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Borg7 Reader of Souls

Joined: 20 Nov 2009 Posts: 942 Skype: borg177
Location: Germoney
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Posted: Fri Nov 25, 2011 9:02 pm Post subject: Subsequences |
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Be (X,d) a metric space and (x_n)_n€N a sequece in X. Prove: (x_n)_n€N converges when the subsequences (x_2n)_n€N, (x_2n+1)_n€N and (x_3n)_n€N converge.
Any ideas? _________________ "I dont believe in anything other than alians"
Dominic Heydon |
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aggsyb Reader of Souls

Joined: 27 Apr 2010 Posts: 886 Skype: ben.agg
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Posted: Fri Nov 25, 2011 9:34 pm Post subject: |
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| Yea stop trying to make yourself look so smart for ego. Your making yourself look gay |
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Borg7 Reader of Souls

Joined: 20 Nov 2009 Posts: 942 Skype: borg177
Location: Germoney
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Posted: Fri Nov 25, 2011 9:38 pm Post subject: |
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Lol, I need to hand in the paper soon but idk how to solve this...it's prob pretty simple for some1 who's good at math _________________ "I dont believe in anything other than alians"
Dominic Heydon |
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Brokerstar This Place Is My Second Home

Joined: 12 Aug 2009 Posts: 2958
Location: England
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aggsyb Reader of Souls

Joined: 27 Apr 2010 Posts: 886 Skype: ben.agg
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Posted: Fri Nov 25, 2011 9:41 pm Post subject: |
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| I think the biggest problem though is that your still coming accross as gay, and its common knowledge that gay people are not liked and in some countries not tolerated. Dont go travelling |
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Borg7 Reader of Souls

Joined: 20 Nov 2009 Posts: 942 Skype: borg177
Location: Germoney
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Posted: Fri Nov 25, 2011 9:45 pm Post subject: |
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| Brokerstar wrote: |
'good at math' or math genius? |
do you remember the sn of the mathematician guy on here? _________________ "I dont believe in anything other than alians"
Dominic Heydon |
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chesslw Reader of Souls
Joined: 17 Jan 2011 Posts: 997 Skype: chesslw
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Posted: Sat Nov 26, 2011 11:33 am Post subject: |
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I might be able to answer this, or at least give you a hint. It really depends how much you have "proved" so far.
But if you know that any subsequence of a convergent sequence converges to the same limit (or if not then use for all epsilon there exists N etc etc it isn't too hard), then consider:
X_6n which is a subsequence of X_2n and X_3n, therefore X_2n and X_3n converges to the same limit. Similarly X_6n+3 is a subsequence of X_2n+1 and X_3n, and so X_2n+1 and X_3n have the same limit.
Therefore all 3 sequences have the same limit, x say, so the whole sequence converges to the same limit.
Of course you should use formal notation, i.e. epsilon-N notation.
Hope that helps!
Edit- missed the v. important bolded part the first time... |
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BetMagicMoney Forum Moderator & Spam Police

Joined: 22 Mar 2010 Posts: 3589 Skype: samuel.woodhams
Location: This Place Is My First Home
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Posted: Sat Nov 26, 2011 11:44 am Post subject: |
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feck knows, i like to think i know math but this is just some high level sh1t bro  _________________ "When everything seems to be going against you, remember that the airplane takes off against the wind, not with it."
Borg7
"Lol he'll get there tho. He is a smart guy"
_red_dog
follow me on twitter @BetMagicMoney |
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aggsyb Reader of Souls

Joined: 27 Apr 2010 Posts: 886 Skype: ben.agg
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Posted: Sat Nov 26, 2011 12:05 pm Post subject: |
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| chesslw wrote: | I might be able to answer this, or at least give you a hint. It really depends how much you have "proved" so far.
But if you know that any subsequence of a convergent sequence converges to the same limit (or if not then use for all epsilon there exists N etc etc it isn't too hard), then consider:
X_6n which is a subsequence of X_2n and X_3n, therefore X_2n and X_3n converges to the same limit. Similarly X_6n+3 is a subsequence of X_2n+1 and X_3n, and so X_2n+1 and X_3n have the same limit.
Therefore all 3 sequences have the same limit, x say, so the whole sequence converges to the same limit.
Of course you should use formal notation, i.e. epsilon-N notation.
Hope that helps!
Edit- missed the v. important bolded part the first time... |
So hold on a sec, what your saying is it has NOTHING to do with being like and or actually being gay? |
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U Cook Socks This Place Is My Second Home

Joined: 13 Mar 2010 Posts: 2756
Location: Walsall
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Posted: Sat Nov 26, 2011 12:16 pm Post subject: |
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| aggsyb wrote: | | chesslw wrote: | I might be able to answer this, or at least give you a hint. It really depends how much you have "proved" so far.
But if you know that any subsequence of a convergent sequence converges to the same limit (or if not then use for all epsilon there exists N etc etc it isn't too hard), then consider:
X_6n which is a subsequence of X_2n and X_3n, therefore X_2n and X_3n converges to the same limit. Similarly X_6n+3 is a subsequence of X_2n+1 and X_3n, and so X_2n+1 and X_3n have the same limit.
Therefore all 3 sequences have the same limit, x say, so the whole sequence converges to the same limit.
Of course you should use formal notation, i.e. epsilon-N notation.
Hope that helps!
Edit- missed the v. important bolded part the first time... |
So hold on a sec, what your saying is it has NOTHING to do with being like and or actually being gay? |
I'm fairly certain it does to be honest. _________________ "Cos that's what I do" |
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