difference of lattice elements

Let $\\mathfrak{A}$ is a lattice with least element $0$ .

Let $a,b\\in\\mathfrak{A}$ . A of $a$ and $b$ is an element $c\\in\\mathfrak{A}$ that $b\\cap c=0$ and $a\\cup b=b\\cup c$ . When there is only one difference of $a$ and $b$ , it is denoted $a\\setminus b$ .

One immediate property is: $0$ is the unique difference of any element $a$ and itself ( $a\\setminus a=0$ ). For if $c$ is such a difference, then $a\\cap c=0$ and $a=a\\cup c$ . So $c\\leq a$ by the second equation, and hence that $c=a\\cap c=0$ by the first equation.

For arbitrary lattices of two given elements do not necessarily exist. For some lattices there may be more than one difference of two given elements.

For a distributive lattice with bottom element $0$ , the difference of two elements, if it exists, must be unique. To see this, let $c$ and $d$ be two differences of $a$ and $b$ . Then

•
$b\\cap c=b\\cap d=0$ , and
•
$a\\cup b=b\\cup c=b\\cup d$ .

So $c=c\\cap(b\\cup c)=c\\cap(b\\cup d)=(c\\cap b)\\cup(c\\cap d)=0\\cup(c\\cap d)=c\\cap d$ . Similarly, $d=d\\cap c$ . As a result, $c=c\\cap d=d\\cap c=d$ .

单词	DifferenceOfLatticeElements
释义	difference of lattice elements Let $\\mathfrak{A}$ is a lattice with least element $0$ . Let $a,b\\in\\mathfrak{A}$ . A of $a$ and $b$ is an element $c\\in\\mathfrak{A}$ that $b\\cap c=0$ and $a\\cup b=b\\cup c$ . When there is only one difference of $a$ and $b$ , it is denoted $a\\setminus b$ . One immediate property is: $0$ is the unique difference of any element $a$ and itself ( $a\\setminus a=0$ ). For if $c$ is such a difference, then $a\\cap c=0$ and $a=a\\cup c$ . So $c\\leq a$ by the second equation, and hence that $c=a\\cap c=0$ by the first equation. For arbitrary lattices of two given elements do not necessarily exist. For some lattices there may be more than one difference of two given elements. For a distributive lattice with bottom element $0$ , the difference of two elements, if it exists, must be unique. To see this, let $c$ and $d$ be two differences of $a$ and $b$ . Then • $b\\cap c=b\\cap d=0$ , and • $a\\cup b=b\\cup c=b\\cup d$ . So $c=c\\cap(b\\cup c)=c\\cap(b\\cup d)=(c\\cap b)\\cup(c\\cap d)=0\\cup(c\\cap d)=c\\cap d$ . Similarly, $d=d\\cap c$ . As a result, $c=c\\cap d=d\\cap c=d$ .
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