Compact Operators and Complete Metrics Spaces

Question

I have a couple of questions about compact operators and compactness in complete metric spaces:

1.I have the following implications: Let $Y$ be a metric space with $A$ a subset of $Y$. $A$ is precompact iff $\bar{A}$ is sequentially compact iff any sequence in $A$ has a convergent subsequence in $Y$.

Do these implications only hold if $Y$ is a complete metric space?

2.The following are characteristics of a bounded compact operator T:

A bounder operator $T$ is compact iff any of the following is true: Image of the unit ball in $X$ under $T$ is relatively comapact in $Y$. Image of any bounded set under $T$ is relatively compact in $Y$. Image of any bounded set under $T$ is totally bounded in $Y$. For any sequence $(x_{n})_{n}$ from the unit ball in $X$, the sequence$(Tx_{n})_{n}$ contains a Cauchy sequence.

Do these implications depend on whether $X$ and $Y$ are Banach Spaces?

3.Are compact operators always defined on Banach Spaces?

Answer 1

Ad 1., yes. A subset of a metric space can be precompact (totally bounded) without its closure being compact (which for metric spaces coincides with being sequentially compact) if the space is not complete. The equivalence that every sequence in $A$ has a convergent (in $Y$) subsequence if and only if $\overline{A}$ is (sequentially) compact also holds in non-complete metric spaces and of course always implies that $A$ is precompact.

Ad 2., it depends on $Y$ being a Banach space. See first part for why.

Ad 3., no, compact operators are also defined on other topological vector spaces, for example Fréchet spaces. One usually wants the target space to be complete when considering compact operators, but that is not much of a restriction, since every (Hausdorff) topological vector space has a completion.