The uniform closure of a unital real function algebra is closed under absolute value, maximum, and minimum

The closure $\bar{A}$ is a real subalgebra. Indeed, fix $w,z\in \bar{A}$, real $\alpha ,\beta$, and a target error $\eta >0$. For each $\delta >0$ there are two witnesses $a,b\in A$ with $\mathrm{\parallel }a-w{\mathrm{\parallel }}_{\mathrm{\infty }}<\delta$ and $\mathrm{\parallel }b-z{\mathrm{\parallel }}_{\mathrm{\infty }}<\delta$. Then $\alpha a+\beta b,ab\in A$, and $\mathrm{\parallel }\alpha a+\beta b-(\alpha w+\beta z){\mathrm{\parallel }}_{\mathrm{\infty }}\le (\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })\delta ,\mathrm{\parallel }ab-wz{\mathrm{\parallel }}_{\mathrm{\infty }}\le \delta (\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}+\delta )\mathrm{.}$ Choose $0<\delta <\min \{1,\eta \mathrm{/}(1+\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }+\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }})\}$ to make both errors less than $\eta$. Constants already belong to $A$. This proves the asserted algebra closure using only finitely many witnesses for each target error.

Fix $\eta >0$ and put $M=\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}+1$. Choose one $a\in A$ with $\mathrm{\parallel }a-u{\mathrm{\parallel }}_{\mathrm{\infty }}<\min \{1,\eta \mathrm{/}3\}$. Then $\mathrm{\mid }a(x)\mathrm{\mid }\le M$ on $K$. By [L2], choose one real polynomial $p$ with ${\sup }_{\mathrm{\mid }t\mathrm{\mid }\le M}\mathrm{\mid }p(t)-\mathrm{\mid }t\mathrm{\mid }\mathrm{\mid }<\eta \mathrm{/}3$. Since $A$ is unital, $p(a)\in A$, including its constant term. The inequality $\mathrm{\mid }\mathrm{\mid }r\mathrm{\mid }-\mathrm{\mid }s\mathrm{\mid }\mathrm{\mid }\le \mathrm{\mid }r-s\mathrm{\mid }$ gives $\mathrm{\parallel }p(a)-\mathrm{\mid }u\mathrm{\mid }{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }p(a)-\mathrm{\mid }a\mathrm{\mid }{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }\mathrm{\mid }a\mathrm{\mid }-\mathrm{\mid }u\mathrm{\mid }{\mathrm{\parallel }}_{\mathrm{\infty }}<2\eta \mathrm{/}3<\eta \mathrm{.}$ Thus $\mathrm{\mid }u\mathrm{\mid }\in \bar{A}$ by the metric definition of closure. No sequence of approximants or diagonal choice is used.

Step 1.1 puts $u-v$ and $u+v$ in $\bar{A}$, and step 1.2 applies to $u-v$. The identities $u\vee v=(u+v+\mathrm{\mid }u-v\mathrm{\mid })\mathrm{/}2$ and $u\wedge v=(u+v-\mathrm{\mid }u-v\mathrm{\mid })\mathrm{/}2$ therefore give the remaining closure.