Answer to Question #330311 in Real Analysis for Sakshi Naik Gaonka

Question #330311

((xn,yn) is convergent iff both (xn}) and (yn) are convergent. In fact, for(x0 ,y0) in R2, we have (xn ,yn) converging to (x0,y0) iff xn converges to x0 and yn converges to y0

1
Expert's answer
2022-04-19T03:35:23-0400

ANSWER

To prove the statement, we use the definitions:

Definition 1 Let {(xn ,yn)}\left\{ \left( { x }_{ n\ },{y}_{n} \right) \right\} is a sequence in R2\R^{2} . We say that (xn,yn)\left( { x }_{ n\quad },{ y }_{ n } \right) convergens to (x0 ,y0)\left( { x }_{0\ },{ y }_{ 0 } \right) and write (xn ,yn)(x0 ,y0)\left( { x }_{ n\ },{ y }_{ n } \right) \rightarrow \left( { x }_{ 0\ },{ y }_{ 0 } \right) if for every ε>0\varepsilon >0\quadthere is an N0NN_{0}\in \N such that for all nN,n\in \N, if n>N0n>N_{0} , then

(xn,yn)(x0,y0)=(xnx0)2+(yny0)2<ε\left\| \left( { x }_{ n\quad },{ y }_{ n } \right) -\left( { x }_{ 0\quad },{ y }_{ 0 } \right) \right\| =\sqrt { { \left( { x }_{ n }-{ x }_{ 0\quad } \right) }^{ 2 }+{ \left( { y }_{ n }-{ y }_{ 0\quad } \right) }^{ 2 } } <\varepsilon

Definition 2 Let {(xn)}\left\{ \left( { x }_{ n }\right ) \right\} is a sequence in R\R . We say that (xn)\left( { x }_{ n } \right) convergens to (x0 )\left( { x }_{0\ } \right) and write xnx0{ x }_{ n } \rightarrow { x }_{ 0 } if for every ε>0\varepsilon >0\quadthere is an NxNN_{x}\in \N such that for all nNn\in\N if n>Nxn>N_{x} , then

xnx0<ε|x_{ n }-x_{ 0 }|<\varepsilon

1) Let (xn ,yn)(x0 ,y0)\left( { x }_{ n\ },{ y }_{ n } \right) \rightarrow \left( { x }_{ 0\ },{ y }_{ 0 } \right) .If ε>0\varepsilon>0 and n>N0n>N_{0} , such that (xn,yn)(x0,y0)<ε\left\| \left( { x }_{ n\quad },{ y }_{ n } \right) -\left( { x }_{ 0\quad },{ y }_{ 0 } \right) \right\| <\varepsilon ,then for all n>N0n>N_{0} :

xnx0=(xnx0)2(xnx0)2+(yny0)2=(xn,yn)(x0,y0)<ε|{ x }_{ n }-{ x }_{ 0\quad }|\quad =\sqrt { { \left( { x }_{ n }-{ x }_{ 0\quad } \right) }^{ 2 }\quad } \le \sqrt { { \left( { x }_{ n }-{ x }_{ 0\quad } \right) }^{ 2 }+{ \left( { y }_{ n }-{ y }_{ 0\quad } \right) }^{ 2 } } =\left\| \left( { x }_{ n\quad },{ y }_{ n } \right) -\left( { x }_{ 0\quad },{ y }_{ 0 } \right) \right\|<\varepsilon

and yny0 =(yny0)2(xnx0)2+(yny0)2=(xn,yn)(x0,y0)<ε|{ y }_{ n }-{ y }_{ 0\ }|\quad =\sqrt { { \left( { y }_{ n }-{ y }_{ 0\quad } \right) }^{ 2 } } \le \sqrt { { \left( { x }_{ n }-{ x }_{ 0\quad } \right) }^{ 2 }+{ \left( { y }_{ n }-{ y }_{ 0\quad } \right) }^{ 2 } } =\left\| \left( { x }_{ n\quad },{ y }_{ n } \right) -\left( { x }_{ 0\quad },{ y }_{ 0 } \right) \right\| <\varepsilon

Therefore, by the definition 2 xnx0{ x }_{ n } \rightarrow { x }_{ 0 } and yny0{y }_{ n } \rightarrow { y }_{ 0 }

2) If xnx0{ x }_{ n } \rightarrow { x }_{ 0 } and yny0{y }_{ n } \rightarrow { y }_{ 0 } , then by the definition 2 , for every ε>0\varepsilon >0\quadthere is an Nx,NyNN_{x}, N_{y}\in \N such that for all nN,n\in \N, if n>Nxn>N_{x} , then

xnx0<ε2|x_{n} -x_{0}|<\frac { \varepsilon }{ \sqrt { 2 } }

and if n>Nyn>N_{y} , then

yny0<ε2|y_{n} -y_{0}|<\frac { \varepsilon }{ \sqrt { 2 } } .

Let N0=max{Nx,Ny}{ N }_{ 0 }=\max { \left\{ N_{ x },\quad N_{ y } \right\} } . Since N0>Nx,N0>NyN_{0}>N_{x}, N_{0}>N_{y} , then for all n>N0n>N_{0} (n>Nx,n>Ny)(\Rightarrow n>N_{x}, n>N_{y})

(xn,yn)(x0,y0)=(xnx0)2+(yny0)2<ε22+ε22=ε\left\| \left( { x }_{ n\quad },{ y }_{ n } \right) -\left( { x }_{ 0\quad },{ y }_{ 0 } \right) \right\| =\sqrt { { \left( { x }_{ n }-{ x }_{ 0\quad } \right) }^{ 2 }+{ \left( { y }_{ n }-{ y }_{ 0\quad } \right) }^{ 2 } } <\sqrt { \frac { { \varepsilon }^{ 2 } }{ 2 } +\frac { { \varepsilon }^{ 2 } }{ 2 } } =\varepsilon .

Therefore, by the definition 1, (xn ,yn)(x0 ,y0)\left( { x }_{ n\ },{ y }_{ n } \right) \rightarrow \left( { x }_{ 0\ },{ y }_{ 0 } \right) .

Hence, 1)2)1)\Leftrightarrow 2) .


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