Making A Mapping Injective
Finding a set of nearly independent objects
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| Wikipedia bio source |
Giuseppe Vitali was the mathematician who famously used the Axiom of Choice, in 1905, to give the first example of a non-measurable subset of the real numbers.
Today I want to discuss another of his results that is a powerful tool.
The existence of a set that cannot properly be assigned a measure was a surprise at the time, and still is a surprise. It is a wonderful example of the power of the Axiom of Choice. See this for details.
We are interested in another of his results that is more a theorem about coverings. It is the Vitali covering theorem–see this. The theorem shows that a certain type of covering—ah, we will explain the theorem in a moment.
The power of this theorem is that it can be used to construct various objects in analysis. There are now many applications of this theorem. It is a powerful tool that can be used to prove many nice results. I do not know of any—many?—applications of the existence of a non-measurable set. Do you know any?
Making A Mapping Injective
Let’s look at an application of the Vitali theorem that may be new. But in any case it may help explain what the Vitali theorem is all about.
Suppose that 
Theorem 1 Let
be a surjective function from
to
. Then there is a subset
of
Proof: For each 
The difficulty with this trivial theorem is that 
How can we do better? The answer is quite a bit better if we assume that
Theorem 2 Suppose that
. Also suppose that
- The complement of
is a null set.
- The map
That is that for all distinct points
and
in
. Moreover the map from
Set Coverings
How can we prove this theorem? An obvious idea is to do the following. Pick an open interval ![{U=[a,b]}](https://s0.wp.com/latex.php?latex=%7BU%3D%5Ba%2Cb%5D%7D&bg=ffffff&fg=000000&s=0&c=20201002)
We can continue in this way and collect many open sets that we add to 
![\displaystyle [k,k+1.1]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Bk%2Ck%2B1.1%5D+&bg=ffffff&fg=000000&s=0&c=20201002)
for 
![{ [k,k+1.1] }](https://s0.wp.com/latex.php?latex=%7B+%5Bk%2Ck%2B1.1%5D+%7D&bg=ffffff&fg=000000&s=0&c=20201002)
![{[k+1,k+1+1.1]}](https://s0.wp.com/latex.php?latex=%7B%5Bk%2B1%2Ck%2B1%2B1.1%5D%7D&bg=ffffff&fg=000000&s=0&c=20201002)
![\displaystyle [k,k+1.1] \cap [k+1,k+1+1.1] \neq \emptyset.](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Bk%2Ck%2B1.1%5D+%5Ccap+%5Bk%2B1%2Ck%2B1%2B1.1%5D+%5Cneq+%5Cemptyset.+&bg=ffffff&fg=000000&s=0&c=20201002)
Vitali’s theorem comes to the rescue. It allows us to avoid his problem, by insisting that intervals have an additional property.
The Vitali Covering Theorem
The trick is to use a refinement of a set cover that allows a disjoint cover to exist for almost all of the target set. The next definition is critical to the Vitali covering theorem.
Definition 3 Let
be a subset of
be intervals over
in some index set
. We say these intervals are a cover of
in
, there is an interval
that contains
.
The Vitali theorem is the following:
Theorem 4 Let
The Vitali theorem can be extended to any finite dimensional space 
Open Problems
Do you see how to prove Theorem 2 from Vitali’s theorem? The insight is now one can set up a Vitali covering of the space
A result of Poincaré says that a surjective morphism of abelian varieties can be restricted to give an isomorphism.Cf Shafarevich Basic Algebraic Geometry. Analogues will be most interesting.
«are a cover of {U} proved {U}» you probably mean «provided»