prove-that-cl-Q-Q-R-2-note-X-d-is-a-metric-space-A-X-x-A-cl-A-r-gt-0-B-r-x-A-

Question Number 204702 by mnjuly1970 last updated on 25/Feb/24

prove that :

cl (Q \times Q) \overset{?}{=} R^{2}

n o t e : (X, d) i s a m e t r i c s p a c e

, A \subseteq X : x \in \bar{A} = cl (A) \Leftrightarrow \forall r > 0, B_{r} (x) \cap A \neq ϕ

Answered by witcher3 last updated on 25/Feb/24

\bar{Q} = R

cl (Q * Q) \subset Cl (\bar{Q} * \bar{Q}) = R^{2}

let x (a, b) \in R^{2} = (x

\Rightarrow \exists U_{n}, V_{n} \in Q

u_{n} \to a; v_{n} \to b

\Rightarrow \forall ϵ > 0 \exists N; \forall n ⩾ N; ∣ U_{n} - a ∣< ϵ; \forall ϵ^{'} > 0 \exists N^{'} \in N; \forall n ⩾ N^{'} ∣ V_{n} - b ∣< 0 \dots . . (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) \cap Cl (Q^{2})

r > 0 withe 1 \exists N_{1}, N_{2}

∣ u_{n} - a ∣< r; \forall n ⩾ N_{1}

∣ u_{n} - b ∣< r; \forall n ⩾ N_{2}

tack N = \max (N_{1}, N_{2}); \forall n ⩾ N

d ((u_{n}, v_{n}); (a, b)) = \max (∣ u_{n} - a ∣; ∣ v_{n} - b ∣) < r

\Rightarrow (a, b) \in cl (Q^{2})

\Rightarrow {IR}^{2} \subset Cl (Q^{2}) \Rightarrow equalitt

Commented by mnjuly1970 last updated on 25/Feb/24

t h a n k s a l o t s i r w h i c h e r

Commented by witcher3 last updated on 25/Feb/24

withe pleasur barak alah Fikoum