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X a subset of R^n is called centrally symmetric if the isometry f_z:R^n->R^n
defined by x|->2z-x for some z in R^n satisifies: f_z(X)=X.
and z is called centre of symmetry.
Now i need to show that:
1. if X is centrally symmetric and f is an isometry then f(X) is also centrally symmetric.
2. prove that if X has two centres of symmetry then it has infinie centres of symmetry.
3. prove that a bounded subset of R^n has at most one centre of symmetry.
Now for 1:
now sure how to do it, i mean from what is given:
fof_z(X)=f(X)
if i can prove that fof_z satisfies fof_z(x)=2z-f(x) then it will suffice, cause fof_z is a composiition of isometries which is itself an isometry.
now we proved that each isometry can be written as an affine transformation, i.e
f(x)=Ax+b where A is an orthogonal matrix (nxn).
now here: fof_z(x)=A(f_z(x))+b=A(2z-x)+b=2Az-Ax+b=2z'-(Ax+b)=2z'-f(x)
where z'=Az+b
will that do, or something is missing?
for the second question i tried to look if we have two centres of symmetry, then
z1,z2, then f_z_i(X)=X
then I thought of looking at f_z3(x)=[f_z2(x)+f_z1(x)]/2=(z2+z1)-x
if we define z3=(z1+z2)/2 then this should be another centre of symmetry, the prblem is that i need to show that: f_z3(X)=X, well perhaps this follows because every point in X is in f_z1(X) and f_z2(X) and then also it's in f_z3(X) cause f_z3(x) is actually a mean value of the other functions, is that even make sense (too mcuh hand wavy)?
for the third i thioght of showing that if it had for example two then from section 2 it will have more and thus get that the set isn't bounded, not sure how to formalise this.
any hints are apprecitaed.
defined by x|->2z-x for some z in R^n satisifies: f_z(X)=X.
and z is called centre of symmetry.
Now i need to show that:
1. if X is centrally symmetric and f is an isometry then f(X) is also centrally symmetric.
2. prove that if X has two centres of symmetry then it has infinie centres of symmetry.
3. prove that a bounded subset of R^n has at most one centre of symmetry.
Now for 1:
now sure how to do it, i mean from what is given:
fof_z(X)=f(X)
if i can prove that fof_z satisfies fof_z(x)=2z-f(x) then it will suffice, cause fof_z is a composiition of isometries which is itself an isometry.
now we proved that each isometry can be written as an affine transformation, i.e
f(x)=Ax+b where A is an orthogonal matrix (nxn).
now here: fof_z(x)=A(f_z(x))+b=A(2z-x)+b=2Az-Ax+b=2z'-(Ax+b)=2z'-f(x)
where z'=Az+b
will that do, or something is missing?
for the second question i tried to look if we have two centres of symmetry, then
z1,z2, then f_z_i(X)=X
then I thought of looking at f_z3(x)=[f_z2(x)+f_z1(x)]/2=(z2+z1)-x
if we define z3=(z1+z2)/2 then this should be another centre of symmetry, the prblem is that i need to show that: f_z3(X)=X, well perhaps this follows because every point in X is in f_z1(X) and f_z2(X) and then also it's in f_z3(X) cause f_z3(x) is actually a mean value of the other functions, is that even make sense (too mcuh hand wavy)?
for the third i thioght of showing that if it had for example two then from section 2 it will have more and thus get that the set isn't bounded, not sure how to formalise this.
any hints are apprecitaed.