- #1
wayneckm
- 68
- 0
Hello all,
I come across this question while reading a proof.
Given a set [tex] A [/tex] with some property related to [tex] f [/tex] which is a function and an element [tex] b [/tex], they may begin with, in the proof, choosing a sequence [tex] a_{n} \in A [/tex] such that [tex] f(a_{n} b) \rightarrow f(b) [/tex], then they keep going until they prove the limit [tex] a_{\infty} [/tex] exists and is also inside [tex] A [/tex].
So how and why can they start with extracting a sequence [tex] a_{n} [/tex] with the desired limiting property, [tex] f(a_{n} b) \rightarrow f(b) [/tex]? I am wondering how and why they can be sure such kind of sequence can be found? Or indeed under what kind of conditions (closure or what else?) this is guaranteed to be true? Apparently this kind of argument/technique appears quite often in some proofs.
Thanks.
Wayne
I come across this question while reading a proof.
Given a set [tex] A [/tex] with some property related to [tex] f [/tex] which is a function and an element [tex] b [/tex], they may begin with, in the proof, choosing a sequence [tex] a_{n} \in A [/tex] such that [tex] f(a_{n} b) \rightarrow f(b) [/tex], then they keep going until they prove the limit [tex] a_{\infty} [/tex] exists and is also inside [tex] A [/tex].
So how and why can they start with extracting a sequence [tex] a_{n} [/tex] with the desired limiting property, [tex] f(a_{n} b) \rightarrow f(b) [/tex]? I am wondering how and why they can be sure such kind of sequence can be found? Or indeed under what kind of conditions (closure or what else?) this is guaranteed to be true? Apparently this kind of argument/technique appears quite often in some proofs.
Thanks.
Wayne