Adjoint of completely continuous operator is completely continuous In the proof of the fact that the adjoint operator $A^\ast$ of a completely continuous linear operator $A:E\to E$ mapping a Banach space into itself is also completely continuous on $E^\ast$ endowed with the strong topology induced by the usual norm, Kolmogorov-Fomin's Introductory Real Analysis proves the fact that the image $A^\ast(S^\ast)$ of the unit sphere $S^\ast=\{f\in E^\ast:\|f\|\le1\}$ is relatively compact.
To prove the relative compactness of $S^\ast$, which is the same as total boundedness in the complete space $E^\ast$ the text establishes an isometry (which I will describe) between the subset $\Phi:=\{\varphi_{\overline{AS}}:\varphi\in S^\ast\subset E^\ast\}$ of the restrictions of the functionals belonging to $S^\ast$  to the closure $\overline{AS}$ (a notation for $\overline{A(S)}$) of the image $A(S)$ of the unit sphere $S\subset E$ through $A$ and $A^\ast(S^\ast)$.
The isometry is between $\Phi\subset C[\overline{AS}]$ as a subset of the space of continuous functions on $\overline{AS}$ endowed with metric $\rho_{C[\overline{AS}]}(f,g):=\sup_{x\in\overline{AS}}|f(x)-g(x)|$ and $A^\ast(S^\ast)\subset E^\ast$ endowed with the norm induced metric: $\forall g_1,g_2\in S^\ast\quad\|A^\ast (g_1)-A^\ast (g_2)\|=\rho_{C[\overline{AS}]}(g_1,g_2)$.
Since $\Phi$ is proved to be totally bounded in $C[\overline{AS}]$ (i.e. it has a finite $\varepsilon$-net in $C[\overline{AS}]$ for any $\varepsilon>0$) and $\Phi$ is isometric to $A^\ast(S^\ast)$, then the text deduces that $A^\ast(S^\ast)$ is totally bounded in $E^\ast$. I do not understand why. In fact I am not sure that the points of the $\varepsilon$-net of $\Phi$, which belong to $C[\overline{AS}]$ (which is not a subset of $E^\ast$), are mapped by $A^\ast$ to an $\varepsilon$-net of $A^\ast(S^\ast)\subset E^\ast$. Can they be chosen in $E^\ast\cap C[\overline{AS}]$? Does anybody understand the argument? Thank you very much for any answer!
 A: The nub of it is that total boundedness is an intrinsic property of a metric [more generally, uniform] space. It does not depend on whether the space is a subspace of some larger space, and if so, what that larger space is, all that matters is the metric.
We have two metric spaces, $X = A^\ast(S^\ast)$, with the metric induced by the norm on $E^\ast$, and $Y = C[\overline{AS}]$, with the uniform metric (induced by the supremum/maximum norm), and an isometric embedding $\iota\colon X \hookrightarrow Y$. Now let $Z = \iota(X)$. And forget about the special structures of $X$ and $Y$, only remember that they are metric spaces and $\iota$ is an isometric embedding of $X$ into $Y$ with image $Z$.
Next, take the definition of total boundedness for subsets of a metric space $Y$. Well, which one? There are several equivalent definitions.
Definition 1: A subset $S\subset Y$ is totally bounded if for every $\varepsilon > 0$ there is a finite subset $F_\varepsilon \subset Y$ such that $$S \subset \bigcup_{y\in F_\varepsilon} B_\varepsilon(y).$$
Definition 2: A subset $S\subset Y$ is totally bounded if for every $\varepsilon > 0$ there is a finite subset $F_\varepsilon \subset S$ such that $$S \subset \bigcup_{y\in F_\varepsilon} B_\varepsilon(y).$$
Definition 3: A subset $S\subset Y$ is totally bounded if for every $\varepsilon > 0$ there is a finite family $\mathscr{F}_\varepsilon$ of subsets of $Y$ with $$S = \bigcup_{A\in \mathscr{F}_\varepsilon} A$$ and $\operatorname{diam}(A) \leqslant \varepsilon$ for all $A\in \mathscr{F}_\varepsilon$.
$\operatorname{diam}(A)$ is the diameter of the set $A$,
$$\operatorname{diam}(A) := \sup \{ d(a,b) : a,b \in A\}.$$
The supremum is taken over a subset of $[0,+\infty)$, hence $\sup \varnothing = 0$ here, or equivalently $\operatorname{diam}(\varnothing) = 0$.
