Weak-star compactness of polar sets

Choose $\epsilon >0$ with $\{x:\parallel x\parallel <\epsilon \}\subseteq U$ and put $r=\epsilon \mathrm{/}2$. Then $r{B}_X\subseteq U$. If $f\in U^{\circ }$ and $\parallel x\parallel \le 1$, then $rx\in U$, whence $r\mathrm{\mid }f(x)\mathrm{\mid }\le 1$. Thus $\parallel f\parallel \le r^{-1}$ and $U^{\circ }\subseteq r^{-1}{B}_{X^{\ast }}$.

The polar is weak-star closed. Indeed, if $f_0\mathrm{\notin }{U}^{\circ }$, some $u\in U$ satisfies $\mathrm{\mid }{f}_0(u)\mathrm{\mid }>1$; the basic neighborhood $\{f:\mathrm{\mid }(f-f_0)(u)\mathrm{\mid }<\mathrm{\mid }{f}_0(u)\mathrm{\mid }-1\}$ misses $U^{\circ }$ by the reverse triangle inequality.

The map $f\mapsto r^{-1}f$ is a weak-star homeomorphism with inverse $g\mapsto rg$, by continuity of scalar multiplication. It carries $B_{X^{\ast }}$ onto $r^{-1}{B}_{X^{\ast }}$, so the latter is compact by [F1]. The ultrafilter lemma enters only through [F1].

By steps 1.1 and 1.2, $U^{\circ }$ is a closed subset of the compact space in step 1.3. Adding its open complement to any open cover of $U^{\circ }$ gives an open cover of that compact space, so deleting the complement from a finite subcover proves that $U^{\circ }$ is compact. For $X=0$ this says that a singleton is compact.