Lattice over Continuous function

Consider $C_{[a,b]}$$C_{[a,b]}$, define a partial order $f⪯g\iff \mathrm{\forall }x\in [a,b],f(x)\le g(x)$$f\preceq g\Leftrightarrow\forall x\in[a,b],f(x)\le g(x)$

Define $\begin{array}{c}max(f,g):=\{\begin{array}{l}f(x),f(x)\ge g(x)\\ g(x),f(x)\le g(x)\end{array}\end{array}$$\max(f,g):=\begin{cases}f(x),f(x)\ge g(x)\\g(x),f(x)\le g(x)\end{cases}$ and duality, $\begin{array}{c}min(f,g):=\{\begin{array}{l}f(x),f(x)\ge g(x)\\ g(x),f(x)\le g(x)\end{array}\end{array}$$\min(f,g):=\begin{cases}f(x),f(x)\ge g(x)\\g(x),f(x)\le g(x)\end{cases}$

Observe that $C_{[a,b]}$$C_{[a,b]}$ is closed under $max,min$$\max,\min$,

In fact, $max(f,g),min(f,g)$$\max(f,g),\min(f,g)$ is the least upper bound and greatest lower bound of $(C_{[a,b]},⪯)$$(C_{[a,b]},\preceq)$

Using the notation of lattice, $max(f,g)=f\vee g,min(f,g)=f\wedge g$$\max(f,g)=f\vee g,\min(f,g)=f\wedge g$

Then we can add $-\mathrm{\infty },\mathrm{\infty }$$-\infty,\infty$ to be the $0,1$$0,1$, $0\vee f=f,1\wedge g=g$$0\vee f=f,1\wedge g=g$

Define duality $f^{\prime }=-f$$f'=-f$, then $(C_{[a,b]},\vee ,\wedge ,0,1{,}^{\prime })$$\left(C_{[a,b]},\vee,\wedge,0,1,'\right)$ is a Boolean Algebra

The associative law, communicative law and idempotent law are obviously,

The distributive law

$f\wedge (g\vee h)=min(f,max(g,h))=max(min(f,g),min(f,h))=(f\wedge g)\vee (f\wedge h)$$f\wedge(g\vee h)=\min(f,\max(g,h))=\max(\min(f,g),\min(f,h))=(f\wedge g)\vee (f\wedge h)$

$f\vee (g\wedge h)=max(f,min(g,h))=min(max(f,g),max(f,h))=(f\vee g)\wedge (f\vee h)$$f\vee(g\wedge h)=\max(f,\min(g,h))=\min(\max(f,g),\max(f,h))=(f\vee g)\wedge (f\vee h)$

The absorb law holds iff $a\vee b=b\iff a\wedge b=a$$a\vee b=b\Leftrightarrow a\wedge b=a$,

That is, the partial order defined by $a\le b$$a\le b$ iff $a\vee b=b$$a\vee b=b$ and $a\le b$$a\le b$ iff $a\wedge b=a$$a\wedge b=a$ is the same

Proof.

if the partial order defined by $a\le b$$a\le b$ iff $a\vee b=b$$a\vee b=b$ and $a\le b$$a\le b$ iff $a\wedge b=a$$a\wedge b=a$ is the same

$a\vee (a\wedge b)=a$$a\vee(a\wedge b)=a$ because $a\wedge b\le a$$a\wedge b\le a$, and $a\wedge (a\vee b)$$a\wedge(a\vee b)$ because $a\le (a\vee b)$$a\le (a\vee b)$

if we have absorb law, then,

$a\vee b=b\Rightarrow a\wedge b=a\wedge (a\vee b)=a$$a\vee b =b\Rightarrow a\wedge b= a\wedge(a\vee b)=a$

$a\wedge b=a\Rightarrow a\vee b=(a\wedge b)\vee b=b$$a\wedge b =a\Rightarrow a\vee b= (a\wedge b)\vee b =b$

The De Morgan Law hold because ${}^{\prime }$$’$ gives an isomorphism between $(L,\le )\cong (L,\ge )$$(L,\le)\cong (L,\ge)$

$(a\le b{)}^{\prime }\iff a^{\prime }\ge b^{\prime }$$(a\le b)'\Leftrightarrow a'\ge b'$, Thus it preserves the lub and glb.

Thus

$(a{\vee }_{\le }b)=a^{\prime }{\vee }_{\ge }{b}^{\prime }=a^{\prime }{\wedge }_{\le }{b}^{\prime }$$(a\vee_{\le} b)=a'\vee_{\ge} b'= a'\wedge_{\le} b'$

$(a{\wedge }_{\le }b)=a^{\prime }{\wedge }_{\ge }{b}^{\prime }=a^{\prime }{\vee }_{\le }{b}^{\prime }$$(a\wedge_{\le} b)=a'\wedge_{\ge} b'= a'\vee_{\le} b'$

And easy to see that $f^{\prime }:=-f$$f':=-f$ gives an isomorphism