Limit point of Sn := {1-1/n} is 1

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SUMMARY

The limit point of the sequence Sn defined as Sn := {1 - 1/n} is conclusively 1. The discussion emphasizes the prohibition of using epsilon-delta definitions, monotonicity, and boundedness in the proof. Instead, it relies on the properties of convergent sequences, specifically that for any open interval containing 1, there exists a positive integer N such that for all n ≥ N, Sn falls within that interval. The participants clarify that the assumption that for every a > 0, there exists a positive integer n such that n^{-1} < a is valid and is part of the foundational axioms.

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Homework Statement




Show that the limit point of Sn:={1-1/n} is 1.

Homework Equations



We are prohibited from using epsilon and delta

The Attempt at a Solution



Let Sn:= {1-1/n} and U be any open interval from (a,b) where a<1<b. Observe that Sn is always \leq 1. Since a<1 is linearly ordered, there is a positive number d between a and 1 such that a<d<1 and 1/d > 1. Then since d < 1,
d-1 < 1-1
d-1 < 0
(d-1)/d < 0 --> (d-1)/d = 1 - \frac{1}{d} and ...

i am stuck, it appears that i shown my Sn is always less then zero making my limit point zero not one. Any help would be appreciated.
 
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kingstrick said:

Homework Statement




Show that the limit point of Sn:={1-1/n} is 1.

Homework Equations



We are prohibited from using epsilon and delta

This appears to be an invitation to prove the result from basic properties of convergent sequences and already known limits.
 
Here are the definitions that we must use. We are prohibited also from using monotonic and bounded.definition:

The statement that the point sequence p1, p2, . . . converges to the point x means that if S is an open interval containing x then there is a positive integer N such that if n is a positive integer and n ≥ N then pn ∈ S.

and

The statement that the sequence p1, p2, p3, . . . converges means that there is a point x such that p1, p2, p3, . . . converges
 
kingstrick said:
Here are the definitions that we must use. We are prohibited also from using monotonic and bounded.


definition:

The statement that the point sequence p1, p2, . . . converges to the point x means that if S is an open interval containing x then there is a positive integer N such that if n is a positive integer and n ≥ N then pn ∈ S.

and

The statement that the sequence p1, p2, p3, . . . converges means that there is a point x such that p1, p2, p3, . . . converges

So, basically, you are using ε and N (not δ, of course, since n is not staying finite).
 
kingstrick said:
Here are the definitions that we must use. We are prohibited also from using monotonic and bounded.


definition:

The statement that the point sequence p1, p2, . . . converges to the point x means that if S is an open interval containing x then there is a positive integer N such that if n is a positive integer and n ≥ N then pn ∈ S.


Are you allowed to assume that for every a &gt; 0 there exists a positive integer n such that n^{-1} &lt; a, or do you have to prove that as well?
 
yes

pasmith said:
Are you allowed to assume that for every a &gt; 0 there exists a positive integer n such that n^{-1} &lt; a, or do you have to prove that as well?

yes. This was one of our axioms
 

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