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Question Number 110219 by mohammad17 last updated on 27/Aug/20

let A and C be two none empety set prove that  A⊆B ∧ C ⊆D iff A×C ⊆B×D?  help me sir

$${let}\:{A}\:{and}\:{C}\:{be}\:{two}\:{none}\:{empety}\:{set}\:{prove}\:{that} \\ $$$${A}\subseteq{B}\:\wedge\:{C}\:\subseteq{D}\:{iff}\:{A}×{C}\:\subseteq{B}×{D}? \\ $$$${help}\:{me}\:{sir} \\ $$

Commented by kaivan.ahmadi last updated on 27/Aug/20

if A⊆B∧C⊆D⇒A×C⊆B×D  otherwise there exist (a,c)∈A×C(a∈A,c∈C)such that  (a,c)∈B×D so a∉B or c∉D  if a∉B then A⊈B and if c∉D then C⊈D  that is a contrudiction.

$${if}\:{A}\subseteq{B}\wedge{C}\subseteq{D}\Rightarrow{A}×{C}\subseteq{B}×{D} \\ $$$${otherwise}\:{there}\:{exist}\:\left({a},{c}\right)\in{A}×{C}\left({a}\in{A},{c}\in{C}\right){such}\:{that} \\ $$$$\left({a},{c}\right)\in{B}×{D}\:{so}\:{a}\notin{B}\:{or}\:{c}\notin{D} \\ $$$${if}\:{a}\notin{B}\:{then}\:{A}\nsubseteq{B}\:{and}\:{if}\:{c}\notin{D}\:{then}\:{C}\nsubseteq{D} \\ $$$${that}\:{is}\:{a}\:{contrudiction}. \\ $$

Commented by kaivan.ahmadi last updated on 27/Aug/20

if A×C⊆B×D then A⊆B∧C⊆D  otherwise A⊈B or C⊈D  if A⊈B then there exist a∈A s.t a∉B so  for each c∈C we have (a,c)∈A×C and  (a,c)∉B×D that is contrudiction with hypothesis.  similarly we can prove for C⊈D.

$${if}\:{A}×{C}\subseteq{B}×{D}\:{then}\:{A}\subseteq{B}\wedge{C}\subseteq{D} \\ $$$${otherwise}\:{A}\nsubseteq{B}\:{or}\:{C}\nsubseteq{D} \\ $$$${if}\:{A}\nsubseteq{B}\:{then}\:{there}\:{exist}\:{a}\in{A}\:{s}.{t}\:{a}\notin{B}\:{so} \\ $$$${for}\:{each}\:{c}\in{C}\:{we}\:{have}\:\left({a},{c}\right)\in{A}×{C}\:{and} \\ $$$$\left({a},{c}\right)\notin{B}×{D}\:{that}\:{is}\:{contrudiction}\:{with}\:{hypothesis}. \\ $$$${similarly}\:{we}\:{can}\:{prove}\:{for}\:{C}\nsubseteq{D}. \\ $$$$ \\ $$

Commented by mohammad17 last updated on 28/Aug/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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