Convergence of Sequences in Bounded Sets: Is the Limit Always the Supremum?

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In summary, the conversation discusses the existence of a sequence (Xn) with X(n) belonging to an infinite subset A of R, where R is bounded above and u:= sup A. It is shown that by the subsequence theorem, if A converges to u, then any sequence belonging to it will also converge to u. The conversation also mentions another theorem stating that the limit is equal to the supremum. Finally, it is proposed that by choosing a sequence of values that are strictly increasing and tending to sup(A), we can show that u = lim(Xn).
  • #1
semidevil
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let A be an infinite subset of R and R is bounded above, and u:= sup A. show that there exist a sequence (Xn) with X(n) belongs to A, such that u = lim(Xn).


ok, so suppose that there does exist a sequence X(n) in A. We know that SupA = u. by the subsequence theorem, if A converges to u, then so will any sequence that belongs to it right? and by another theorem, the limit is the supremum...correct?

I don't know...maybe too easy? I feel I didn't cover everything
 
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  • #2
semidevil said:
let A be an infinite subset of R and R is bounded above, and u:= sup A. show that there exist a sequence (Xn) with X(n) belongs to A, such that u = lim(Xn).


ok, so suppose that there does exist a sequence X(n) in A. We know that SupA = u. by the subsequence theorem, if A converges to u, then so will any sequence that belongs to it right?

Are you sure this is what the subsequence theorem says?
 
  • #3
Let e>0, then there exists a(1) in A such that supA - e <= a(1) <=Sup(A)

Now let e be one half SupA -a(1)

Pick and a(2) in the range sup(A) - e to Sup A.

repeat and get a strictly increasing sequence tending to sup(A).
 

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