Re: Characterization of an Open Set in the Reals

From: tinyurl.com/uh3t (rem642b_at_Yahoo.Com)
Date: 12/31/04 

Date: Thu, 30 Dec 2004 17:46:13 -0800

> From: Dave Seaman <dseaman@no.such.host>
> And on the real line, "convex" turns out to be the same as "path
> connected", and both are equivalent to "connected."

Yes. By the way, I looked up the definition of "path connected" to make
sure I remembered the correct definition:
  http://open-site.org/Science/Mathematics/Topology/
       A space is path-connected if for any two points x,y there is a
    continuous map f from the unit interval [0,1] with the usual metric
                  topology such that f(0)=x and f(1) = y.
I assume [0,1] refers to a closed interval of the **reals**. So that
definition is intimately tied to the reals. A path is a mapping from a
closed interval of **reals**. So testing whether a subset of the reals
is path-connected or not is usually trivial, involving nothing more
than the question fitting a linear function f(x)=a*x+b to two points
with range of that function restricted to the given subset of reals.
Accordingly it's trivial that a subset of reals is path-connected iff
it's convex. And your missing-point proof shows convex iff connected.
All of that can be shown with high-school algebra, even if the full
axiomization or construction of the reals has been forgotten or set
aside or even not learned in the first place.

However I consider the word "path" to be a misnomer, and likewise
"path-connected", since these are functions from *real* intervals, they
should really be called "real-path" and "real-path-connected" to
emphasize how they are defined in terms of domain equalling interval of
*reals*. So "path-connected" i.e. "real-path-connected" isn't a
topological property per se, it's rather a property of how a given
topological space relates to the reals per whatever your favorite
construction or axiomization of the reals happens to be, or maybe it
*requires* the usual axiomization of the reals or something equivalent
to it, so the constructable reals or the
nonstandard-analysis-with-infinitesimals reals are excluded from that?

Anyway, we could take any ordered topological space as the basis for
such a definition, taking intervals from that space instead of from the
reals. For example, suppose we take intervals of the rationals. Then we
get a definition of "rational-path-connected". By the same arithmetic
interpolation (fitting of linear function between two points), we can
show that the rationals are rational-path-connected, and that a subset
of rationals is convex (as subset of rationals) iff it's
rational-path-connected.

> I claim that what I actually showed is that a subset of
> R that is not path connected (or convex) is not connected.

I'll go with "convex" in that claim. If it's not convex, then from the
definition of "convex" there are two points in A such that the line
segment between the two points contains some point missing from A, it's
obvious that line segment is simply the interval between the two
numbers, so then you demonstrate two open subsets of A, neither empty,
that comprise all of A, QED. The corresponding argument with "path
connected" instead of "convex" is not so trivial, and is irrelevant to
the overall proof since "convex" is the property I used to prove the
four types of intervals are the only possible such subsets of R, so
there's really no need to bother introducing "path-connected" in any
lemma.

So I claim the overall proof that the only connected open subsets of R
are in fact the four kinds of open intervals, breaks down into a chain
of two lemmas:
  A has property connected open (in R as topological space)
     iff
  A has property convex (in R as geometric space)
     iff
  A is exactly one of the four types of intervals of R (as ordered set)
So with that slight extension of your proof, you did half the job, and
my addenda did the other half, OK? (Note "geometric space" means the
only definitions we use are points, lines, betweenness, and from them
we define finite line segments and directed rays. So the four kinds of
intervals are actually three geometrically: finite open segment, open
directed ray, and whole-line. For example, if a and b are two distinct
points, the directed open ray from a towards b may be defined as the
union of all finite open intervals (b,x) which do not include a, and
the directed open ray from a away from b may be defined as the set of
all x such that a is between b and x. But alas, the bottom line of the
iff chain uses R as an ordered set, not a geometric space, so we can't
really express the final conclusion in only 3 cases except by
abstracting the two cases (a,posinf)={x|a<x} and (neginf,b)={x|x<b}
into a single definition of "half-bounded interval" or somesuch
jargon.)

> Yes, the statement that every path-connected subset of R is an
> interval requires justification, but it's completely trivial.

Are you sure you meant to say "path-connected" instead of "convex"
there? I outlined the quick/easy proof for "convex", making use of the
fact that if any subset of reals has an upper bound then it has a least
upper bound (within the reals!). But the corresponding proof for
"path-connected" would take a bit more work and I'd rather ignore that
issue. Obviously any interval is path-connected, for any two points use
the linear function as a path between them. But the converse would seem
to require application of the bounded value theorem (*) to show that the
image of any path is a finite interval, and then some additional logic
I can't quite figure out to show that the union of all such paths is a
single interval. I really think it's best if we stick with "convex" as
the link between "connected" and "exactly an interval", OK?

* (Is that the correct name: For any continuous real function with
domain being a closed interval, the range of the function is bounded?)

> In particular, I do not consider a proof in this newsgroup to be
> "incomplete" if it involves taking the sup of a bounded set of real
> numbers without justification. You are supposed to know what the
> real numbers are, after all.

But you left out the entire section of proof where you show that either
there is or is not an upper bound and in the former case you apply the
theorem/axiom that there's a sup. It's ok to sketch the proof where you
mention the theorem/axiom is used, but if you don't even mention the
theorem/axiom or the outline of the section where it's used then I
claim something was missing.

By the way, it occurs to me that the logic used here might have some
application in that horrible thread where a troll is claiming that
999... is not equal to 1. First, the rationals are "missing" some
points in the sense you outlined. There are sets A and B, each open,
disjoint, such that A union B is *all* the rationals. Each of A and B
is a half-bounded convex set, yet there's no rational number which is
the common boundary point between A and B. For example, B is the set of
positive rationals whose square is greater than 2, and A is the union
of the negative rationals, singleton zero, and positive rationals with
square less than 2. There's no such thing as sqrt(2) in the rationals.
In fact between *any* two distinct rationals, there's such as missing
place. The rationals ar full of "holes" like that. But then we use
Dedekind cuts to "fill in the holes", so that the reals don't have any
missing places whatsoever. For example of that crap/troll thread:
  http://groups.google.co.in/groups?selm=20041219153018.06125.00001358%40mb-m23.aol.com
  ... there is a space existing between, .999... and 1 ...

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