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

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Homework Help Overview

The problem involves demonstrating that the limit point of the sequence Sn := {1 - 1/n} is 1, with specific constraints on the methods allowed for the proof.

Discussion Character

  • Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definitions of convergence and limit points, noting the prohibition against using epsilon and delta arguments, as well as monotonic and bounded definitions. Some express confusion about the implications of these restrictions on their reasoning.

Discussion Status

There is an ongoing exploration of the definitions and properties that can be used to approach the problem. Some participants have provided insights into the axioms that can be assumed, while others are questioning the implications of these assumptions on their attempts to prove the limit point.

Contextual Notes

Participants are required to adhere to specific definitions and constraints, which may limit traditional approaches to proving convergence. The discussion reflects a need to navigate these limitations while still attempting to establish the limit point.

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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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