# Equivalent definitions of convergence

1. Sep 11, 2013

### stripes

1. The problem statement, all variables and given/known data

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

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

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

2. Relevant equations

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3. 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!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 11, 2013

### vela

Staff Emeritus
If $\lvert x_n - x \rvert \le \rho$, then surely $\lvert x_n - x \rvert < \rho$ is satisfied, right?

3. Sep 11, 2013

### stripes

That information is helpful to show D2 ==> D1 right?

Last edited: Sep 11, 2013
4. Sep 13, 2013

### brmath

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)... $$|x - x_n| < \epsilon$$.

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: Sep 13, 2013
5. Sep 14, 2013

### stripes

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!

6. Sep 14, 2013

### brmath

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.

7. Sep 14, 2013

### stripes

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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