Chapter I.6

Operations on Cardinal Numbers

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Problem I.6.1

Prove that for arbitrary cardinalities mm and nn exactly one of the conditions m=nm = n, m<nm < n, or n<mn < m holds (trichotomy).

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Problem I.6.2

Prove that the cardinal numbers are linearly ordered by the relation ≤\leq.

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Problem I.6.3

Prove that there is no greatest cardinal number.

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Problem I.6.4

Prove that:

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(a)

3+5=83 + 5 = 8;

(b)

n+ℵ0=ℵ0n + \aleph_{0} = \aleph_{0}, where nn is finite;

(c)

ℵ0+ℵ0=ℵ0\aleph_{0} + \aleph_{0} = \aleph_{0};

(d)

ℵ0+c=c\aleph_{0} + c = c;

(e)

c+c=cc + c = c.

Problem I.6.5

Prove that:

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(a)

for any sets A1,A2A_{1}, A_{2} there exist sets B1,B2B_{1}, B_{2} such that A1∼B1A_{1} \sim B_{1}, A2∼B2A_{2} \sim B_{2}, and B1∩B2=∅B_{1} \cap B_{2} = \emptyset;

(b)

the sum of two cardinal numbers always exists.

Problem I.6.6

Prove for arbitrary cardinal numbers:

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(a)

n1+n2=n2+n1n_{1} + n_{2} = n_{2} + n_{1};

(b)

n1+(n2+n3)=(n1+n2)+n3n_{1} + (n_{2} + n_{3}) = (n_{1} + n_{2}) + n_{3};

(c)

n+0=nn + 0 = n.

Problem I.6.7

Let AA, BB, CC, A1,…,AnA_{1}, \ldots , A_{n} be finite sets. Prove that:

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(a)

∣A∪B∣=∣A∣+∣B∣−∣A∩B∣\left|A \cup B\right| = \left|A\right| + \left|B\right| - \left|A \cap B\right|;

(b)

∣A∪B∪C∣=∣A∣+∣B∣+∣C∣−∣A∩B∣−∣A∩C∣−∣B∩C∣+∣A∩B∩C∣\left|A \cup B \cup C\right| = \left|A\right| + \left|B\right| + \left|C\right| - \left|A \cap B\right| - \left|A \cap C\right| - \left|B \cap C\right| + \left|A \cap B \cap C\right|;

(c)

∣⋃i=1nAi∣=∑i=1n∣Ai∣+…+(−1)k−1∑i1,…,ik=1i1<…<ikn∣Ai1∩…∩Aik∣+…;\left|\bigcup_{i=1}^{n} A_{i}\right| = \sum_{i=1}^{n} \left|A_{i}\right| + \ldots + (-1)^{k-1} \sum_{\substack {i_{1}, \ldots , i_{k} = 1 \\ i_{1} < \ldots < i_{k}}}^{n} \left|A_{i_{1}} \cap \ldots \cap A_{i_{k}}\right| + \ldots ;

(d)

∣A1∖A2∖…∖An∣=∑i=1n∣Ai∣+…+(−2)k−1∑i1,…,ik=1i1<…<ikn∣Ai1∩…∩Aik∣+…\left|A_{1} \setminus A_{2} \setminus \ldots \setminus A_{n}\right| = \sum_{i=1}^{n} \left|A_{i}\right| + \ldots + (-2)^{k-1} \sum_{\substack {i_{1}, \ldots , i_{k} = 1 \\ i_{1} < \ldots < i_{k}}}^{n} \left|A_{i_{1}} \cap \ldots \cap A_{i_{k}}\right| + \ldots

Problem I.6.8
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(a)

Prove that n<m⇒n+1≤mn < m \Rightarrow n + 1 \leq m.

(b)

Give an example of cardinal numbers nn and mm such that n+1≤mn + 1 \leq m, but m≤nm \leq n.

Problem I.6.9

Prove that n≤mn \leq m if and only if there exists n1n_{1} such that n+n1=mn + n_{1} = m. Show that such an n1n_{1} is not uniquely determined.

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Problem I.6.10

Prove that:

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(a)

3⋅5=153 \cdot 5 = 15;

(b)

ℵ0⋅ℵ0=ℵ0\aleph_{0} \cdot \aleph_{0} = \aleph_{0};

(c)

ℵ0⋅c=c\aleph_{0} \cdot c = c;

(d)

c⋅c=cc \cdot c = c.

Problem I.6.11

Prove that:

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(a)

if A∼BA \sim B and C∼DC \sim D, then A×C∼B×DA \times C \sim B \times D;

(b)

the product of two cardinal numbers always exists.

