# Is the natural numbers implicit in the statement?

1. Aug 9, 2009

### jaejoon89

In class we defined convergence as

\forall\varepsilon>0\;\;\;\exists\mathrm{N}\epsilon\mathbf{N}\;\;\;\forall\mathrm{n}\geq\mathrm{N}\;\;\;\left|a_{n} \right|<\varepsilon

So if a sequence {a_n} of real numbers converge to 0 if for every ε > 0 there is N s.t. |x_n| < ε for n ≥ N...

Is the natural numbers implicit in the statement? or do you have to change it somehow for the reals? How would you use the symbolic logic to negate it? (i.e. if it did not converge to 0)?

So... to understand better... when it is converging to a particular number, is that whatever the epsilon is greater than?

2. Aug 9, 2009

### snipez90

Re: Convergence

Note you need to add latex tags around a formula to express it correctly. Click on the formula in the quote box below for an example.

In the statement, n and N are implicitly understood to be natural numbers. A sequence is a function from the naturals to the reals, so n really serves as an index. The values x_n are real numbers and the sequence itself could be entirely arbitrary, without any discernible pattern.

A sequence {x_n} does not converge to L if there exists an $$\varepsilon > 0$$ such that for all N (N a natural number), there exists n > N such that $$|x_n - L| \geq \varepsilon$$.

When the sequence is converging to a particular number L, then we want the distance between the terms of our sequence and the number L to be arbitrarily small provided that our index is sufficiently large. On the real line, this distance is |x_n - L|. So we want |x_n - L| to be less than an arbitrarily chosen epsilon provided that n is sufficiently large.

3. Aug 9, 2009

### jaejoon89

Re: Convergence

I don't understand why a sequence has to be from the naturals to the reals. Can't you have a complex sequence? Shouldn't it be specified somewhere that here {x_n} contains real numbers?

Also, if you were negating the statement (saying that it doesn't converge to a particular number) which part would you negate?

4. Aug 9, 2009

### snipez90

Re: Convergence

I never said that a sequence had to be from the naturals to the reals. From the definition you gave (which only specifies convergence to 0), I assumed that the sentence right after that was a restatement and took {a_n} to be a real sequence, as you specified.

Yes, you can have a complex sequence. The definition is essentially the same except we take absolute values to mean the modulus of a number, which specifies the distance between two complex numbers on the complex plane.

What do you mean which part would be negated?

5. Aug 9, 2009

### Elucidus

Re: Convergence

In general a sequence can be defined to be from any well-ordered set to any set.

i.e. $s: T \rightarrow S$ is a sequence if T is well-ordered.

$\mathbb{Z}^+, \mathbb{Q}^+ \text{ and } \mathbb{R}^+$ are examples of well-ordered sets.

But, typically sequences are defined as functions on either $\mathbb{Z}^+ \text{ or } \mathbb{N}$ or close variations of these. And additionally one's first exposure is real-valued sequences, but this is not required - complex-valued sequences are entirely possible and are frequently studied in more advanced courses.

Convergence as you've described it is the metric form. The codomain of the sequence has to have some concept of "distance" between two points (called a metric - and such a set is called a metric space). Since both the real numbers and complex numbers have such a metric, convergence of sequences is definable is such sets.

Assuming $\{a_n\}_{n=0}^\infty$ is an M-valued sequence, where M is a set where for all u, v in M, the distancde between u and v is denoted |u - v| then the definitions for convergence (to a limit) and divergence are:

CONVERGENCE
The sequence $\{a_n\}_{n=0}^\infty$ is said to converge to a limit L (in M) if for any given $\varepsilon > 0$ there exists a natural number N so that for all n > N we have $|a_n - L|<\varepsilon$.

Using quantifiers one has:

$$(\exists L \in M)(\forall\varepsilon>0)(\exists N\in\mathbb{N})(\forall n>N)(|a_n - L|<\varepsilon)$$

DIVERGENCE
The sequence $\{a_n\}_{n=0}^\infty$ is said to diverge if for any L in M there exists an $\varepsilon > 0$ such that for all natural numbers N there exists n0 > N so that $|a_{n_0} - m| \geq \varepsilon$.

