Why is this negation regarding rational Cauchy series true?

Question

$(x_n)$ as Cauchy sequence

Suppose the preposition below:

${(*)}$ $|x_n| \geq r$ holds for all $n \geq n_0$ and some rational $r>0$.

my professor states that the negation of ${(*)}$ is:

for each $N \in \mathbb{N}$ so there exists $n_N \in \mathbb{N}$ such that $\displaystyle|x_{n_{N}}|<1/k$

I didn't understand this, why did he used subsequencies?

Answer 1

This is not quite the direct logical negation but it is certainly equivalent to the negation.

The logical negation:

For any $r>0$, any $n_0$, there exists $n\ge n_0$ with $|x_n|<r$.

This is equivalent to what your professor said (at least, what I think they said: what you’ve written doesn’t make sense). I’ll show that the negation implies what your professor said: you can try to show the opposite implication.

Assume the negation. Because it quantifies over all (rational) $r>0$, for any natural $k$ I may set $r=1/k$. Let $n_0=1$ and $k=1$. There exists $n>n_0$ - call it $n_1$ - with $|x_{n_1}|<r=1$. Now let $n_0=n_1+1$ and $k=2$. There exists an $n_2\ge n_0=1+n_1>n_1$, with $|x_{n_2}|<r=1/2$. And so on. I can inductively find $n_1<n_2<\cdots<n_k<\cdots$ with $|x_{n_k}|<1/k$ for every natural $k$.