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

Question

((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

Answer 1

ANSWER

To prove the statement, we use the definitions:

Definition 1 Let $\left\{ \left( { x }_{ n\ },{y}_{n} \right) \right\}$ is a sequence in $\R^{2}$ . We say that $\left( { x }_{ n\quad },{ y }_{ n } \right)$ convergens to $\left( { x }_{0\ },{ y }_{ 0 } \right)$ and write $\left( { x }_{ n\ },{ y }_{ n } \right) \rightarrow \left( { x }_{ 0\ },{ y }_{ 0 } \right)$ if for every $\varepsilon >0\quad$there is an $N_{0}\in \N$ such that for all $n\in \N,$ if $n>N_{0}$ , then

$\left\| \left( { x }_{ n\quad },{ y }_{ n } \right) -\left( { x }_{ 0\quad },{ y }_{ 0 } \right) \right\| =\sqrt { { \left( { x }_{ n }-{ x }_{ 0\quad } \right) }^{ 2 }+{ \left( { y }_{ n }-{ y }_{ 0\quad } \right) }^{ 2 } } <\varepsilon$

Definition 2 Let $\left\{ \left( { x }_{ n }\right ) \right\}$ is a sequence in $\R$ . We say that $\left( { x }_{ n } \right)$ convergens to $\left( { x }_{0\ } \right)$ and write ${ x }_{ n } \rightarrow { x }_{ 0 }$ if for every $\varepsilon >0\quad$there is an $N_{x}\in \N$ such that for all $n\in\N$ if $n>N_{x}$ , then

$|x_{ n }-x_{ 0 }|<\varepsilon$

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

$|{ x }_{ n }-{ x }_{ 0\quad }|\quad =\sqrt { { \left( { x }_{ n }-{ x }_{ 0\quad } \right) }^{ 2 }\quad } \le \sqrt { { \left( { x }_{ n }-{ x }_{ 0\quad } \right) }^{ 2 }+{ \left( { y }_{ n }-{ y }_{ 0\quad } \right) }^{ 2 } } =\left\| \left( { x }_{ n\quad },{ y }_{ n } \right) -\left( { x }_{ 0\quad },{ y }_{ 0 } \right) \right\|<\varepsilon$

and $|{ y }_{ n }-{ y }_{ 0\ }|\quad =\sqrt { { \left( { y }_{ n }-{ y }_{ 0\quad } \right) }^{ 2 } } \le \sqrt { { \left( { x }_{ n }-{ x }_{ 0\quad } \right) }^{ 2 }+{ \left( { y }_{ n }-{ y }_{ 0\quad } \right) }^{ 2 } } =\left\| \left( { x }_{ n\quad },{ y }_{ n } \right) -\left( { x }_{ 0\quad },{ y }_{ 0 } \right) \right\| <\varepsilon$

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

2) If ${ x }_{ n } \rightarrow { x }_{ 0 }$ and ${y }_{ n } \rightarrow { y }_{ 0 }$ , then by the definition 2 , for every $\varepsilon >0\quad$there is an $N_{x}, N_{y}\in \N$ such that for all $n\in \N,$ if $n>N_{x}$ , then

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

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

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

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

$\left\| \left( { x }_{ n\quad },{ y }_{ n } \right) -\left( { x }_{ 0\quad },{ y }_{ 0 } \right) \right\| =\sqrt { { \left( { x }_{ n }-{ x }_{ 0\quad } \right) }^{ 2 }+{ \left( { y }_{ n }-{ y }_{ 0\quad } \right) }^{ 2 } } <\sqrt { \frac { { \varepsilon }^{ 2 } }{ 2 } +\frac { { \varepsilon }^{ 2 } }{ 2 } } =\varepsilon$ .

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

Hence, $1)\Leftrightarrow 2)$ .