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 $\left(1,\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\ldots\right)=\left(2^{-n}\right)_{n\in \mathbb{N}}$ which converges to zero. The proof is that $\forall \,\varepsilon > 0 \,\,\exists \,N(\varepsilon)= \left[\log_2 \frac{1}{\varepsilon }\right]+1 \in \mathbb{N}$ such that
$$|2^{-n}-0 |=2^{-n}<2^{-N(\varepsilon )}=2^{\left[\log_2 \varepsilon \right]-1}< \varepsilon$$ for all $n>N(\varepsilon ).$
(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 $\exists \,N \in \mathbb{N}\,\, \forall \,\varepsilon > 0\,\,\forall \, n>N \, : \,|2^{-n}|<\varepsilon$ but such a number $N$ does not exist for $\varepsilon = 2^{-N-2}$ and $n=N+1.$

 

[fresh_42]

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