3
$\begingroup$

Def. Convergent Sequence

We say that a sequence $\{a_j\}$ in $\mathbb{R}^n$ converges to the limit $L∈\mathbb{R}^n$, if $∀ε>0,∃J>0$ such that if $j≥J$ then $|a_j−L|<ε$.

(I think that might be a mathematical def. which based on FOL)

I'm trying to understand what this means, is it saying:

$$\forall\varepsilon>0,\exists J> 0,s.t.(j\ge J\rightarrow|a_j−L|<ε)$$


Update: $\color{lightgrey}{\text{(Thanks @Michael to point out, there are some mistakes in those steps)}}$

Steps:

apply $p\rightarrow q\Leftrightarrow \neg p\vee q$

$$\forall\varepsilon>0,\exists J> 0,(j<J\vee |a_j−L|<ε)$$

distribute the quantifiers $\color{lightgrey}{\text{(which is not valid)}}$

$$(\forall\varepsilon>0,\exists J> 0,s.t. j< J)\vee (\forall\varepsilon>0,\exists J> 0,s.t. |a_j−L|<ε)$$

Therefore $$(\exists J> 0, s.t. j< J)\vee (\forall\varepsilon>0, |a_j−L|<ε)$$


equivalent to

$$\color{blue}{(\forall J> 0,j\ge J)}\rightarrow \color{orange}{\forall\varepsilon>0,|a_j−L|<ε}$$

$\underline{\text{What's the quantifier for $j$ here}}$


My thoughts

Since $j$ is an index, which we might have $j\in\mathbb{N}$, the definition might want to say when $j$ goes to $\infty$, something happens

I'll try to write the def. by myself first maybe, which might help to understand it, I guess it want to say something like

$$j\text{ approach } \infty\rightarrow a_j=L$$

$$\Leftrightarrow \color{blue}{j \text{ approach } \infty}\rightarrow \color{orange}{a_j-L=0}$$

I think the orange part are equivalent, so we have

$$\forall\varepsilon>0,|a_j−L|<ε\leftrightarrow a_j-L=0$$

The blue part might be equivalent, then we have

$$\forall J> 0,j\ge J\leftrightarrow j \text{ approach } \infty$$

If this is the case, the definition is somehow making sense

But I don't understand how is $(\forall J> 0,j\ge J\leftrightarrow j \text{ approach } \infty)$ hold

What the word approach really means, i know that $$\forall c\in\mathbb{R},c<\infty$$

And I can understand that

$$(\forall J> 0,j>J)\leftrightarrow j\equiv\infty$$

Since $\forall J>0, \infty\neq J$

So in the blue part, I think it's fine to replace $\ge$ with $>$, we have

$$\forall J>0,j> J\rightarrow a_j=L$$

$$\Leftrightarrow j\equiv \infty\rightarrow a_j=L$$

However, I personally feel there is some difference between, $j\equiv\infty$ and $j$ approach to $\infty$

And if this is the case, what I wrote above would make no sense.

Then I will stop for now $\dots$

Could someone expain this to me in details.

Thanks for your help.

$\endgroup$
  • 3
    $\begingroup$ The way to write it formally would be $$ (\forall \epsilon>0)(\exists J\in \mathbb N)(\forall j\ge J)(|a_j-L|<\epsilon)$$ (Still this isn't quite in bare first order logic since we're using abbreviations in "bounding" the quantifiers. For instance $(\forall \epsilon>0) X$ is an abbreviation for $\forall \epsilon(\epsilon>0\to X)$ But it quickly becomes hard to read.) $\endgroup$ – spaceisdarkgreen Oct 10 '19 at 0:49
  • 1
    $\begingroup$ This first thing is the definition: $$\forall\varepsilon>0,\exists J> 0,s.t.(j\ge J\rightarrow|a_j−L|<ε)$$ This second thing is certainly not equivalent: $$\color{blue}{(\forall J> 0,j\ge J)}\rightarrow \color{orange}{\forall\varepsilon>0,|a_j−L|<ε}$$ why are you trying to move the parentheses? If $|a_j-L|<\epsilon$ for all $\epsilon>0$ then $a_j=L$. And stand-alone $(\forall J >0, j \geq J)$ seems to mean that you have $j=\infty$. $\endgroup$ – Michael Oct 10 '19 at 1:13
  • $\begingroup$ Agree, I changed the order of the quantifier which might be a mistake @Michael $\endgroup$ – Manx Oct 10 '19 at 1:27
  • $\begingroup$ It does seem pertinent to specify what sets the variables come from. J and j are naturals. epsilon is a real. The sequence a_j and the limit L dont have to be reals, or even numbers, as long as a suitable metric defined by the absolute values is chosen. But that might be getting a little too topological. $\endgroup$ – SquishyRhode Oct 10 '19 at 2:43
  • $\begingroup$ I see, make sense @MauroALLEGRANZA $\endgroup$ – Manx Oct 10 '19 at 7:31
1
$\begingroup$

Def. Convergent Sequence

We say that a sequence $\{a_j\}$ in $\mathbb{R}^n$ converges to the limit $L∈\mathbb{R}^n$, if $∀ε>0,∃J>0$ such that if $j≥J$ then $|a_j−L|<ε$.

What's the quantifier for $j$ here ?

We need an universal quantifier for $j$ :

$$∀ε>0 \ ∃J>0 \ ∀j \ (j≥J→|a_j −L|<ε).$$


The symbol $\lim_{n \to \infty}$ is a symbol that we cannot "split" into parts: there is no $\infty$ value such that $n=\infty$.

See e.g. Terence Tao, Analysis I (Springer, 3rd ed. 2016), page 129 :

We sometimes use the phrase “$a_n \to x$ as $n \to \infty$” as an alternate way of writing the statement “$(a_n)^{\infty}_{n=m}$ converges to $x$”. Bear in mind, though, that the individual statements $a_n \to x$ and $n \to \infty$ do not have any rigorous meaning; this phrase is just a convention, though of course a very suggestive one.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.