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Question Number 204702 by mnjuly1970 last updated on 25/Feb/24
prove that : cl(Q×Q )=^? R^2 note: (X ,d ) is a metric space , A ⊆ X : x∈ A^( −) =cl(A) ⇔ ∀ r >0 , B_r (x) ∩ A ≠ φ
$$ \\ $$$$\:\:\:\mathrm{prove}\:\mathrm{that}\:: \\ $$$$\:\:\:\:\:\:\:\mathrm{cl}\left(\mathbb{Q}×\mathbb{Q}\:\right)\overset{?} {=}\:\mathbb{R}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{note}:\:\:\:\left({X}\:,{d}\:\right)\:{is}\:{a}\:{metric}\:{space} \\ $$$$\:\:\:\:\:\:\:\:\:\:,\:\:\:{A}\:\subseteq\:{X}\::\:\:\:\:\:{x}\in\:\overset{\:\:−} {{A}}=\mathrm{cl}\left({A}\right)\:\Leftrightarrow\:\forall\:{r}\:>\mathrm{0}\:,\:{B}_{{r}} \:\left({x}\right)\:\cap\:{A}\:\neq\:\phi \\ $$
Answered by witcher3 last updated on 25/Feb/24
Q^− =R cl(Q∗Q)⊂Cl(Q^− ∗Q^− )=R^2 let x(a,b)∈R^2 =(x ⇒∃ U_n ,V_n ∈Q u_n →a;v_n →b ⇒∀ε>0 ∃N;∀n≥N ;∣U_n −a∣<ε;∀ε′>0 ∃N′∈N ;∀n≥N′∣V_n −b∣<0.....(1) since we are in finite dimensilm lets defind A metric?over R^2 d((x,y);(x′,y′))=max(∣x−x′∣;∣y−y′∣) let r>0 β_r (x) ∩Cl(Q^2 ) r>0 withe 1 ∃N_1 ,N_2 ∣u_n −a∣<r;∀n≥N_1 ∣u_n −b∣<r;∀n≥N_2 tack N=max(N_1 ,N_2 );∀n≥N d((u_n ,v_n );(a,b))=max(∣u_n −a∣;∣v_n −b∣)<r ⇒(a,b)∈cl(Q^2 ) ⇒IR^2 ⊂Cl(Q^2 )⇒equalitt
$$\overset{−} {\mathrm{Q}}=\mathbb{R} \\ $$$$\mathrm{cl}\left(\mathrm{Q}\ast\mathrm{Q}\right)\subset\mathrm{Cl}\left(\overset{−} {\mathrm{Q}}\ast\overset{−} {\mathrm{Q}}\right)=\mathbb{R}^{\mathrm{2}} \\ $$$$\mathrm{let}\:\mathrm{x}\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{R}^{\mathrm{2}} =\left(\mathrm{x}\right. \\ $$$$\Rightarrow\exists\:\mathrm{U}_{\mathrm{n}} ,\mathrm{V}_{\mathrm{n}} \:\in\mathrm{Q} \\ $$$$\mathrm{u}_{\mathrm{n}} \rightarrow\mathrm{a};\mathrm{v}_{\mathrm{n}} \rightarrow\mathrm{b} \\ $$$$\Rightarrow\forall\epsilon>\mathrm{0}\:\exists\mathrm{N};\forall\mathrm{n}\geqslant\mathrm{N}\:;\mid\mathrm{U}_{\mathrm{n}} −\mathrm{a}\mid<\epsilon;\forall\epsilon'>\mathrm{0}\:\exists\mathrm{N}'\in\mathbb{N}\:;\forall\mathrm{n}\geqslant\mathrm{N}'\mid\mathrm{V}_{\mathrm{n}} −\mathrm{b}\mid<\mathrm{0}…..\left(\mathrm{1}\right) \\ $$$$\mathrm{since}\:\mathrm{we}\:\mathrm{are}\:\mathrm{in}\:\mathrm{finite}\:\mathrm{dimensilm} \\ $$$$\mathrm{lets}\:\mathrm{defind}\:\mathrm{A}\:\mathrm{metric}?\mathrm{over}\:\mathbb{R}^{\mathrm{2}} \\ $$$$\mathrm{d}\left(\left(\mathrm{x},\mathrm{y}\right);\left(\mathrm{x}',\mathrm{y}'\right)\right)=\mathrm{max}\left(\mid\mathrm{x}−\mathrm{x}'\mid;\mid\mathrm{y}−\mathrm{y}'\mid\right) \\ $$$$\mathrm{let}\:\mathrm{r}>\mathrm{0}\:\:\:\beta_{\mathrm{r}} \left(\mathrm{x}\right)\:\cap\mathrm{Cl}\left(\mathrm{Q}^{\mathrm{2}} \right) \\ $$$$\mathrm{r}>\mathrm{0}\:\mathrm{withe}\:\mathrm{1}\:\exists\mathrm{N}_{\mathrm{1}} ,\mathrm{N}_{\mathrm{2}} \\ $$$$\mid\mathrm{u}_{\mathrm{n}} −\mathrm{a}\mid<\mathrm{r};\forall\mathrm{n}\geqslant\mathrm{N}_{\mathrm{1}} \\ $$$$\mid\mathrm{u}_{\mathrm{n}} −\mathrm{b}\mid<\mathrm{r};\forall\mathrm{n}\geqslant\mathrm{N}_{\mathrm{2}} \\ $$$$\mathrm{tack}\:\mathrm{N}=\mathrm{max}\left(\mathrm{N}_{\mathrm{1}} ,\mathrm{N}_{\mathrm{2}} \right);\forall\mathrm{n}\geqslant\mathrm{N} \\ $$$$\mathrm{d}\left(\left(\mathrm{u}_{\mathrm{n}} ,\mathrm{v}_{\mathrm{n}} \right);\left(\mathrm{a},\mathrm{b}\right)\right)=\mathrm{max}\left(\mid\mathrm{u}_{\mathrm{n}} −\mathrm{a}\mid;\mid\mathrm{v}_{\mathrm{n}} −\mathrm{b}\mid\right)<\mathrm{r} \\ $$$$\Rightarrow\left(\mathrm{a},\mathrm{b}\right)\in\mathrm{cl}\left(\mathrm{Q}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{IR}^{\mathrm{2}} \subset\mathrm{Cl}\left(\mathrm{Q}^{\mathrm{2}} \right)\Rightarrow\mathrm{equalitt} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 25/Feb/24
thanks alot sir whicher
$${thanks}\:{alot}\:{sir}\:{whicher} \\ $$
Commented by witcher3 last updated on 25/Feb/24
withe pleasur barak alah Fikoum
$$\mathrm{withe}\:\mathrm{pleasur}\:\mathrm{barak}\:\mathrm{alah}\:\mathrm{Fikoum} \\ $$

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