Equivalent definitions of convergence

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Homework Help Overview

The discussion revolves around the equivalence of two definitions of convergence for sequences, specifically focusing on the formal statements D1 and D2. Participants are exploring how to demonstrate that D1 implies D2 and vice versa, while grappling with the nuances of strict versus non-strict inequalities in the context of convergence.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Some participants attempt to show the implication from D1 to D2 by manipulating the definitions, but express uncertainty about how to handle the inequalities correctly. Others question whether the strict inequality in D1 can be reconciled with the non-strict inequality in D2. There are also discussions about the implications of choosing different values for M and K in the definitions.

Discussion Status

Participants are actively engaging with the problem, sharing their thoughts and attempts. Some have offered insights that may help clarify the relationship between the definitions, while others express frustration at their inability to progress. There is a recognition of the complexity involved in the definitions, but no explicit consensus has been reached regarding the proof.

Contextual Notes

Participants note the challenge of working with definitions that involve both strict and non-strict inequalities, and some express concern about the timing of their inquiries relative to assignment deadlines. There is an acknowledgment of the potential for confusion in the problem's formulation.

stripes
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Homework Statement



\mathbf{D1:}\forall\varepsilon>0,\exists K\epsilon\mathbb{N},\forall n\epsilon\mathbb{N},n\geq K\Longrightarrow|x_{n}-x|<\varepsilon

\mathbf{D2:}\forall\rho>0,\exists M\epsilon\mathbb{N},\forall n\epsilon\mathbb{N},n>M\Longrightarrow|x_{n}-x|\leq\rho

Show these two definitions of convergence (of a sequence xn) are equivalent by showing both D1 implies D2, and D2 implies D1.

Homework Equations



--

The Attempt at a Solution



To show D1 ==> D2, i just took M = K - 1, where K was from D1. But it didn't seem to get me anywhere since i can show

n>M\Longrightarrow|x_{n}-x|< \rho,

but this does not encompass the possibility of ≤ ρ. So basically i have no idea how to do this.

Similarly, i need something to get me started on D2 ==> D1. How do I go about this?

I'm sure I'm overthinking it as I've done much more complicated proofs before, and I've done all other questions on this assignment without help. I just honestly don't know how to do this one. Thanks in advance!


 
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stripes said:

Homework Statement



\mathbf{D1:}\forall\varepsilon>0,\exists K\epsilon\mathbb{N},\forall n\epsilon\mathbb{N},n\geq K\Longrightarrow|x_{n}-x|<\varepsilon

\mathbf{D2:}\forall\rho>0,\exists M\epsilon\mathbb{N},\forall n\epsilon\mathbb{N},n>M\Longrightarrow|x_{n}-x|\leq\rho

Show these two definitions of convergence (of a sequence xn) are equivalent by showing both D1 implies D2, and D2 implies D1.

Homework Equations



--

The Attempt at a Solution



To show D1 ==> D2, i just took M = K - 1, where K was from D1. But it didn't seem to get me anywhere since i can show

n>M\Longrightarrow|x_{n}-x|< \rho,

but this does not encompass the possibility of ≤ ρ. So basically i have no idea how to do this.

Similarly, i need something to get me started on D2 ==> D1. How do I go about this?

I'm sure I'm overthinking it as I've done much more complicated proofs before, and I've done all other questions on this assignment without help. I just honestly don't know how to do this one. Thanks in advance!
If ##\lvert x_n - x \rvert \le \rho##, then surely ##\lvert x_n - x \rvert < \rho## is satisfied, right?
 
That information is helpful to show D2 ==> D1 right?
 
Last edited:
These are definitions of convergence as n \rightarrow \infty of the sequence x_n to a particular value x. The concept is that the bigger n gets, the closer x_n gets to x.

Now to state it formally we have to choose an \epsilon, which supposedly is a small number like 1/n or 1/n^2 and show that if n is big enough

(1)... |x - x_n| &lt; \epsilon.

When we say n is big enough we mean that the inequality (1) is true for n and every higher integer -- n+1, n+2,... etc. In other words, once we pick the \epsilon there is some point from M on where the inequality holds.

Of course, we said any \epsilon so if you make it smaller you'll need a bigger M (or K or whatever). The important thing is that once you pick the \epsilon you can always find some M that works.

Note we said there is some point M, not that M is necessarily the smallest cutoff that will work. If you're not sure M is adequately large, you can always pick a bigger one. So the difference between saying there is an M such that n \ge M makes (1) work, and there is a K such that n> K makes (1) work is almost non-existent. Just pick an N which is bigger than both M and K and let n be larger than that, and it won't matter whether you write < or less than or =.

Once you understand why convergence is defined in terms of n > N, it really doesn't matter whether you have strict inequality or not.

I hope you can use these thoughts to construct a proof your teacher will like.

If I may editorialize, I think this is a poor choice of problem for a student. The strict or not strict inequality is essentially meaningless, and takes your eye off the ball, which is what convergence really means. However, do what you have to to get a good grade.
 
Last edited:
Thanks brmath. I do understand this on an intuitive level. I'm not particularly great at this stuff but I've taken a few junior and even a senior analysis/proof oriented course before, and I have proved much more complex stuff. That's why I'm surprised I couldn't get started here. But I definitely understand and your explanation helped.

Because I asked this question the night before it was due, I didn't really have time to do much so i just threw down an "explanation". Hopefully I'll get something out of it, but the remainder of the assignment was completed, so I hope it will not affect my grade too much.

Thanks!
 
I remember being a student and doing things the night before they were due. Computer programming cured me of that (missed deadlines where money was involved). My theory became that if I start early and get something done early, this could not possibly hurt. If I start early and something goes wrong (which is practically always) then there is time to recover. I hope your explanation will be satisfactory to your teacher.
 
Normally i too start my assignments late (i.e., i start my assignments the night before). this one, however, i started early :) hopefully my habits change and continue this way this semester.
 

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