[FOM] Question on the number line

Ron Rood ron.rood at planet.nl
Thu Nov 17 11:56:53 EST 2005


Dean Buckner wrote on, 15 nov 2005 at 21:34 (Europe/Amsterdam):

> Suppose one takes a line segment of length 1 and
> break it in half (assuming we can break it "in half").  Intuitively, 
> you
> would imagine you get two identical segments of length 0.5.  But if you
> "identify" that line segment with the reals from 0.0 to 1.0, you have 
> to
> ask what happens to the point at 0.5 -- which "half" does it belong to?
>
> If it stays with only one side, then you don't have two identical line
> segments.  Instead, you have one segment that is closed at both ends 
> and
> another segment that is open at one end.
>


Here's an attempt at a solution to the puzzle. I hope that it does make 
sense:

 From a purely metrical point of view, one considers the line as a 
subset of
a space endowed with a metric. The question " are the line segments are 
identical?"
can be taken to mean "do the line segments have identical length?" To 
which half
the point at 0.5 belongs makes no difference in this case. (Something 
along these
lines appears to motivate Rob Arthan's posting from november 15, 
although he considers
a space endowed with a measure instead of a metric.)

A difference *does* arise when you consider the original question from 
a topological point
of view, that is, when you consider the line as a subset of a space 
endowed with
a topology. In that case, one half of the line is open at one end, the 
other half of the line
is closed (assuming that the original line was closed), i.e., they are 
topologically different.

The puzzle arises when you conflate the metrical perspective with the 
topological. Metrically
they are identical, topologically they are not. (Of course, the two 
segments are always different
when considered merely as *sets* without any sort of structure, be it 
metrical or topological.)

Of course, the metrical perspective naturaly gives rise to a topology: 
as a basis for a
topology, take all the "open balls" in the original metric space, that 
is, all sets of
the form B(x, r) = {y : d(x, y) < r} (x any point of the space, r a 
nonnegative real, denoting
the metric as "d"). But that doesn't harm the point--or so it seems to 
me.

Ron Rood



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