proof of Banach-Alaoglu theorem

For any $x\in X$, let $D_x=\{z\in ℂ:|z|\le \parallel x\parallel \}$ and $D={\mathrm{\Pi }}_{x\in X}{D}_x$. Since $D_x$ is a compact subset of $ℂ$, $D$ is compact in product topology by Tychonoff theorem.

We prove the theorem by finding a homeomorphism that maps the closed unit ball$B_{X^{*}}$ of $X^{*}$ onto a closed subset of $D$. Define ${\mathrm{\Phi }}_x:B_{X^{*}}\to D_x$ by${\mathrm{\Phi }}_x(f)=f(x)$ and $\mathrm{\Phi }:B_{X^{*}}\to D$ by $\mathrm{\Phi }={\mathrm{\Pi }}_{x\in X}{\mathrm{\Phi }}_x$, so that$\mathrm{\Phi }(f)={(f(x))}_{x\in X}$. Obviously, $\mathrm{\Phi }$ is one-to-one, and a net $(f_{\alpha })$ in $B_{X^{*}}$ converges to $f$ in weak-* topology of $X^{*}$ iff $\mathrm{\Phi }(f_{\alpha })$ converges to $\mathrm{\Phi }(f)$ in product topology, therefore $\mathrm{\Phi }$ is continuous and so is its inverse ${\mathrm{\Phi }}^{-1}:\mathrm{\Phi }(B_{X^{*}})\to B_{X^{*}}$.

It remains to show that $\mathrm{\Phi }(B_{X^{*}})$ is closed. If $(\mathrm{\Phi }(f_{\alpha }))$ is a netin $\mathrm{\Phi }(B_{X^{*}})$, converging to a point $d={(d_x)}_{x\in X}\in D$, we can define a function$f:X\to ℂ$ by $f(x)=d_x$. As ${lim}_{\alpha }\mathrm{\Phi }(f_{\alpha }(x))=d_x$ for all $x\in X$ by definition of weak-* convergence, one can easily see that $f$ is a linear functional in $B_{X^{*}}$ and that $\mathrm{\Phi }(f)=d$. This shows that $d$ is actually in $\mathrm{\Phi }(B_{X^{*}})$ and finishes the proof.