neighborhood base for the Mackey topology I'm reading the proof of a theorem due to Mackey in a note of functional analysis:

I don't see why the first sentence is clear. By the definition of neighborhood base and locally convex topology, to show the underscored sentence, it suffices to show that 

given a closed absolutely convex neighborhood $U$ of zero in $X$, there exists $B\subset Y$ absolutely convex and $\sigma(Y,X)$-compact such that $^oB\subset U$. 

Could anyone show me why such $B$ exists?
[Edited:]
Thanks to the comment, denoting the collection of sets of the form $^oB$ (with $B$ satisfying the properties) as $\mathcal{B}$, I should show the following instead:


*

*Each set $V\in\mathcal{B}$ is balanced (trivially true) and absorbing;

*For every $V\in\mathcal{B}$ there exists $U\in\mathcal{B}$ such that $U+U\subset V$; 

*Given $U,V\in\mathcal{B}$, there exists $W\in\mathcal{B}$ such that $W\subset U\cap V$.


To show that $^oB$ is absorbing, I think one somehow needs to use the compactness of $B$. But I don't see how:
Let $B\subset Y$ be absolutely convex and $\sigma(Y,X)$-compact. 
$$
^oB=\{x\in X\mid |\langle x ,y\rangle|\leq 1,\ \hbox{for all}\ y\in B\}.
$$
For $x\in X$, by the definition of "absorbing", how can I find $t>0$ such that
$$
\lambda x\in\ ^oB
$$
for all $|\lambda|\leq t$? 
 A: Let's take a look at some properties of polars. If we have a dual pairing $\langle\,\cdot\,,\,\cdot\,\rangle \colon X \times Y \to \mathbb{K}$, where $\mathbb{K}$ can be either of $\mathbb{R}$ or $\mathbb{C}$, we define the polar of a set $A \subset Y$ as
$${}^{\Large\circ}A := \bigl\{ x \in X : \bigl(\forall y\in A\bigr)\bigl(\lvert \langle x,y\rangle\rvert \leqslant 1\bigr)\bigr\}.$$
Writing $\lambda_y = \langle\,\cdot\,,y\rangle$ for $y\in Y$ and $\overline{\mathbb{D}} = \{ z \in \mathbb{K} : \lvert z\rvert \leqslant 1\}$, we can also write
$${}^{\Large\circ}A = \bigcap_{y\in A} \lambda_y^{-1}\bigl(\overline{\mathbb{D}}\bigr).\tag{1}$$
Since the $\lambda_y$ are linear and $\sigma(X,Y)$-continuous, it follows immediately from $(1)$ that ${}^{\Large\circ}A$ is absolutely convex and $\sigma(X,Y)$-closed.
Next, we have the inclusion-reversion property, $A \subset B \implies {}^{\Large\circ}B \subset {}^{\Large\circ}A$. For subsets of a specific type ($\sigma(Y,X)$-closed and absolutely convex) we have an equivalence, but generally we have only a one-directional implication. Also, directly from $(1)$ we have ${}^{\Large\circ}(A\cup B) = {}^{\Large\circ}A \cap {}^{\Large\circ}B$.
And polars have a simple scaling behaviour, for all $\alpha \neq 0$ we have ${}^{\Large\circ}(\alpha\cdot A) = \alpha^{-1}\cdot({}^{\Large\circ}A) = \lvert\alpha\rvert^{-1}\cdot({}^{\Large\circ}A)$.
Finally, $x\in X$ is absorbed by ${}^{\Large\circ}A$ if and only if $\{ \langle x,y\rangle : y \in A\}$ is bounded in $\mathbb{K}$, or, put in different words, $x$ is bounded on $A$ (more correctly, $\mu_x = \langle x,\,\cdot\,\rangle$ is bounded on $A$). Namely, for $s\neq 0$ we have
$$sx \in {}^{\Large\circ}A \iff \bigl(\forall y \in A\bigr)\bigl(\lvert\langle sx,y\rangle\rvert \leqslant 1\bigr) \iff \bigl(\forall y\in A\bigr)\bigl(\lvert\langle x,y\rangle\rvert \leqslant \lvert s\rvert^{-1}\bigr).$$
Hence ${}^{\Large\circ}A$ is absorbing if and only if $A$ is $\sigma(Y,X)$-bounded.
Now let us consider $\mathcal{K}(Y) = \{ B \subset Y : B \text{ is absolutely convex and } \sigma(Y,X)\text{-compact}\}$ and the family $\mathcal{B} = \{ {}^{\Large\circ}B : B \in \mathcal{K}(Y)\}$ of its polars.
Since every $\sigma(Y,X)$-compact set is $\sigma(Y,X)$-bounded, $\mathcal{B}$ consists of ($\sigma(X,Y)$-closed) absolutely convex absorbing sets. Hence the first bullet point.
Further, the family $\mathcal{K}(Y)$ is scaling-invariant - $B \in \mathcal{K}(Y) \iff t\cdot B \in \mathcal{K}(Y)$ for every $t > 0$ - and thus the family $\mathcal{B}$ is also scaling-invariant. Since the members of $\mathcal{B}$ are also convex, for every $V\in \mathcal{B}$ we have $\frac{1}{2}V + \frac{1}{2}V = V$, hence the second bullet point.
For the last bullet point, we know that ${}^{\Large\circ}A \cap {}^{\Large\circ}B = {}^{\Large\circ}(A\cup B)$, and if we had $A \cup B \in \mathcal{K}(Y)$ for $A,B\in \mathcal{K}(Y)$, that would settle it. But the union of two convex sets need not be convex (the union of two balanced sets is again balanced, and the union of two $\sigma(Y,X)$-compact sets is again $\sigma(Y,X)$-compact, so it's only convexity that can throw a spanner here), so we need something else.
We need an absolutely convex $\sigma(Y,X)$-compact set that contains $A\cup B$. If $A = \varnothing$, we have ${}^{\Large\circ}A = X = {}^{\large\circ}\{0\}$, so we can assume that neither of $A$ and $B$ is empty. Then we have $A\cup B \subset A + B$, since every nonempty balanced set contains $0$. And the sum of two balanced resp. convex sets is again balanced resp. convex, so $A+ B$ is absolutely convex. And the sum of two $\sigma(Y,X)$-compact sets is $\sigma(Y,X)$-compact, since addition is $\sigma(Y,X)$-continuous. Thus, for nonempty $A,B \in \mathcal{K}(Y)$ we have ${}^{\Large\circ}(A + B) \subset {}^{\Large\circ}A \cap {}^{\Large\circ}B$, which gives us the third bullet point.
Thus there is a unique vector space topology $\tau(X,Y)$ (the Mackey topology) on $X$ for which $\mathcal{B}$ is a neighbourhood base of $0$. This vector space topology is Hausdorff, since it is finer than $\sigma(X,Y)$.
