| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
A recent usenet article stated without proof that
a) There exists a continuous bijective function f from
the closed upper half-space of R^3 (i.e., z >= 0) onto
R^3.
b) There does not exist a continuous bijective function
g from the closed upper half-plane of R^2 (i.e., y >= 0)
onto R^2.
c) There does not exist a continuous bijective function
h from the closed half-line [?] of R (i.e., x >= 0) onto
R.
[A bijective function is one that is both 1-1 and onto.
Continuity is taken relative to the standard Euclidean
topology of R^n. The inverse of f in part (a) will not
not be continuous.]
Discuss.
Dan
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1190.1 | AITG::DERAMO | Dan D'Eramo, nice person | Tue Feb 13 1990 22:24 | 15 | |
Hints:
A subset of R^n is compact if and only if it is closed
and bounded in the Euclidean topology.
A continuous image of a compact set is compact.
A continuous image of a connected set is connected.
A subset of R is connected if and only if it is convex.
These are enough to prove part (c).
Dan
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| 1190.2 | ABSZK::KOMATSU | Existentialist | Thu Feb 15 1990 13:45 | 16 | |
Projective Plan in n-dim Sphere :
in 1-dim : image of one point is just an point.
it devides R.
in 2-dim : image of the circle is just an circle.
(extend Boundary(S(2)) -> 1-dim sphere
extend R(2) -> 2-dim sphere)
in higher dimension you can find projective plan.
interesting thing in 1-dim is,
if you extend the map in the same way as higher dimension,
it is like this...
f : {0,infinity} --> 1-dim sphere
images of 2 points are 2 points. it will devide 1-dim sphere.
reverse image will give the direct sum of 2 open set for connective space.
...
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| 1190.3 | AITG::DERAMO | Dan D'Eramo, nice person | Thu Feb 15 1990 20:32 | 14 | |
(c) Let A be the set of nonnegative reals, A = {x | x >= 0}.
Assume f:A -> R is a continuous bijective function.
-1 -1
Let y1 = f(0) - 1, y2 = f(0) + 1. Let x1 = f (y1) and x2 = f (y2).
Neither x1 nor x2 is 0, and x1 =/= x2, so we have 0 < x1 < x2
or 0 < x2 < x1. In the first case let B = [x1,x2] and in the
second case let B = [x2,x1], i.e., B is the closed interval
between x1 and x2. Now B is connected, so the image f[B] is
connected, therefore convex, therefore it must contain every
real between y1 and y2, including f(0). Thus there is a (non-zero)
x between x1 and x2 such that f(x) = f(0) which contradicts
the choice of f (bijective means "1-1 and onto").
Dan
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