Sequences Eventually and Frequently in a Set

Question

A sequence $(a_n)$ is eventually in an set $A \subseteq \mathbb R$ if there exists an $N \in \mathbb N$ such that $a_n \in A$ for all $n \ge N.$

A sequence $(a_n)$ is frequently in an set $A \subseteq \mathbb R$ if, for every $N \in \mathbb N$ there exists an $n \ge N$ such that $a_n \in A$.

Question: Which definition is stronger? Does frequently imply eventually or eventually imply frequently?

Intuitively I am inclined to say that eventually implies frequently because from examples I reason that the sequence ${\lfloor \frac{5}{n}\rfloor}$ is eventually in $\{0\}$ and also frequently in $\{0\}$ but a sequence like $(-1)^n$ which is frequently in $\{1\}$ but not eventually.

However from looking at the quantifiers it seems to me that the converse should be true for I reason

$$\forall N \exists n(n \ge N \Rightarrow a_n \in A)$$ should imply $$\exists N \exists n(n \ge N \Rightarrow a_n \in A)$$ (i.e. frequnctly implies eventually)?

Am I making a mistake with the quantifiers? Which one is correct? Thanks in advance.

Answer 1

Eventually implies frequently. If $a_n \in A$ for all $n \ge N$, then for every $n$ you can take $m = \max(n,N)$ and have $m \ge n$ and $a_m \in A$.

Answer 2

Expression for eventually:$$\exists N\forall n\geq N[a_n\in A]$$

Equivalent is the statement: $$\liminf1_A(a_n)=1$$

Expression for frequently:$$\forall N\exists n\geq N[a_n\in A]$$

Equivalent is the statement: $$\limsup1_A(a_n)=1$$

We always have: $$\liminf1_A(a_n)\leq\limsup1_A(a_n)\leq1$$so eventually implies frequently.