- #1

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Is this well founded, d = {{x},{x,y},{x,y,z}}

because there exists an element of d i.e. {x} = e

such that d n e = 0 ???

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- Thread starter Mikemaths
- Start date

- #1

- 23

- 0

Is this well founded, d = {{x},{x,y},{x,y,z}}

because there exists an element of d i.e. {x} = e

such that d n e = 0 ???

- #2

vici10

Well-foundness is generalization of well-order. The difference is that well-order is linear and well-found is not necessary linear, but is partial order.

The definition is:

The relation E on set P is well-founded if any non-empty subset has E-minimal element.

Now, very often the natural order(relation) for sets is belonging [tex]\in[/tex].

So in this case, in your example d is well-founded since any non-empty subset:

1){{x}},2) {{x,y}},3) {{x,y,z}}, 4){{x},{x,y}}, 5){{x},{x,y,z}}, 6){{x,y}, {x,y,z}} , 7){{x},{x,y},{x,y,z}} has a minimal elements such as:

1){x}, 2){x,y}, 3){x,y,z}, 4){x}, {x,y} (since {x} [tex]\notin[/tex] {x,y} 5) {x}, {x,y,z} 6) {x,y} , {x,y,z} 7) {x}, {x,y} {x,y,z}

Notice that [tex]\in[/tex] is partial order on your set d. That is why 4), 5) 6) and 7) have several minimal elements.

So d is well-founded but not well-ordered, since it is not linear ordered by [tex]\in[/tex].

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