I've been taught that the definition of compact operator is:
(1) Let $X,Y$ be normed spaces and let $T:X \rightarrow Y$ be a linear operator. Then $T$ is said to be compact if the closure of $T(B_X)$ is compact.
And it can be proved that an equivalent condition for (1) is (2):
(2) For any bounded sequence $(x_n)_n \subseteq X$, there exists a subsequence of $(Tx_n)_n$ which is convergent.
Here's what I don't understand about (2): this condition is usually stated as above, without explicitly indicating where the subsequence of $(Tx_n)_n$ must converge. Is it because it does not matter where the limit of the subsequence is contained? Wouldn't it be necessary to check that the limit of the converging subsequence belongs to a certain set, let's say the closure of $T(B_X)$ or any other?
I've read the proof of (1) is equivalent to (2), but from that I couldn't surmise wether it does or does not matter wether the limit of the convergent subsequence belongs to any set in particular. I'm inclined to think it does matter, but if that were the case it should be explicitly stated in (2), right?
Thanks in advance!
The first important point is, as observed by @OğuzhanKılıç, that the equivalence between (1) and (2) is only valid if $Y$ is complete, which we'd better assume here.
However I believe that your question is not really related to this but to where the limit is supposed to be, right?
The short answer is that it really doesn't matter where the limit will be.
The main points are perhaps that:
a subset $C\subseteq Y$ is compact iff every sequence in $C$ admits a subsequence converging to a point in $C$.
a subset $C\subseteq Y$ is relatively compact (that is, it's closure is compact) iff every sequence in $C$ admits a subsequence converging to a point anywhere in the ambient space $Y$.
The crux of the matter is that, in (2), the limit will necessarily lie in the closure of $C$ since the closure is precisely the set of limits of convergent sequences in $C$.