Existence of sequence

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  • #1
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I asked this question before but I totally misunderstood what it was asking. Basically, I need to find that there exists a sequence {a_k} such that it converges to x for some x in R.

Since the real numbers are equivalence classes of convergent Cauchy sequences the result seems fairly obvious, but I was thinking of instead of dealing with equivalence classes I could use the definition of the limit of a sequence and have x be the limit of {a_k}.

I'm having some trouble starting up, though. Do I have to find a sequence that must converge to some arbitrary x to prove its existence?

Thanks!
 

Answers and Replies

  • #2
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Take the constant sequence [itex]a_k=x[/itex]. Surely, the sequence

[tex]x,~x,~x~,...[/tex]

converges to x...
 
  • #3
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Sorry, forgot to mention that the sequence is in Q.
 
  • #4
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Sorry, forgot to mention that the sequence is in Q.
Well, what can you go from?? Can you use that Q is dense in R?? Or must you use the definition of R?
 
  • #5
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does the fact that Q is dense in R tell me that if I take any member of {s_n} it can exist in an interval of, say, (x-1/n,x+1/n) ?
 
  • #6
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does the fact that Q is dense in R tell me that if I take any member of {s_n} it can exist in an interval of, say, (x-1/n,x+1/n) ?
Q being dense says that for each x in R, there exists a rational in (x-1/n,x+1/n). You can use this to construct a rational sequence.
 
  • #7
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Sorry, forgot to mention that the sequence is in Q.
Can you just take successive truncations of x's decimal expansion? For example

3, 3.1, 3.14, 3.141, 3.1415, ... is a sequence of rationals converging to pi.
 

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