Quantifiers.

[bloodhound]

Here are two statments, the second one with a different order of the quantifiers. i understand what the first one says. can someone say what the second one means.

 

First

 

[math]\forall_{\varepsilon>0}\exists_{N\in\mathbb{N}}\forall_{n\ge N} |x-x_{n}|<\varepsilon[/math]

 

Second.

[math]\exists_{N\in\mathbb{N}}\forall_{\varepsilon>0}\forall_{n\ge N} |x-x_{n}|<\varepsilon[/math]

[matt grime]

In the first one we impliclty are saying that N depends on epsilon, in the second we are saying that N is independent of e, so in effect we are saying that x_n=x for all n>N

[bloodhound]

yes. my lecturer said the same thing. but i am still having trouble visualising it... to me it reads like the first one..

[matt grime]

Lets do an example, let x_n = 1/n, which converges to 0 and satisfies the first one.

 

is the second one of these true (using x=0):

 

there is an N such that for all e>0 and all n> N, |1/n| < e

 

 

given that this N exists (and is fixed), what if I now pick e=1/2N?

[bloodhound]

argghhhhh... still dont get it. doesnt that give u a contradiction, if u fix N and pick e=1/2N as that will give u statement

 

for all n>N , 1/n < 1/2N which is not true.

[matt grime]

Yes, so the idea that the first and second are the same we can clearly see is wrong by this example.

 

At the risk of simply repeating myself, the secoond one states that given a sequence x_n and some number x that

 

we can find AN integer N such that for all e>0 and all n>N we have that |x_n-x|<e

 

 

Consider the negation instead if that helps:

 

for all N there is an e>0 and an n>N such that |x_n-x| => e.

 

this implies the sequence is not eventually constant and equal to x since it states that no matter what N is ,there is a term after the N'th which is some positive distance from x. that is the sequence is not after any point always equal to x, agreed?

[bloodhound]

yes i got the last part of ur last post. I am still having troulbe understanding how the second statement implies the sequence is constant for n>N. what i am currently seeing ,looking at that statement , is that all of the x_n's are in the neighbourhood of x for n>N. but i dont see why x_n has to be constant.

[matt grime]

Eventually constant.

 

If x_n is not eventually constant at x, then i can pick an infinite subsequence y_m of the x_n such that non of the y_n are equal to x: if it isn't constant I can find a first place where x_n differs, and then a second and so on...

 

(so I can assume that non of the x_n are equal to x by passing to a subsequence if i wihsed)

 

Let N be any integer, and let e=|y_{N+1}-x|/2

 

Then for this N I know that the assertion: for all e>0 and for all n>N |y_n-x|<e is false.

 

but this N was completely arbitrary so i've shown that if the sequence is not eventually constant at x then the condition is false.

 

or equivalently if the codnition is true then the sequence is eventually constant at x.

[matt grime]

Or if you prefer you know that after N evey element of x_n is in EVERY neighbourhood of x, but the only way for that to happen is for all the x_n after the N'th to be equal to X. Note it is for ALL e>0

[bloodhound]

the last post m akes it much more clear. ill have to work on it... its a bit like nested intevals whose diameter tends to 0.

[matt grime]

Note, we're picking N first, right, so i can take a new sequence y_m where y_r=x_{N+r}

 

then that sequence has the property that for all e>0 and for all n |y_n_x|<e

[bloodhound]

ok. cheers a lot for the help.