Every lingering sequence has a convergent subsequence/Proof
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< Every lingering sequence has a convergent subsequence
Revision as of 17:26, 6 December 2015 by Alec (Talk | contribs) (Created page with "{{Todo|Write proof}} '''Proof outline:''' # Take {{M|k_1}} to be the index of any point of the sequence in {{M|B_1(x)}} # Take {{M|k_2}} to be any index AFTER {{M|k_1}} of the...")
TODO: Write proof
Proof outline:
- Take [ilmath]k_1[/ilmath] to be the index of any point of the sequence in [ilmath]B_1(x)[/ilmath]
- Take [ilmath]k_2[/ilmath] to be any index AFTER [ilmath]k_1[/ilmath] of the sequence in the ball [ilmath]B_\frac{1}{2}(x)[/ilmath]
- ...
- Show the sequence [ilmath](x_{k_n})_{n=1}^\infty[/ilmath] converges to [ilmath]x[/ilmath]
We have exhibited a convergent subsequence, we're done.
