Question #329494

Given a sequence ((xn,yn)) is R2 .prove that if ((xn,yn)) is bounded ,then (xn) and (yn) are bounded.

1
Expert's answer
2022-04-18T00:28:31-0400

ANSWER

By the definition, the sequence {(xn,yn)}\left\{ \left( { x }_{ n }{ ,y }_{ n } \right) \right\} is bounded in R2R^2 if there exists M>0M>0 such that

(xn,yn)=(xn)2+(yn)2M\left\| \left( { x }_{ n },{ y }_{ n } \right) \right\| =\sqrt { { \left( { x }_{ n } \right) }^{ 2 }+{ \left( { y }_{ n } \right) }^{ 2 } } \le M for all nNn\in N .


Therefore

xn=(xn)2(xn)2+(yn)2M\left| { x }_{ n } \right| =\sqrt { { \left( { x }_{ n } \right) }^{ 2 }\quad } \le \sqrt { { \left( { x }_{ n } \right) }^{ 2 }+{ \left( { y }_{ n } \right) }^{ 2 } } \le M for all nNn\in N .

and

yn=(yn)2(xn)2+(yn)2M\left| { y }_{ n } \right| =\sqrt { { \left( { y }_{ n } \right) }^{ 2 }\quad } \le \sqrt { { \left( { x }_{ n } \right) }^{ 2 }+{ \left( { y }_{ n } \right) }^{ 2 } } \le M

Equivalent to

MxnM-M\leq x_{n}\leq M . for all nNn\in N

and

MynM-M\leq y_{n}\leq M for all nNn\in N

So, the sequences {(xn)},\left\{ \left( { x }_{ n } \right) \right\} , {(yn)}\left\{ \left( {y }_{ n } \right) \right\} are bounded in RR .


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