Negation of Content 0

Question

I would like to know how to negate the definition of the following (Spivak, Calculus on Manifolds):

A subset A of $R^n$ has (n-dimensional) content 0 if for every $\epsilon >0$ there is a finite cover {$U_1 ,..., U_n $} of A by closed rectangles such that $$\sum_{i=1} ^n v(U_i) < \epsilon.$$

A small breakdown of these compound negations would also be helpful! Thanks !

Answer 1

To negate a sentence with quantifiers, you just need to flip the type of each one - "for every" becomes "there exists", and "there exists" becomes "for every". Doing this one step at a time:

A subset $A$ of $\mathbb{R}^n$ does not have $n$-dimensional content $0$ if it is not the case that for every $\epsilon > 0$ there is a finite cover $\{U_1,\ldots,U_n\}$ of $A$ by closed rectangles so that $\sum_{i = 1}^nv(U_i) < \epsilon$.

is equivalent to

A subset $A$ of $\mathbb{R}^n$ does not have $n$-dimensional content $0$ if there is an $\epsilon > 0$ so that it is not the case that there is a finite cover $\{U_1,\ldots,U_n\}$ of $A$ by closed rectangles with $\sum_{i = 1}^nv(U_i) < \epsilon$.

is equivalent to

A subset $A$ of $\mathbb{R}^n$ does not have $n$-dimensional content $0$ if there is an $\epsilon > 0$ so that for every finite cover $\{U_1,\ldots,U_n\}$ of $A$ by closed rectangles it is not the case that $\sum_{i = 1}^nv(U_i) < \epsilon$.

is equivalent to

A subset $A$ of $\mathbb{R}^n$ does not have $n$-dimensional content $0$ if there is an $\epsilon > 0$ so that for every finite cover $\{U_1,\ldots,U_n\}$ of $A$ by closed rectangles, $\sum_{i = 1}^nv(U_i) \geq \epsilon$.

Answer 2

The negation is that for some $\varepsilon,$ there is no such cover. That means that there is an $\varepsilon>0$ such that for any finite cover by closed rectangles we have $$\sum_{i=1} ^n v(U_i) \ge \varepsilon.$$

Another way of looking at it is that the infimum of $\sum_{i=1} ^n v(U_i)$ over all such covers is positive.