$(x_n)$ 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?

  • $\begingroup$ I created a new account since I didn't get know how figure this out from anywhere, I'll try respect $\endgroup$ Sep 16, 2022 at 20:33

1 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 $k$.

  • $\begingroup$ is for any the same as for all in this case? $\endgroup$ Sep 18, 2022 at 0:26
  • $\begingroup$ @DaviAmérico Yes. They are used mathematically equivalently in my experience. I suppose “for all” has the tone: [I have checked each and every single possibility] whereas “for any” has the tone: [if you give me an example, I’ll always be able to prove the conclusion]. So I sometimes prefer to use one or the other. But it doesn’t matter $\endgroup$
    – FShrike
    Sep 18, 2022 at 7:42

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