There are certainly more possibilities to define total boundedness, but these seem to be the most common ones.
If definition 2 or definition 3 is used, it is direct to see that in the situation above, $X$ is totally bounded (as a subset of $X$, if one so wishes) if and only if $Z$ is totally bounded as a subset of $Y$. For, if we use definition 2, we observe $B^X_\varepsilon(\iota^{-1}(y)) = \iota^{-1}(B^Y_\varepsilon(y)\cap Z) = \iota^{-1}(B^Y_\varepsilon(y))$ and thus see that there is a finite subset $F^X_\varepsilon$ of $X$ such that the $\varepsilon$-balls with centre in $F^X_\varepsilon$ cover $X$ if and only if there is a finite subset $F^Z_\varepsilon$ of $Z$ such that the $\varepsilon$-balls with centre in $F^Z_\varepsilon$ cover $Z$. If we use definition 3, we observe that $\operatorname{diam}(A) = \operatorname{diam}(\iota(A))$, and hence we can cover $X$ by finitely many sets of diameter at most $\varepsilon$ if and only if we can cover $Z$ by finitely many sets of diameter at most $\varepsilon$.
Only for definition 1 is there something to show.
We can do that directly. Suppose that $Z$ is totally bounded as a subset of $Y$ according to definition 1. Prescribe an arbitrary $\varepsilon > 0$. Now, by the definition of total boundedness, choose a finite subset $F_{\varepsilon/2} \subset Y$ such that
$$Z \subset \bigcup_{y\in F_{\varepsilon/2}} B_{\varepsilon/2}(y).$$
Let $G = \{ y\in F_{\varepsilon/2} : B_{\varepsilon/2}(y) \cap Z \neq \varnothing\}$. Then $G$ is a finite set, and
$$Z \subset \bigcup_{y\in G} B_{\varepsilon/2}(y).$$
Next, for every $y\in G$, choose a $z_y \in B_{\varepsilon/2}(y)\cap Z$. Let $H = \{ z_y : y \in G\}$. Then $H$ is a finite subset of $Z$, and
$$Z\subset \bigcup_{z\in H} B_\varepsilon(z),$$
since by the triangle inequality we have $B_{\varepsilon/2}(y) \subset B_\varepsilon(z_y)$ for every $y\in G$. Then, since $\iota$ is an isometry, $K = \iota^{-1}(H)$ is a finite subset of $X$ with
$$X = \bigcup_{x\in K} B_\varepsilon(x).$$
$\varepsilon > 0$ was arbitrary, hence we conclude that $X$ is totally bounded. The other direction, that total boundedness of $X$ implies total boundedness of $Z$ is more direct: If $F_\varepsilon$ is a finite subset of $X$ with
$$X = \bigcup_{x\in F_\varepsilon} B_\varepsilon(x),$$
then $G_\varepsilon = \iota(F_\varepsilon)$ is a finite subset of $Y$ (even of $Z$, but that's not important for definition 1) with
$$Z\subset \bigcup_{y\in G_\varepsilon} B_\varepsilon(y).$$
Or we could appeal to the fact that definitions 1-3 are equivalent.
We would need to prove that then, of course. A direct proof that definition 1 implies definition 2 is very much like the above proof that total boundedness of $Z$ according to definition 1 implies total boundedness of $X$. But with the three definitions, we can prove the equivalence by showing $1 \implies 3 \implies 2 \implies 1$, and each of these steps is shorter.
$1\implies 3$: For $\varepsilon > 0$, choose $F_{\varepsilon/2}$ as per definition 1. Since $\operatorname{diam}(B_r(y)) \leqslant 2r$, the family $$\mathscr{F}_\varepsilon = \{ B_{\varepsilon/2}(y) \cap Z : y \in F_{\varepsilon/2}\}$$ satisfies the requirements of definition 3.
$3\implies 2$: For $\varepsilon > 0$, choose $\mathscr{F}_{\varepsilon/2}$ according to definition 3. Assume that $\varnothing \notin \mathscr{F}_{\varepsilon/2}$. For every $A\in \mathscr{F}_{\varepsilon/2}$, choose $s_A \in A$. Then $F_\varepsilon = \{ s_A : A\in \mathscr{F}_{\varepsilon/2}\}$ satisfies the requirements of definition 2.
$2\implies 1$: That is immediate, since $S\subset Y$.
