Re: A topological property

In article <1147863837.863703.200900@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, cliomseerg@xxxxxxxxxxxxxxxxxxxxxxxxx says...


Hello again,

Sorry to reincarnate this topic.. I have a few things that is not quite
unrelated to this one. So I think I will post it in this thread.

I have something new which is worth sharing and is related to this
topic. Not at all complicated, would make a nice exercise in
combinatoric theory... here goes...

Let X be a topological space
A family of open set {U_i} indexed by I is said to be intertwined (we
also roughly say that the open sets are intertwined) iff for any two
nonempty partition J,K of I we have

\/_{j in J} U_j /\ \/_{k in K} U_k =/= emptyset

I used a new definition namely, "intertwining".. which I originally
intended when trying to define pseudoconnectedness. I don't know maybe
there is already a term for this (just like pseudoconnectedness, I have
to confirm this with you people :) )


....
Thus in our discussions.. a set is pseudoconnected iff any open
covering of it is intertwined.
----

And here is a remark which I am almost done proving.

Let I be a subset of the natural numbers and {U_i}_{i in I} be an
intertwined family of open sets in a topological space X. Then there is
a surjective map
f: N -> I
such that U_f(i) /\ U_f(i+1) =/= emptyset

Sketch of proof:
wlog let 1 be in I, then set f(1)=U_1
if |I|>1 set f(2) to be an open set (which must exist because of
intertwineness)
which intersects with U_1, wlog let it be U_2
If |I|>2 set f(3) be
- if U_2 doesnt intersect with any open set other than U_1, let it be
U_2 and then set f(4) to be U_3 which is an open set that intersect
with U_1 that isnt U_2

continue in the same manner for the rest of the natural numbers. If I
is finite, this will end somewhere at say f(n)=U_k, then set
f(n+1)=U_k, f(n+2)=U_k .. etc.
If I is infinite I will leave it to you to show that f is indeed
surjective :) (if it where not, there will arrive a certain
contradiction with regards to intertwining)!

[snip]



I don't know how useful this may be to you, but I think it makes the
intuitive picture clearer. Given sets {U_i, i in I} you can form a
graph G whose vertices are the U_i's and whose edges are just the pairs
U_i U_j that have a nonempty intersection. Then the U_i's are
intertwined iff G is connected, and the existence of your map f:N -> I
is a consequence of the fact that a finite connected graph admits
an infinite walk that touches every vertex.



Robert Sheskey


.

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