Quantification

[Nine Divines]

Could someone provide an example how quantification that's not in the right order can mean something totally different

 

[fresh_42]

Consider the sequence [imath] \left(1,\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\ldots\right)=\left(2^{-n}\right)_{n\in \mathbb{N}} [/imath] which converges to zero. The proof is that [imath] \forall \,\varepsilon > 0 \,\,\exists \,N(\varepsilon)= \left[\log_2 \frac{1}{\varepsilon }\right]+1 \in \mathbb{N}[/imath] such that
[math] |2^{-n}-0 |=2^{-n}<2^{-N(\varepsilon )}=2^{\left[\log_2 \varepsilon \right]-1}< \varepsilon [/math] for all [imath] n>N(\varepsilon ). [/imath]
(The plus one is a safety margin since this ways we don't need to bother wether the rounding goes up or down.)

If we change the order of the quantifiers, we get [imath]\exists \,N \in \mathbb{N}\,\, \forall \,\varepsilon > 0\,\,\forall \, n>N \, : \,|2^{-n}|<\varepsilon [/imath] but such a number [imath] N [/imath] does not exist for [imath] \varepsilon = 2^{-N-2} [/imath] and [imath] n=N+1. [/imath]

 

[fresh_42]

The key is that in the first version, [imath] N [/imath] can be chosen according to a given value of [imath] \varepsilon [/imath] whereas the second version requires an [imath] N [/imath] that covers all cases.