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Making A Mapping Injective

February 22, 2019


Finding a set of nearly independent objects

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 {F:X \rightarrow Y}. We can make the map surjective if we restrict {Y} to be equal to {F(X)}. It is not so simple to make the map injective, but we can in general do that also.

Theorem 1 Let {F} be a surjective function from {X} to {Y}. Then there is a subset {S} of {X} so that {F} is injective from {S} to {Y}.

Proof: For each {y} in {Y} select one {x} from the set {F^{-1}(y)} and place it into {S}. Recall {F^{-1}(y)} is the set of {z} so that {F(z)=y}.This of course uses the Axiom of Choice to make the choices of which {x} to choose. Then clearly {S} is the required set. \Box

The difficulty with this trivial theorem is that {S} cannot be controlled easily if it is constructed via the Axiom of Choice. It could be a very complicated set. Our goal is to see how well we can control {S} if we assume that the mapping {F} is smooth.

How can we do better? The answer is quite a bit better if we assume that {F} is a “nice” function. We give up surjectivity onto {Y} but only by a null set.

Theorem 2 Suppose that {F} is a surjective smooth map from {X} to {Y} where {X} and {Y} are open subsets of {{\mathbb R}}. Also suppose that {F} locally is invertible. Then there is a subset {S} of {X} so that

  1. The complement of {F(S)} is a null set.

  2. The map {F} is injective from {S} to {F(S)}.

That is that for all distinct points {\boldsymbol{a}} and {\boldsymbol{b}} in {S}, {F(\boldsymbol{a}) \neq F(\boldsymbol{b})}. Moreover the map from {F(S)} to {S} is smooth.

Set Coverings

How can we prove this theorem? An obvious idea is to do the following. Pick an open interval {U=[a,b]} in {X} so that {F(U) = V} for an open set in {V} and so that {F} is injective from {U} to {V}. Setting {S} to {U} clearly works: the map {F} is injective on {S}. This is far from the large set that we wish to have, but it is a start. The intuition is to select another open interval {U'} that is disjoint from {U} so that again {F} is injective from {U'} to {V'}. We can then add {U'} to our {S}.

We can continue in this way and collect many open sets that we add to {S}. Can we arrange that the union of these sets yield a {S} so that {F(S)} is most of {Y}? In general the answer is no. Suppose that the intervals are the following:

\displaystyle  [k,k+1.1]

for {k=0,1,2,\dots} Roughly we can only get about half of the space that the intervals cover and keep the chosen intervals disjoint. If we select { [k,k+1.1] } then we cannot select {[k+1,k+1+1.1]} since

\displaystyle  [k,k+1.1] \cap [k+1,k+1+1.1] \neq \emptyset.

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 {U} be a subset of {{\mathbb R}}. Let {[a_{\lambda},b_{\lambda}]} be intervals over {\lambda} in some index set {I}. We say these intervals are a cover of {U} proved {U} is a subset of the union of all the intervals. Say the intervals also are a Vitali cover of {U} provided for all points {x} in {U} and all {\epsilon > 0}, there is an interval {[c,d]} that contains {x} and {0 < d-c < \epsilon}.

The Vitali theorem is the following:

Theorem 4 Let {U} be a subset of {{\mathbb R}}. Let {[a_{\lambda},b_{\lambda}]} be intervals for {\lambda} in some index set {I}. Assume that the family is a Vitali cover of {U}. Then there is a countable subfamily of disjoints intervals in the family so that they cover all of {U} except for possibly a null set.

The Vitali theorem can be extended to any finite dimensional space {{\mathbb R}^{n}}. Then intervals become disks and so on.

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 {Y}.

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3 Comments leave one →
  1. Dr. S Srinivas Rau permalink
    February 23, 2019 8:39 pm

    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.

  2. arthurrainbow permalink
    March 9, 2019 10:53 pm

    «are a cover of {U} proved {U}» you probably mean «provided»

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