What is the Intuitive Explanation for the Definition of Convergence?

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The discussion centers on the definition of convergence for sequences, highlighting confusion over its completeness. Convergence is defined by the condition that for every ε > 0, there exists a natural number N such that for all n > N, the absolute difference between the sequence term and the limit is less than ε. An example provided is the sequence a_n = 1/n, which converges to 0 as n approaches infinity, illustrating the definition through a proof. The conversation also explores the intuitive understanding of convergence using the concept of open balls, emphasizing that all but a finite number of sequence elements must fit within any chosen neighborhood around the limit. This geometric perspective aids in grasping the epsilon definition and its broader topological implications.
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I'm a bit confused about how my book defines convergence.

Definition: A sequence {an} convergences to l if for every ε > 0 there is a natural number N such that, for all natural numbers n, if n > N, then l a,-l l < ε

note, l a,-l l = the absolute value

Maybe someone could give me an example? The definition seems incomplete. This is essentially like an epsilon-delta proof for limits, right?
 
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It's quite similar to epsilon-delta proofs, except there's no delta because the domain of a sequence is limited to integer values of ##n##, so there is no notion of choosing points in the domain arbitrarily close to some ##n##. But we can still talk about what happens when ##n## becomes large.

A simple example would be ##a_n = 1/n##. This converges to the limit ##L = 0## as ##n \rightarrow \infty##. To prove this using the definition, let ##\epsilon > 0##. Since ##\epsilon## is positive, I can get a number as large as I like by multiplying ##\epsilon## by a sufficiently large integer. In particular, there is some integer ##N## for which ##N \epsilon > 1##. Dividing both sides by ##N##, this is equivalent to ##\frac{1}{N} < \epsilon##. Furthermore, for any ##n \geq N##, we have ##\frac{1}{n} \leq \frac{1}{N} < \epsilon##. Therefore,
$$|a_n - L| = \left| \frac{1}{n} - 0 \right| = \left| \frac{1}{n} \right| = \frac{1}{n} < \epsilon$$
for all ##n \geq N##. We conclude that ##\lim_{n \rightarrow \infty} a_n = L##, i.e. ##\lim_{n \rightarrow \infty} \frac{1}{n} = 0##.
 
Thanks, I understand the how but I'm still a bit lost on the why. I'll review epsilon delta proofs again.

I understand the mechanism now, but the intuition is evading me. Hm, I'll have to give it some more thought.
 
One intuition I like to use is to think of it in terms of open balls (in fact in topology, where you have no metric, the definition of convergence is given entirely in terms of neighborhoods and it makes it more intuitive imo). Let's say our sequence is ##(x_n)## and ##x_n\rightarrow x\in \mathbb{R}##. So what is that definition of convergence really saying?

Well first take any ##\epsilon >0## and consider the open ball ##B(x,\epsilon )##. We should be able to fit all but a finite number of elements of the sequence into this open ball i.e. there should exist an ##N\in \mathbb{N}## such that for all ##n\geq N##, ##x_n\in B(x,\epsilon )##. So what does it really mean for this sequence to converge to ##x## then? It means that no matter how small an open ball you make around ##x##, I can always fit in all but a finite number of elements of the sequence into this open ball. So you can picture making the open ball smaller and smaller and smaller still but always being able to fit in all but a finite number of said elements into the open ball. This, for me, makes it geometrically clear what the usual epsilon definition is saying. It also motivates the more general topological definition.
 
WannabeNewton said:
One intuition I like to use is to think of it in terms of open balls (in fact in topology, where you have no metric, the definition of convergence is given entirely in terms of neighborhoods and it makes it more intuitive imo). Let's say our sequence is ##(x_n)## and ##x_n\rightarrow x\in \mathbb{R}##. So what is that definition of convergence really saying?

Well first take any ##\epsilon >0## and consider the open ball ##B(x,\epsilon )##. We should be able to fit all but a finite number of elements of the sequence into this open ball i.e. there should exist an ##N\in \mathbb{N}## such that for all ##n\geq N##, ##x_n\in B(x,\epsilon )##. So what does it really mean for this sequence to converge to ##x## then? It means that no matter how small an open ball you make around ##x##, I can always fit in all but a finite number of elements of the sequence into this open ball. So you can picture making the open ball smaller and smaller and smaller still but always being able to fit in all but a finite number of said elements into the open ball. This, for me, makes it geometrically clear what the usual epsilon definition is saying. It also motivates the more general topological definition.

Aha, I think I get you. Although I had to look up what a "ball" was. Many thanks!
 
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