is an empty set or A is a nonempty set.

For A = ∅, then the cardinality of A is |∅| = 0. The power set of A is then ℘(A) = {∅}. And so the cardinality of ℘(A) is |{∅}| = 1. Hence, |A| < |℘(A)| holds true so far.

For A ≠ ∅, let f: A -> ℘(A) be defined by f(a) = {a} for each a ∈ A. Since the function f is one-to-one, |A| = |f(A)| ≤ |℘(A)|.

To show that |A| ≠ |℘(A)|, it remains to be proven that there is no function g: A -> ℘(A) that is onto.

Let g: A -> ℘(A) and B = {a | a ∈ A ᴧ a ∉ g(a)}, and so B = g(a). There is an element a inside A such that it is not in the range of g(a), and so g(a) = B ⊆ A.

Since B ∈ ℘(A), then if g is an onto function, there must exist an element a’ ∈ A such that g(a’) = B. However, when a’ ∈ g(a’) = B, then from the definition of B, a’ ∉ g(a’), which forms a contradiction that a’ ∈ g(a’) and a’ ∉ g(a’). When a’ ∈ g(a’), then a’ ∈ B, but B = g(a’). This is a contradiction.

Therefore, there is no such a’ ∈ A where g(a’) = B, and so the function g cannot be onto. Hence, |A| ≠ |℘(A)|, yet |A| < |℘(A)|.

**Sizes of Infinity**

As a consequence to the above proof, it is found that there is no infinite cardinal number that is the largest, where if A is any finite set, then |A| < |℘(A)| < |℘(℘(A))| < …

However, the smallest cardinal number for countably infinite sets is the size of ℤ, ℤ⁺, ℚ, or any set A, such that the function f: ℤ⁺ -> A is a bijection. This smallest infinity is denoted as א₀, read as, “aleph zero,” where א₀ = |ℤ|, |ℤ⁺|, |ℚ|, |A|, where f: ℤ⁺ -> A is a bijection or |ℤ⁺| = |A|.

Additionally, the next larger cardinality for uncountable sets, such as ℝ, is denoted as א₁, or c for continuum.

Therefore, for any countably infinite set A, |A| = |ℤ| = |ℤ⁺| = |ℚ| = א₀ is the smallest infinity for a cardinality to be.

PDF reference: 887

Page A-32. Question 1.

a) true

b) false, the subset (1,2] in ℝ⁺ is uncountable, and therefore ℝ⁺ is uncountable.

d) true

e) false, let A = ℤ ⋃ (0,1] and B = ℤ ⋃ (1,2]. A and B are uncountable sets, yet A ⋂ B = ℤ, which is a countable set.

f) true

g) false, let A = ℤ⁺ ⋃ (0,1] and B = (0,1]. A and B are uncountable sets, yet A ⧹ B = {2,3,4,…} is a countable set.

A pair of F-2A AF the Japan self-Defense with b/n 03-8507 and 13-8521 “Viper Zero” from the 3rd Tactical Fighter Squadron. Air Base Misawa.

F-2A air force-Defense of Japan from the 6th Tactical Fighter Squadron. The Base Zouki.

Japanese formation flight Sunday. Mitsubishi F-2A and F-2B

自衛隊は、災害派遣要請を受諾する以前から自主的に情報収集活動を開始。地震発生から21分後の21時47分には航空自衛隊築城基地（福岡県）からF-2A戦闘機が2機、緊急発進（スクランブル）し、次いで九州北部の各基地・駐屯地から陸海空自衛隊のUH-1Jヘリコプター2機、UH-60/SH-60ヘリコプター5機、P-3C哨戒機1機、U-125A救難捜索機2機を相次いで発進させました。

真っ先に離陸した自衛隊機がF-2戦闘機であることに疑問を抱いた方もおられるかもしれませんが、これは全ての戦闘機基地にて、発進命令後5分以内に2機が離陸できる即時発進体制「5分待機」を維持していることに由来します。

5分待機は本来、領空に接近する所属不明機を迎撃するために行われており、災害発生時の緊急発進という場合にも、F-2（場合によってF-4やF-15）はミサイルを携行したまま離陸。戦闘機の高い機動性を活かして真っ先に被災地上空を飛行し、パイロットは目視で地上を確認、無線通信によって第一報を送ることを任務とします。

今回の地震発生は夜間でした。F-2は対艦攻撃など、夜間の低高度飛行が必要な任務に用いる赤外線前方監視装置を搭載可能ですが、5分待機に就く機体には装備されません。よって今回、緊急発進したF-2パイロットは、ほとんど何も見えず帰投したはずです。しかし、例えば「大規模な火災は確認できない」といった「何も見えなかった」という報告そのものが、貴重な情報となりえます。

真っ先に離陸した自衛隊機がF-2戦闘機であることに疑問を抱いた方もおられるかもしれませんが、これは全ての戦闘機基地にて、発進命令後5分以内に2機が離陸できる即時発進体制「5分待機」を維持していることに由来します。

5分待機は本来、領空に接近する所属不明機を迎撃するために行われており、災害発生時の緊急発進という場合にも、F-2（場合によってF-4やF-15）はミサイルを携行したまま離陸。戦闘機の高い機動性を活かして真っ先に被災地上空を飛行し、パイロットは目視で地上を確認、無線通信によって第一報を送ることを任務とします。

今回の地震発生は夜間でした。F-2は対艦攻撃など、夜間の低高度飛行が必要な任務に用いる赤外線前方監視装置を搭載可能ですが、5分待機に就く機体には装備されません。よって今回、緊急発進したF-2パイロットは、ほとんど何も見えず帰投したはずです。しかし、例えば「大規模な火災は確認できない」といった「何も見えなかった」という報告そのものが、貴重な情報となりえます。