Prove that infinite integers that are not fibonatic For $ a $ positive integer, define $ F ^ {(a)} _1 $, $ F ^ {(a)} _2 = a $ and, for $ n> 2 $, $ F ^ {(a)} _n = F ^ {(a)} _ {n-1} + F ^ {(a)}_ {n-2} $. A positive integer is Fibonatic when it equals $ F ^ {(a)} _ {n} $ for some $ a $ positive integer and some $ n> 3 $. Prove that there are infinitely many integers that are not Fibonatic.
Attemp: I would like you to evaluate and correct my solution by topology, please!
Let $a > 2_i \in A=\left\{\left[ F_i ^{(a)}; F^{(a)}_{j+1}\right] |i = 1,2,3,....\right\}$
Let's take the following lema:
Lema $ _1 $: Let $ M $ be a metric space, be two balls $ B (a, r) \in B (b, r) $ disjunct, then $ d (a, b) <r $
Proofs: Let $ 0 <r \leq d (a, b)$, then take $ x \in B (a, r) \cap B (b, a)$, then $ d (x, a) <r $ and $ d (b, x) <r $, then $ d (a, b) \leq d (b, x) <2r \iff r> \frac {d (a, b)} {2} $ Abs'. So $ B (a, r) \cap B (b, r) = \emptyset $. We know that $ \mathbb {Z} $ is discrete, in turn, $ \mathbb {N} $ is discrete, that is, there is no $ x \in X $ element such that $ B (a, r) \cap X$  It is other than $\emptyset $.
If we take $ E> 0 $ and centered on each $ F_i ^ {(a)}$, it follows that $ B (F_i ^{(a)}, E) \cap B (F_j^{(a)}, E) \neq \emptyset $, so $ E> \frac{d (F_i ^ {(a)}, F_j ^ {(a)}}{2} $. So as we take $ a> z $, it follows that at least one integer $ C \in \left[F_j ^ {(a)}, F_ {i + 1} ^ {(a)} \right] $. Therefore, there are infinite integers that do not satisfy the condition of being fibonatic
I took a different approach, so I wanted a correction
 A: Revised to address the OP’s argument in much more detail.
Your argument makes no sense that I can see. First, your set $A$ is not well-defined, since there is no explanation or definition of $j$; the context suggests that you may have wanted the right endpoint to be $F_{\color{red}i+1}^{(a)}$. Then, after the introductory bit showing that if $B(a,r)\cap B(b,r)\ne\varnothing$, then $r>\frac{d(a,b)}2$, you apply that bit to an undefined set that appears to be $\left\{F_n^{(a)}:n\ge 1\right\}$. After that you ‘take $a>z$’, which makes no sense at all, since you have not defined $z$.
It also indicates that you have misunderstood what you have to prove. It’s not enough to show that for each positive integer $a$ there are infinitely many positive integers that are not of the form $F_n^{(a)}$ for any $n>3$: you must show that there are infinitely many positive integers $m$ such that for all positive integers $a$ and all integers $n>3$, $a\ne F_n^{(a)}$. That is, there are infinitely many positive integers that cannot be written as $F_n^{(a)}$ no matter what $a\in\Bbb Z^+$ and $n>3$ you try. Thus, you do not have any freedom to choose $a$: any argument that you make must work for all positive integers $a$ simultaneously.
Finally, it’s not enough to get an integer in a closed interval with endpoints $F_j^{(a)}$ and $F_{i+1}^{(a)}$, even if it were clear how $i$ and $j$ were chosen or what the relationship between them is: that integer could be one of the endpoints and hence a fibonatic number.
You really need a very different approach.
HINT: Show that $F_n^{(a)}=aF_n$ for all $a,n\in\Bbb Z^+$. Then consider, for instance, the numbers $11^k$ for $k\in\Bbb Z^+$.
