Re: Would it matter if ZF was inconsistent?

In article
<e70aec30-c798-4a65-a5c7-1157c47f4fb8@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
LudovicoVan <julio@xxxxxxxxxxxxx> wrote:

On 17 July, 18:07, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
On Jul 17, 4:04 am, LudovicoVan <ju...@xxxxxxxxxxxxx> wrote:
On 16 July, 22:55, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:

There are uncountably many [paths]. We prove it. You don't prove
otherwise.

My 2c: Since a bijection can be given between paths and leaf nodes
over the transfinite tree,

There ARE NO leaf nodes.

Thanks for the question!

Below I have tried some transfinite-inductive definitions, leveraging
the infinite cartesian product to define the infinite case. For
illustration, there is also an implementation in Prolog that works for
the finite cases, with some sample output.

The predicate nodes(n) defines the set (actually, the list) of nodes
at level (depth) n in the binary tree, i.e. those nodes with n-1
parents up to (but excluding) the root. In particular, each node
encodes the whole path from the root to it, i.e. paths and nodes are
equivalent. For instance, an element of nodes(3) is [1, [0, [1, []]]],
corresponding to the path R-1-0-1.

The predicate tree(n) defines the set (actually, the list) of nodes in
a tree of level (depth) n. In particular, we will call *leaf nodes*
that subset of tree(n) that is the set nodes(n). And the set paths(n)
in tree(n) (i.e. the set of paths of length n in the tree of depth n)
is equivalent to the set nodes(n), i.e. the set of leaf nodes in tree
(n).

Extending the definitions to the transfinite, we have in particular
that the set of infinite paths in tree(w) corresponds to the set nodes
(w): i.e., again, the set of leaf nodes.

Let mathematics triumph!!


According to your definition, there is no difference between nodes and
paths, but in the definitions everyone else is using, there is an
essential difference: a node is not a set of nodes, but a path is a set
of nodes of a certain type.

Thus what may be the case in your trees need not be the case in ours and
vice versa.

And according to our definitions, e.g., see below, the paths in a
maximal complete infinite binary tree cannot have leaf nodes.



Model for a maximal infinite complete binary tree:
The set of nodes is the (1-origin) set of naturals, N = {1,2,3,...} with
the usual arithmetic.
1 is the root node.
For each n in N, 2*n_+0 is its left child and 2*n+1 is its right child.
A path is a subset, P, of N such that
1 is in P and
for each n in P one and only one of 2*n+0 and 2*n+1 is in P.

In THIS model, paths do not have leaf nodes. Further, no node is a path
nor any path a node.

--
Virgil
.

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