Using quantifiers:

$$(\forall L \in M)(\exists \varepsilon >0)(\forall N \in \mathbb{N})(\exists n_0>N)(|a_{n_0}-L|\geq\varepsilon)$$

One can derive the other using the logical identities:

$\neg(\forall x) P(x) = (\exists x)(\neg P(x))$
$\neg(\exists x)P(x) = (\forall x)(\neg P(x))$

Side note: it is also possible to define convergence in a topological space (without a metric):

A sequence $\{a_n\}_{n=0}^\infty$ is said to converge to L in S (a topological space) if for every open set U containing L there exists a natural number N such that for all n > N we have $a_n \in U$.

--Elucidus

6. Aug 12, 2009

### jaejoon89

Re: Convergence

Thanks, but my question just has to do with understanding the terminology not actually showing whether it's convergent or not.

So if a sequence {a_n} of real numbers converge to 0 if for every ε > 0 there is N s.t. |x_n| < ε for n ≥ N...

If you want to use quantifiers to write the statement "{a_n} converge to 0", is it

$$\forall\varepsilon>0\;\;\;\exists N\in\mathbb{N}\;\;\;\forall\mathrm{n}\geq\mathrm{N}\; \;\;\left|a_{n} \right|<\varepsilon$$

You don't need to specify that it is of real numbers? The natural numbers in the definition is still ok here?

Also, then to negate it and say the statement "{a_n} doesn't converge to 0", is it

$$\forall\varepsilon>0\;\;\;\exists\mathrm{N}\epsilo n\mathbb{N}\;\;\;\forall\mathrm{n}\geq\mathrm{N}\; \;\;\left|a_{n} \right|\not<\varepsilon$$

Not showing that it doesn't converge to 0, just understanding how to write the statement using formalism...

7. Aug 12, 2009

### jaejoon89

Re: Convergence

Sorry, that should be N epsilon Natural numbers, I don't know why it isn't printing...

8. Aug 12, 2009

### Staff: Mentor

Re: Convergence

To negate a proposition with for any/for all/for each and "there exists" quantifiers, you also have to switch "for any" quantifiers to "there exists" and vice-versa.

For example, to negate that statement "For every real number x >= 2, x2 + x - 6 >= 0",
you would get
"There exists a real number x >= 2 such that x2 + x - 6 < 0"

9. Aug 12, 2009

### jaejoon89

Re: Convergence

Why isn't it... "There exists a real number x < 2..."

I thought you have to negate the proposition, too.

---

So for not converging to 0, is it

there exists epsilon greater than 0
for all N in N (negating the N in N would be weird, so that can't be right)
there exists n less than N (?)
|a_n| >= epsilon

10. Aug 12, 2009

### Staff: Mentor

Re: Convergence

You don't negate everything in the statement. Here's the whole technique, taken from "How to Read and Do Proofs 2/ed" by Daniel Solow, pp. 126 and 127.
1. Put the word NOT in front of the entire statement.
2. If the word NOT appears to the left of a quantifier, move the word NOT to the right of the quantifier and place it just before the something that happens. As you do so, change the quantifier to its opposite--"for all" becomes "there is" and vice versa.
3. When all the quantifiers appear to the left of the NOT, elimiate the NOT by incorporating it into the statement that appears immediately to its right.

Using the example from before, the steps are:
1. NOT[For every real number x >= 2, x2 + x - 6 >= 0]
2. There is a real number x >= 2 such that NOT[x2 + x - 6 >= 0]
3. There is a real number x >= 2 such that x2 + x - 6 < 0]

The quantifiers referred to here are the "for every" and "there exists," not the specific property (e.g. x >= 2) that the quantifier applies to.
Hope that's clear.

11. Aug 12, 2009

### Elucidus

Re: Convergence

To show that something does not converge to a specific number (in your case, 0) then L is already specified. If you scroll back to the definition of Divergence, just let L = 0 and remove the first Quantifier (The "For all L.."). You should be good to go.

--Elucidus