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Question Number 82792 by M±th+et£s last updated on 24/Feb/20
show that if A⊂R^m  and B⊂R^n  are   compact sets.   then A×B={(a,b)∈R^(m+n) :a∈A and b∈B}
$${show}\:{that}\:{if}\:{A}\subset\mathbb{R}^{{m}} \:{and}\:{B}\subset\mathbb{R}^{{n}} \:{are}\: \\ $$$${compact}\:{sets}.\: \\ $$$${then}\:{A}×{B}=\left\{\left({a},{b}\right)\in\mathbb{R}^{{m}+{n}} :{a}\in{A}\:{and}\:{b}\in{B}\right\} \\ $$
Commented by M±th+et£s last updated on 24/Feb/20
is compact
$${is}\:{compact} \\ $$
Answered by mind is power last updated on 24/Feb/20
since A is a subset of IR^n   dim IR^n =n  let N_(1 ) bee a norme i dont know english nam of   norme definitionn N(x)=0⇔x=0  N(x+y)≤N(x)+N(y)  N(kx)=∣k∣N(x)   let N_2  norme of  IR^m   N_3 (x)=N_1 (a)+N_2 (b) is norme over A×B  x=(a,b)∈A×B  since A is compact ∀U_n ∈A ∃ U_(ϕ(n)) →l∈A  ⇒N_1 (U_(ϕ(n)) −l)→0  since B is Compact  ∀V_n ∈B ∃l_2 ∈B and∃V_(δ(n)) such  N_2 (V_(δ(n)) −l_2 )→0  we check that  ∀(U_n ,V_n )∈A×B   (U_(ϕ(n)) ,V_(δ(n)) ) Cv  (l,l_2 )  N((U_(ϕ(n)) ,V_(δ(n)) );(l,l_2 ))=N_1 (U_(ϕ(n)) ,l)+N_2 (V_(δ(n)) ,l_2 )→0  ⇒∀(U_n ,V_n )∈A×B  has a cv subsequences ⇒A×B is compact set
$${since}\:{A}\:{is}\:{a}\:{subset}\:{of}\:{IR}^{{n}} \:\:{dim}\:{IR}^{{n}} ={n} \\ $$$${let}\:{N}_{\mathrm{1}\:} {bee}\:{a}\:{norme}\:{i}\:{dont}\:{know}\:{english}\:{nam}\:{of}\: \\ $$$${norme}\:{definitionn}\:{N}\left({x}\right)=\mathrm{0}\Leftrightarrow{x}=\mathrm{0} \\ $$$${N}\left({x}+{y}\right)\leqslant{N}\left({x}\right)+{N}\left({y}\right) \\ $$$${N}\left({kx}\right)=\mid{k}\mid{N}\left({x}\right)\: \\ $$$${let}\:{N}_{\mathrm{2}} \:{norme}\:{of}\:\:{IR}^{{m}} \\ $$$${N}_{\mathrm{3}} \left({x}\right)={N}_{\mathrm{1}} \left({a}\right)+{N}_{\mathrm{2}} \left({b}\right)\:{is}\:{norme}\:{over}\:{A}×{B} \\ $$$${x}=\left({a},{b}\right)\in{A}×{B} \\ $$$${since}\:{A}\:{is}\:{compact}\:\forall{U}_{{n}} \in{A}\:\exists\:{U}_{\varphi\left({n}\right)} \rightarrow{l}\in{A} \\ $$$$\Rightarrow{N}_{\mathrm{1}} \left({U}_{\varphi\left({n}\right)} −{l}\right)\rightarrow\mathrm{0} \\ $$$${since}\:{B}\:{is}\:{Compact}\:\:\forall{V}_{{n}} \in{B}\:\exists{l}_{\mathrm{2}} \in{B}\:{and}\exists{V}_{\delta\left({n}\right)} {such}\:\:{N}_{\mathrm{2}} \left({V}_{\delta\left({n}\right)} −{l}_{\mathrm{2}} \right)\rightarrow\mathrm{0} \\ $$$${we}\:{check}\:{that} \\ $$$$\forall\left({U}_{{n}} ,{V}_{{n}} \right)\in{A}×{B}\:\:\:\left({U}_{\varphi\left({n}\right)} ,{V}_{\delta\left({n}\right)} \right)\:{Cv}\:\:\left({l},{l}_{\mathrm{2}} \right) \\ $$$${N}\left(\left({U}_{\varphi\left({n}\right)} ,{V}_{\delta\left({n}\right)} \right);\left({l},{l}_{\mathrm{2}} \right)\right)={N}_{\mathrm{1}} \left({U}_{\varphi\left({n}\right)} ,{l}\right)+{N}_{\mathrm{2}} \left({V}_{\delta\left({n}\right)} ,{l}_{\mathrm{2}} \right)\rightarrow\mathrm{0} \\ $$$$\Rightarrow\forall\left({U}_{{n}} ,{V}_{{n}} \right)\in{A}×{B}\:\:{has}\:{a}\:{cv}\:{subsequences}\:\Rightarrow{A}×{B}\:{is}\:{compact}\:{set} \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 24/Feb/20
nice solution sir thank you
$${nice}\:{solution}\:{sir}\:{thank}\:{you} \\ $$
Commented by mind is power last updated on 24/Feb/20
withe pleasur
$${withe}\:{pleasur} \\ $$

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