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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