Problem I.6.12

Prove for arbitrary cardinal numbers:

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(a)

n1⋅n2=n2⋅n1n_{1} \cdot n_{2} = n_{2} \cdot n_{1};

(b)

n1⋅(n2⋅n3)=(n1⋅n2)⋅n3n_{1} \cdot (n_{2} \cdot n_{3}) = (n_{1} \cdot n_{2}) \cdot n_{3};

(c)

n1⋅(n2+n3)=(n1⋅n2)+(n1⋅n3)n_{1} \cdot (n_{2} + n_{3}) = (n_{1} \cdot n_{2}) + (n_{1} \cdot n_{3});

(d)

n⋅1=nn \cdot 1 = n;

(e)

n⋅0=0n \cdot 0 = 0;

(f)

n⋅m=nn \cdot m = n, if mm is finite and nn is an infinite cardinal number;

(g)

n⋅ℵ0=nn \cdot \aleph_{0} = n, if nn is an infinite cardinal number.

Problem I.6.13

Prove that n2=nn^{2} = n, if nn is an infinite cardinal number.

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Problem I.6.14

Prove that if nn and mm are cardinal numbers and one of them is infinite, then n⋅m=n+m=max⁡(n,m)n \cdot m = n + m = \max (n, m), provided n≠0n \neq 0 and m≠0m \neq 0.

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Problem I.6.15

Prove that:

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(a)

2ℵ0=c2^{\aleph_{0}} = c;

(b)

ℵ0ℵ0=c\aleph_{0}^{\aleph_{0}} = c;

(c)

cℵ0=cc^{\aleph_{0}} = c.

Problem I.6.16

Prove that for two cardinal numbers nn and mm, nmn^{m} always exists.

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Problem I.6.17

Prove for arbitrary cardinal numbers mm, nn, and pp:

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(a)

mn+p=mn⋅mpm^{n+p} = m^{n} \cdot m^{p};

(b)

(m⋅n)p=mp⋅np(m \cdot n)^{p} = m^{p} \cdot n^{p};

(c)

(mn)p=mn⋅p(m^{n})^{p} = m^{n \cdot p};

(d)

m1=mm^{1} = m;

(e)

1m=11^{m} = 1.

Problem I.6.18

Prove that ∣P(A)∣=2∣A∣\left|P(A)\right| = 2^{\left|A\right|}.

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Problem I.6.19
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
Problem I.6.20

Prove for arbitrary cardinal numbers:

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(a)

if 2m≥ℵ02^{m} \geq \aleph_{0}, then 2m≥2ℵ02^{m} \geq 2^{\aleph_{0}};

(b)

if mn=ℵ0m^{n} = \aleph_{0}, then m=ℵ0m = \aleph_{0}, and nn is finite.

Problem I.6.21

Prove for arbitrary cardinal numbers:

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(a)

if n≥ℵ0n \geq \aleph_{0}, then 2n=nn2^{n} = n^{n};

(b)

if 1<m≤n1 < m \leq n, ℵ0≤n\aleph_{0} \leq n, then mn=nnm^{n} = n^{n}.

Problem I.6.22

Prove for arbitrary cardinal numbers:

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(a)

if ∣I∣=m\left|I\right| = m, ni=nn_{i} = n for all i∈Ii \in I, then m⋅n=∑i∈Inim \cdot n = \sum_{i \in I} n_{i};

(b)

if m+p=n+pm + p = n + p and pp is finite, then m=nm = n;

(c)

if 2⋅n1=2⋅n22 \cdot n_{1} = 2 \cdot n_{2}, then n1=n2n_{1} = n_{2}.

Problem I.6.23

Prove that

∣⋃i∈IAi∣≤∑i∈I∣Ai∣. \left|\bigcup _{i \in I} A_{i}\right| \leq \sum _{i \in I} \left|A_{i}\right|.
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Problem I.6.24
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(a)

Let {mi}\left\{ m_{i}\right\} (i∈I)(i \in I) be a family of cardinal numbers, with mi=0m_{i} = 0 for i∈J⊆Ii \in J \subseteq I. Prove that

∑i∈Imi=∑i∈I∖Jmi. \sum _{i \in I} m_{i} = \sum _{i \in I \setminus J} m_{i}.
(b)

Let {mi}\left\{ m_{i}\right\} (i∈I)(i \in I) be a family of cardinal numbers, with mi=1m_{i} = 1 for i∈J⊆Ii \in J \subseteq I. Prove that

∏i∈Imi=∏i∈I∖Jmi. \prod _{i \in I} m_{i} = \prod _{i \in I \setminus J} m_{i}.
(c)

Prove that ∏i∈Imi=0\prod_{i \in I} m_{i} = 0 if and only if there exists i0∈Ii_{0} \in I such that mi0=0m_{i_{0}} = 0.

