Sujet : Re: WM and end segments...
De : ben (at) *nospam* bsb.me.uk (Ben Bacarisse)
Groupes : sci.mathDate : 22. Jul 2024, 00:13:38
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <878qxu72zh.fsf@bsb.me.uk>
References : 1 2
User-Agent : Gnus/5.13 (Gnus v5.13)
Ben Bacarisse <
ben@bsb.me.uk> writes:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>
For some damn reason when I hear end segments from WM I think of a
tree. Take the following infinite 2-ary tree that holds the positive
integers:
___________________________________________
0
/ \
/ \
/ \
/ \
1 2
/ \ / \
/ \ / \
3 4 5 6
/ \ / \ / \ / \
.........................
___________________________________________
>
this goes on and on for infinity... We all can see how this can go for
infinity, right WM? Wrt trees there are only leaves in a finite view of
it. However, the "infinite view" of the tree has no leafs because it never
ends... Fair enough? Or too out there?
>
That's a can of worms in WMaths. WM has written 734,342,120 nonsense
posts about binary trees over the years. It's one of his favourite
examples to use to bamboozle his poor students.
>
The infinite binary tree -- simply a graph with node set N and edge set
(n, 2n+2) (in your numbering) -- is a particular puzzle for WM because
Correction, there are two such edges of course: (n, 2n+1) and (n, 2n+2).
the node and edge sets are countable but the path set isn't.
>
Can you see a proof that the infinite rooted paths can be mapped, one to
one, with an uncountable subset of R?
>
... The infinite one has no leaves.
>
If you consider graphs in general, they do not have to be infinite to
have no leaves.
-- Ben.