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

• kingstrick
In summary: in class, that for any positive real number a, there exists a positive integer n such that n^{-1} < a.
kingstrick

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.

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 $a > 0$ there exists a positive integer $n$ such that $n^{-1} < a$, or do you have to prove that as well?

yes

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

yes. This was one of our axioms

1. What is the definition of a limit point?

A limit point is a point in a set such that every neighborhood of the point contains infinitely many other points from the set.

2. How do you determine the limit point of a sequence?

To determine the limit point of a sequence, you can either use the formula for limit points or graph the sequence to observe the behavior as n approaches infinity.

3. What is the formula for finding the limit point of a sequence?

The formula for finding the limit point of a sequence is lim n→∞ Sn, where Sn is the nth term in the sequence.

4. Why is the limit point of Sn := {1-1/n} equal to 1?

The limit point of Sn := {1-1/n} is equal to 1 because as n approaches infinity, the sequence becomes {1-1/∞}, which simplifies to just 1. Therefore, 1 is the limit point of the sequence.

5. How is the limit point of a sequence related to the concept of convergence?

The limit point of a sequence is closely related to the concept of convergence. If a sequence has a limit point, it means that the sequence is converging to that point. However, if a sequence does not have a limit point, it does not necessarily mean that the sequence is diverging.

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