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Question Number 1616 by 123456 last updated on 27/Aug/15
lets two sets A,B and take ∣X∣ the number of elements of the set X, them proof or give a counter example that if ∣A∪B∣=∞ and ∣A∩B∣=∞ then ∣A∣=∞ and ∣B∣=∞
$$\mathrm{lets}\:\mathrm{two}\:\mathrm{sets}\:\mathrm{A},\mathrm{B}\:\mathrm{and}\:\mathrm{take}\:\mid\mathrm{X}\mid\:\mathrm{the}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{elements}\:\mathrm{of}\:\mathrm{the}\:\mathrm{set}\:\mathrm{X},\:\mathrm{them} \\ $$$$\mathrm{proof}\:\mathrm{or}\:\mathrm{give}\:\mathrm{a}\:\mathrm{counter}\:\mathrm{example}\:\mathrm{that} \\ $$$$\mathrm{if}\:\mid\mathrm{A}\cup\mathrm{B}\mid=\infty\:\mathrm{and}\:\mid\mathrm{A}\cap\mathrm{B}\mid=\infty\:\mathrm{then}\:\mid\mathrm{A}\mid=\infty\:\mathrm{and}\:\mid\mathrm{B}\mid=\infty \\ $$
Commented by 112358 last updated on 27/Aug/15
A∪B=A+B−A∩B ⇒∣A∪B∣=∣A∣+∣B∣−∣A∩B∣ ∴ ∣A∩B∣+∣A∪B∣=∣A∣+∣B∣ By logic, p→q≡∽q→∽p. So, propose the following statements p:∣A∩B∣ and ∣A∪B∣ are non−finite. q: ∣A∣ and ∣B∣ are non−finite. Then one may find it easier to show that ∽q→∽p rather than p→q directly. Let ∣A∣ and ∣B∣ be finite. ⇒ ∣A∣=n , ∣B∣=m where n,m∈Z^+ and n,m are finite. ⇒∣A∪B∣+∣A∩B∣=m+n ∵ n and m arefinite ⇒m+n is finite ⇒∣A∩B∣+∣A∪B∣ is finite ∴ ∣A∩B∣=x,∣A∪B∣=y so that x+y=m+n with x,y∈Z^+ and are finite. Thus, ∣A∪B∣ and ∣A∩B∣ are finite. ∵ ∽q→∽p then we have that p→q. Hence, ∣A∩B∣=∞ and ∣A∪B∣=∞ implies that ∣A∣=∞ and ∣B∣=∞.
$${A}\cup{B}={A}+{B}−{A}\cap{B} \\ $$$$\Rightarrow\mid{A}\cup{B}\mid=\mid{A}\mid+\mid{B}\mid−\mid{A}\cap{B}\mid \\ $$$$\therefore\:\mid{A}\cap{B}\mid+\mid{A}\cup{B}\mid=\mid{A}\mid+\mid{B}\mid \\ $$$${By}\:{logic},\:{p}\rightarrow{q}\equiv\backsim{q}\rightarrow\backsim{p}.\:{So}, \\ $$$${propose}\:{the}\:{following}\:{statements} \\ $$$${p}:\mid{A}\cap{B}\mid\:{and}\:\mid{A}\cup{B}\mid\:{are}\:{non}−{finite}. \\ $$$${q}:\:\mid{A}\mid\:{and}\:\mid{B}\mid\:{are}\:{non}−{finite}. \\ $$$${Then}\:{one}\:{may}\:{find}\:{it}\:{easier}\:{to}\: \\ $$$${show}\:{that}\:\backsim{q}\rightarrow\backsim{p}\:{rather}\:{than} \\ $$$${p}\rightarrow{q}\:{directly}. \\ $$$${Let}\:\mid{A}\mid\:{and}\:\mid{B}\mid\:{be}\:{finite}. \\ $$$$\Rightarrow\:\:\:\:\:\:\mid{A}\mid={n}\:\:\:\:\:,\:\:\:\:\mid{B}\mid={m} \\ $$$${where}\:{n},{m}\in\mathbb{Z}^{+} \:{and}\:{n},{m}\:{are}\:{finite}. \\ $$$$\Rightarrow\mid{A}\cup{B}\mid+\mid{A}\cap{B}\mid={m}+{n} \\ $$$$\because\:{n}\:{and}\:{m}\:{arefinite}\: \\ $$$$\Rightarrow{m}+{n}\:{is}\:{finite} \\ $$$$\Rightarrow\mid{A}\cap{B}\mid+\mid{A}\cup{B}\mid\:{is}\:{finite} \\ $$$$\therefore\:\mid{A}\cap{B}\mid={x},\mid{A}\cup{B}\mid={y}\:{so}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}+{y}={m}+{n} \\ $$$${with}\:{x},{y}\in\mathbb{Z}^{+} \:{and}\:{are}\:{finite}.\:{Thus}, \\ $$$$\mid{A}\cup{B}\mid\:{and}\:\mid{A}\cap{B}\mid\:{are}\:{finite}. \\ $$$$\because\:\backsim{q}\rightarrow\backsim{p}\:{then}\:{we}\:{have}\:{that} \\ $$$${p}\rightarrow{q}.\:{Hence},\:\mid{A}\cap{B}\mid=\infty\:{and}\:\mid{A}\cup{B}\mid=\infty \\ $$$${implies}\:{that}\:\mid{A}\mid=\infty\:{and}\:\mid{B}\mid=\infty. \\ $$
Commented by 123456 last updated on 28/Aug/15
nice :D thanks
$$\mathrm{nice}\::\mathrm{D} \\ $$$$\mathrm{thanks} \\ $$
Commented by Rasheed Ahmad last updated on 28/Aug/15
Appreciations! Good approach!
$${Appreciations}!\:{Good}\:{approach}! \\ $$

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