Problem I.6.25

Prove that if φ\varphi is a permutation of the set II, then:

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(a)

∑i∈Imi=∑i∈Imφ(i)\sum_{i \in I} m_{i} = \sum_{i \in I} m_{\varphi (i)};

(b)

∏i∈Imi=∏i∈Imφ(i)\prod_{i \in I} m_{i} = \prod_{i \in I} m_{\varphi (i)}.

Problem I.6.26

Let {Aλ}λ∈L\left\{ A_{\lambda }\right\}_{\lambda \in L} be a partition of the set JJ into nonempty pairwise disjoint sets. Prove that:

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(a)

∑i∈Imi=∑λ∈L(∑i∈Aλmi)\sum_{i \in I} m_{i} = \sum_{\lambda \in L} \left(\sum_{i \in A_{\lambda }} m_{i}\right);

(b)

∏i∈Imi=∏λ∈L(∏i∈Aλmi)\prod_{i \in I} m_{i} = \prod_{\lambda \in L} \left(\prod_{i \in A_{\lambda }} m_{i}\right).

Problem I.6.27

Prove that:

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(a)

n⋅∑i∈Imi=∑i∈I(n⋅mi)n \cdot \sum_{i \in I} m_{i} = \sum_{i \in I} (n \cdot m_{i});

(b)

∏i∈I(∑j∈Jimij)=∑f∈K∏i∈Imif(i)\prod_{i \in I} \left(\sum_{j \in J_{i}} m_{ij}\right) = \sum_{f \in K} \prod_{i \in I} m_{i f(i)}, where K=∏i∈IJiK = \prod_{i \in I} J_{i}.

Problem I.6.28

Prove that if mi≤nim_{i} \leq n_{i} for all i∈Ii \in I, then:

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(a)

∑i∈Imi≤∑i∈Ini\sum_{i \in I} m_{i} \leq \sum_{i \in I} n_{i};

(b)

∏i∈Imi≤∏i∈Ini\prod_{i \in I} m_{i} \leq \prod_{i \in I} n_{i}.

Problem I.6.29

Let J⊆IJ \subseteq I. Prove that:

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(a)

∑i∈Jmi≤∑i∈Imi\sum_{i \in J} m_{i} \leq \sum_{i \in I} m_{i};

(b)

∏i∈Jmi≤∏i∈Imi\prod_{i \in J} m_{i} \leq \prod_{i \in I} m_{i}, where mi≠0m_{i} \neq 0 for i∈I∖Ji \in I \setminus J.

Problem I.6.30

Let {mi}i∈I\left\{ m_{i}\right\}_{i \in I} and {ni}i∈I\left\{ n_{i}\right\}_{i \in I} be two families of cardinal numbers, with ni≥2n_{i} \geq 2 for all i∈Ii \in I. Prove that:

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(a)

if mi≤nim_{i} \leq n_{i} for all i∈Ii \in I, then ∑i∈Imi≤∏i∈Ini\sum_{i \in I} m_{i} \leq \prod_{i \in I} n_{i};

(b)

if mi<nim_{i} < n_{i} for all i∈Ii \in I, then ∑i∈Imi<∏i∈Ini\sum_{i \in I} m_{i} < \prod_{i \in I} n_{i}.

Problem I.6.31

Let 0<m0<m1<…0 < m_{0} < m_{1} < \ldots Prove that ∑n∈Nmn<∏n∈Nmn\sum_{n \in \mathbb {N}} m_{n} < \prod_{n \in \mathbb {N}} m_{n}.

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Problem I.6.32

Let n≥ℵ0n \geq \aleph_{0}, {ni}i∈I\left\{ n_{i}\right\}_{i \in I} be a family of cardinal numbers not exceeding nn, ∣I∣≤n\left|I\right| \leq n. Prove that ∑i∈Ini≤n\sum_{i \in I} n_{i} \leq n.

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Problem I.6.33

Let α\alpha and β\beta be cardinal numbers, ∣I∣=β\left|I\right| = \beta, and αi=α\alpha_{i} = \alpha for every i∈Ii \in I. Prove that αβ=∏i∈Iαi\alpha^{\beta } = \prod_{i \in I} \alpha_{i}.

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Problem I.6.34

Prove that:

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(a)

m(∑i∈Ini)=∏i∈Imnim^{\left(\sum_{i \in I} n_{i}\right)} = \prod_{i \in I} m^{n_{i}};

(b)

(∏i∈Imi)n=∏i∈Imin\left(\prod_{i \in I} m_{i}\right)^{n} = \prod_{i \in I} m_{i}^{n}.

Problem I.6.35

Prove that for any cardinal number mm, mℵ0m^{\aleph_{0}} cannot be represented in the form ∑i∈Nmi\sum_{i \in \mathbb {N}} m_{i}, where mi<mi+1m_{i} < m_{i+1}.

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