- #1
transgalactic
- 1,395
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regarding this defintion
http://img515.imageshack.us/img515/5666/47016823jz1.gif
i was told that
Remember that if [itex]x_n[/itex] is bounded then [itex]\limsup x_n = \lim \left( \sup \{ x_k | k\geq n\} \right)[/itex].
The sequence, [itex]\sup \{ x_k | k\geq n\}[/itex] is non-increasing, therefore its limits is its infimum.
Thus, [itex]\limsup x_n = \inf \{ \sup\{ x_k | k\geq n\} | n\geq 0 \} [/itex][/quote]
i can't understand the first part
why he is saying that
[itex]\sup \{ x_k | k\geq n\}[/itex]
is not increasing.
you are taking a bounded sequence and you get one number
which is SUP (its least upper bound)
thats it.
no more members
??
http://img515.imageshack.us/img515/5666/47016823jz1.gif
i was told that
Remember that if [itex]x_n[/itex] is bounded then [itex]\limsup x_n = \lim \left( \sup \{ x_k | k\geq n\} \right)[/itex].
The sequence, [itex]\sup \{ x_k | k\geq n\}[/itex] is non-increasing, therefore its limits is its infimum.
Thus, [itex]\limsup x_n = \inf \{ \sup\{ x_k | k\geq n\} | n\geq 0 \} [/itex][/quote]
i can't understand the first part
why he is saying that
[itex]\sup \{ x_k | k\geq n\}[/itex]
is not increasing.
you are taking a bounded sequence and you get one number
which is SUP (its least upper bound)
thats it.
no more members
??
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