- #1
CGandC
- 326
- 34
Summary:: x
Let ## \{ a_{n} \} ## be a sequence.
Prove: If for all ## N \in { \bf{N} } ## there exists ## n> N ## such that ## a_{n} \leq L ## , then there exists a subsequence ## \{ a_{n_{k}} \} ## such that ##
a_{n_{k}} \leq L ##
My attempt:
Suppose that for all ## N \in {\bf{N}} ## there exists ## n> N ##, such that ## a_{n} \leq L ##. Since ## \{ a_{n} \} ## is a sequence, there exists a subsequence ## \{ a_{n_{k}} \}## such that there exists a monotonically increasing sequence of natural numbers ## \{ n_{k} \} ## such that ## \forall k \in {\bf{N}}, n_{k} \geq k ## . Let ## k \in {\bf{N}} ## be arbitrary so there exists ## n_{k} \geq k ##. Thus ( by universal instantiation of ## n_{k} ## in the statement " for all ## N \in {\bf{N}} ## there ... " ) there exists ## n > n_{k} > k ## such that ## a_{n} \leq L ##
However ,I don't know if ## a_{n_{k}} ## is less than or greater than ## a_{n} ## . So I'm stuck, how am I supposed to proceed from here to show that ## a_{n_{k}} \leq L ## ?
Let ## \{ a_{n} \} ## be a sequence.
Prove: If for all ## N \in { \bf{N} } ## there exists ## n> N ## such that ## a_{n} \leq L ## , then there exists a subsequence ## \{ a_{n_{k}} \} ## such that ##
a_{n_{k}} \leq L ##
My attempt:
Suppose that for all ## N \in {\bf{N}} ## there exists ## n> N ##, such that ## a_{n} \leq L ##. Since ## \{ a_{n} \} ## is a sequence, there exists a subsequence ## \{ a_{n_{k}} \}## such that there exists a monotonically increasing sequence of natural numbers ## \{ n_{k} \} ## such that ## \forall k \in {\bf{N}}, n_{k} \geq k ## . Let ## k \in {\bf{N}} ## be arbitrary so there exists ## n_{k} \geq k ##. Thus ( by universal instantiation of ## n_{k} ## in the statement " for all ## N \in {\bf{N}} ## there ... " ) there exists ## n > n_{k} > k ## such that ## a_{n} \leq L ##
However ,I don't know if ## a_{n_{k}} ## is less than or greater than ## a_{n} ## . So I'm stuck, how am I supposed to proceed from here to show that ## a_{n_{k}} \leq L ## ?