Re: Functions and Curves

quasi wrote:
On Sat, 30 Jun 2007 17:11:21 -0400, David Bernier
<david250@xxxxxxxxxxxx> wrote:

quasi wrote:
On Fri, 29 Jun 2007 18:30:43 -0400, David Bernier
<david250@xxxxxxxxxxxx> wrote:

David Bernier wrote:
[...]
As a robustness test for all these ideas and tricks, I'd appreciate any feedback on the following
tentative claim:

(Tentative Claim):
There exists a continuous, injective map h: R-> R^2 such that Qx {0} is included in h(R).

My reasoning starts from constructing a bijection from Z to Q and proceeds from there following
quasi's path-glueing approach and then thinking along the lines of Wade's
"Now P cannot separate A ..." argument.
Going further, with basically the same ideas,

Does there exist a continuous, injective map g: R-> R^2
such that QxQ is a subset of g(R) ?
Yes -- see my other reply.
That article is not on the list from where I post, but I quote below from another
Usenet feed.

[quasi]
In fact, by the same construction, for _any_ countable subset S of
R^2, there exists an injective continuous map h:R->R2 such that h(R)
contains S.
[quasi, over]

I'll assume for now S is unbounded, just like QxQ.

We could call S the set of stops. If we arrange our bijection, say we call it s,
s: Z -> S
so that s(N) is unbounded, and s(Z-N) is also unbounded, then when we extend, we can
always find a point not in the range of the n'th path, since the range of the n'th path
is bounded. This is done so that the end result will be an injective map.

In case S is bounded, I haven't thought about what one could do.

The boundedness or unboundedness of S is irrelevant.

For this dicussion, "polygonal path" we will mean a path between 2
distinct points consisting of finitely many line segments such no 2
line segments intersect or overlap.

Assume S is countably infinite. Partition S arbitrarily into two
countably infinite subsets, S1 and S2.

Build an open-ended path containing S as follows ...

The idea is to use a sequence of polygonal paths, alternately
extending each endpoint.

For the initial polygonal path, connect the first point of S1 to the
first point of S2. Label the endpoints L1 and L2 respectively.

Then extend L1 to the next point of S1 and relabel L1 to represent the
new unused endpoint.

Similarly, extend L2 to the next point of S2 and relabel L2.

Continue forever. The union of all the resulting line segments gives
an open-ended path containing S.

Remarks:

It's not necessary to avoid other points of S1 and S2 as you extend.
Just so long as you leave infinitely many points of each unused.
However it's conceptually simpler just to avoid all points of S except
the new target point as you extend.

I agree that it's almost at the point where one could write
an algorithm or procedure for this.

I'd just like to give an example where the rules so far still seem to me
to require a bit of insight, so a robot might fail.

If S=({1/n : n an integer >0} union {0}) x {0},
then starting from (0,0) and going along a straight line to (1/3,0) would leave two target points: (1,0) and (1/2,0).

But I guess after reaching those two points, one could extend both ways parallel to the
y-axis. However, extending the second time from (0,0) the other way,
I know and you know we can take a circular arc to (1/2,0). But that's
a rule the robot might not know. He might be programmed to try a straight line,
which won't work. The robot needs a new rule...


At any stage of the construction, there are only finitely many line
segments, and no loops, so there is no disconnection of R^2.

quasi


.

Relevant Pages

  • Re: Functions and Curves
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  • Re: Functions and Curves
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