Alternative proof Bolzano–Weierstrass theorem












































because halve length of interval @ each step limit of interval s length zero. there number



x


{\displaystyle x}

in each interval




i

n




{\displaystyle i_{n}}

. show,



x


{\displaystyle x}

accumulation point of



(

x

n


)


{\displaystyle (x_{n})}

.


take neighbourhood



u


{\displaystyle u}

of



x


{\displaystyle x}

. because length of intervals converges zero, there interval




i

n




{\displaystyle i_{n}}

subset of



u


{\displaystyle u}

. because




i

n




{\displaystyle i_{n}}

contains construction infinitely many members of



(

x

n


)


{\displaystyle (x_{n})}

,




i

n



u


{\displaystyle i_{n}\subseteq u}

,



u


{\displaystyle u}

contains infinitely many members of



(

x

n


)


{\displaystyle (x_{n})}

. proves,



x


{\displaystyle x}

accumulation point of



(

x

n


)


{\displaystyle (x_{n})}

. thus, there subsequence of



(

x

n


)


{\displaystyle (x_{n})}

converges



x


{\displaystyle x}

.







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