# Divergent Complex sequence with distinct points and without accumulation point tends to $\infty$?

Is it true that any Divergent complex sequence with distinct points and without accumulation point tends to $\infty$ ? (One can also replace distinctness condition by condition finitely repeating terms)

I think it's true as my intuition, but tried to prove this as follows :

By contradiction, suppose $\{z_i\}$ are bounded, that is: $\forall n\,:\,|z_n|\in D_M$. Now as sequence doesn't have accumulation point and $D_M$ is open, all points of the sequence (which their cardinality is infinite) and for each $i$, there exists $r_i$ such that : $$\forall i\neq j\,:\,D_{r_i}(z_i)\cap\{z_j\}=\emptyset\quad\wedge\quad \bigcup_{i=1}^\infty D_{r_i}(z_i)\subset D_M$$ From distintness, we know that the set of all these neighborhoods are infinitely countable.

Any help is appreciate for the rest.

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Divergence is necessary to be determined which I have not used it yet !

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You need to prove more than the fact the $z_i$ are not bounded. Try to show that any disk with centre the origin has only finitely many points of the sequence. That will basically get you there. –  André Nicolas Jun 7 at 6:12
Yes you're right. Thanks –  Fardad Pouran Jun 7 at 6:24
You are welcome. You added a question about divergence. That is not needed, a convergent sequence will have an accumulation point. –  André Nicolas Jun 7 at 6:27
Ok I meant I had not considerd this fact and i wanted to emphasize on it. –  Fardad Pouran Jun 7 at 6:33
@AndréNicolas, Could you please give another hint ? –  Fardad Pouran Jun 7 at 17:58

Hint: Take a large number $B$. If the sequence $(z_n)$ has an infinite number of terms in the disk with radius $B$, then the sequence has an accumulation point (Bolzano-Weierstrass).
It follows that for any $B$, there are only finitely many $n$ such that $|z_n|\le B$. Therefore there is an $N=N_B$ such that if $n\gt N$ then $|z_n|\gt B$. This is precisely what it means for $z_n$ to go to infinity.
Indeed it was suffices for my idea to discuss on $\bar{D}_M$ instead of $D_M$, which compactness is perfect property for sequences :) –  Fardad Pouran Jun 8 at 5:19