The boundary of
is the set of points
which satisfy condition
. Let
be as in condition
; then the same
works for every point in
such that
. In
particular, each such
is in
. Further,
also is map
which shows that condition
is satisfied for each such
. So
is a manifold of dimension
, and because those points which don't satisfy
must satisfy
, it follows that
is a manifold of
dimension
.
Following the hint, consider
defined by
Let
,
,
,
. Then condition
holds except for part (3) since
Since
is open, each of its points satisfies condition
with
.
Let
. Then
satisfies
with
, say with
the function
. Let
be one of the half-planes
or
Suppose there is a sequence
of points of
such that the
all lie in
and converge to
. If there is no open neighborhood
of
such that
, then there is a sequence
of points
of
such that the sequence converges to
. But then
the line segments from
to
must contain a point o the boundary
of
, which is absurd since the points of U in the boundary of
all
map to points with last coordinate 0. It follows that h restricted to
an appropriately small open subset of
either satisfies condition
or condition
. This proves the assertion.
The generalization to manifolds is proved in the same way, except you need
to restrict attention to a coordinate system around
. By working in
the set
of condition
, one gets back into the case where one is
contained within
, and the same argument applies.
Let
be as in condition
applied to
,
,
and
be defined by
.
Then the function
satisfies all the desired conditions.
Let
be a basis for the subspace, and choose
so that all the
together form a basis for
. Define a
map
by
. One can verify that
satisfies the condition
.
If
is differentiable, the map
defined by
is easily verified to be a coordinate
system around all points of the graph of f; so the graph is a manifold of
dimension n.
Conversely, suppose
is as in
condition
for some point
in the graph. Let
be
the projection on the last
coordinates. Then apply the Implicit function
theorem to
. The differentiable
function
obtained from this theorem must be none other than
since
the graph is the set of points which map to zero by
.
Consider the case where n = 3. If
is defined in some open set by
, then
is defined by
. The Jacobian
is
Since either
or
is non-zero, it is easy to see that the Jacobian
has the proper rank.
In the case where n > 3, it is not obvious what one means by ``rotate".
For each
, one has condition
holding for some function
. Let
be the domain of one of these functions, where
we can choose
to be a ball with center at rational coordinates and
rational radius. Then
is a countable
cover of
. Now each
maps points of
in
to points with
the last
coordinates 0. Take a thin plate including the image of
; its inverse image has volume which can be bounded by
where
is the volume of the plate (by the change of
variables formula). By choosing the thickness of the plate sufficiently
small, we can guarantee that this value is no more than
for the
element of the cover. This shows the result.
Clearly, every element of
is in the boundary of
by the
condition
. If
is in the boundary of
, then
since
is closed. So if
, it must satisfy condition
.
But then
is in the interior of
because the dimension of
is n.
The open unit interval in
is a counter-example if we do not
require
to be closed.
By part (b), the boundary of
is
. By Problem 5-1,
is an
-dimensional manifold contained in
. By part (a),
it follows that
is of measure 0. Finally, since
is bounded,
the definition of Jordan measurable is satisfied.