Direct Image / Reciprocal Image of a Part

Some notions on the scientific CPGE program are less easy to assimilate than others. If a list of the 10 concepts that pose the greatest difficulty for students is drawn up, I can say from experience that the notion of direct image or mutual image would figure prominently. Direct Image

The purpose of this note is to help clarify this issue and to show some of the main uses at this level.


  1. Notion of application,
  2. Image of an element of the starting set,
  3. Possible history of an element of the arrival set. Direct Image

1 – Direct image: definition and first examples

Consider an application f:X\rightarrow Y and a part (= a subset) A of X .

We call a direct image of A by f all :

    \[ \boxed{f\left\langle A\right\rangle =\left\{ y\in Y;\thinspace\exists x\in A,\thinspace f\left(x\right)=y\right\} }\]

f\left\langle A\right\rangleso is the set of elements of Y which are the image of an element of A .

Rephrase …

f\left\langle A\right\rangleis the set of elements of Y having an antecedent in A .

We can write just as well:

    \[ \boxed{f\left\langle A\right\rangle =\left\{ f\left(x\right);\thinspace x\in A\right\} }\]

which reads: f\left\langle A\right\rangle is the set of images by f elements of A .

If the above is not perfectly clear, the following examples may help …

Example 1.

We start, it is unavoidable, by an example with “potatoes”


If we note f the application represented by the diagram opposite and X,Y his start and finish sets, then:

f\left\langle X\right\rangle =Y-\left\{ C\right\}

f\left\langle \left\{ a,b,c\right\} \right\rangle =\left\{ A,B\right\}

f\left\langle \left\{ c,d,e\right\} \right\rangle =\left\{ B,D\right\}

Example 2.

For f:\mathbb{N}\rightarrow\mathbb{N},\thinspace n\mapsto2n+1 , it’s clear that f\left\langle \left\{ 0,2,7\right\} \right\rangle =\left\{ 1,5,15\right\} and f\left\langle \mathbb{N}\right\rangle is the set of odd natural numbers.

Let’s go back to the general case and make three small remarks about an application f:X\rightarrow Y.

The condition f\left\langle X\right\rangle =Y when filled, expresses the surjectivity of f (any element of Y has at least one antecedent f ). If x\in X , then f\left\langle \left\{ x\right\} \right\rangle =\left\{ f\left(x\right)\right\}. Finally : f\left\langle \emptyset\right\rangle =\emptyset. Let’s continue our exploration now, with new examples … Direct Image

Example 3.

To g:\mathbb{R}\rightarrow\mathbb{R},\thinspace x\mapsto x^{2}, it’s clear that g\left\langle \mathbb{R}\right\rangle \subset\left[0,+\infty\right[.

In fact : g\left\langle \mathbb{R}\right\rangle =\left[0,+\infty\right[, But why ? And would you know how to determine g\left\langle \left[-1,2\right]\right\rangle?

Answers detailed in the APPENDIX at the end of the article.

Example 4.

Let’s note E=\mathbb{N}-\left\{ 0,1\right\} and consider the application

    \[ P:E^{2}\rightarrow\mathbb{N},\thinspace\left(a,b\right)\mapsto ab\]

So P\left\langle E^{2}\right\rangle is the set of natural numbers that can be written as the product of two strictly positive integers, but in a non-trivial way (since none of the two factors can be worth 1). We see that P\left\langle E^{2}\right\rangle is the set of compound numbers (that is, non-primes).

2 – We pause to think about the notation

You may be wondering why I persist in using the notation f\left\langle A\right\rangle whereas we find, in a lot of books, the notation f\left(A\right). The reason is simple. It is essential to avoid confusion between:

  • on the one hand, the image of an element
  • on the other hand, the direct image of a part

No risk, you will say to me: an element and a part, it is not the same thing! The context must therefore make it possible to know what we are talking about.

It must be recognized that it is – in practice – often the case, but not always …

Consider the set:

    \[ X=\left\{ 0,1,2,\left\{ 0,1\right\} \right\} \]

Obviously, the integers 0 and 1 are elements of X , so the whole \left\{ 0,1\right\} is a part of X . But it’s also an element of X …! In these circumstances, if f:X\rightarrow Y is an application (no matter the arrival set Y), it is essential to note differently the image of the element \left\{ 0,1\right\} and the direct image of the part \left\{ 0,1\right\}.

Not clear ? Let’s go after this example by defining completely f :

    \[ f\left(0\right)=3\qquad f\left(1\right)=4,\quad f\left(2\right)=5,\quad f\left(\left\{ 0,1\right\} \right)=6\]

We see that :

    \[ f\left\langle \left\{ 0,1\right\} \right\rangle =\left\{ 3,4\right\} \neq6=f\left(\left\{ 0,1\right\} \right)\]

I hope I have convinced you of the necessity of distinguishing, by different notations, the image of an element of the direct image of a part.

3 – Two more elaborate examples of direct images

Example 5.

Consider the application Direct Image

    \[ \varphi:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2},\thinspace\left(a,b\right)\mapsto\left(a+b,ab\right)\]

and try to determine \varphi\left\langle \mathbb{R}^{2}\right\rangle. Let’s start by observing that if \left(x,y\right)\in\varphi\left\langle \mathbb{R}^{2}\right\rangle, then there is \left(a,b\right)\in\mathbb{R}^{2} such as :

    \[ \left\{ \begin{array}{ccc} x & = & a+b\\ y & = & ab\end{array}\right.\]

which imposes:

    \[ x^{2}-4y=\left(a+b\right)^{2}-4ab=\left(a-b\right)^{2}\geqslant0\]

Noting V=\left\{ \left(x,y\right)\in\mathbb{R}^{2};\thinspace x^{2}-4y\geqslant0\right\}, So we have \varphi\left\langle \mathbb{R}^{2}\right\rangle \subset V.

Let’s see now what is the opposite inclusion …

Is \left(x,y\right) a couple of real numbers checking x^{2}-4y\geqslant0. The discriminator of the equation of unknown T :

    \[ T^{2}-xT+y=0\]

being positive or zero, it has two roots a and b realities (possibly confused) and we know (relations between coefficients and roots) that: Direct Image

    \[ a+b=x\qquad\text{et}\qquad ab=y\]

If the “relationship between coefficients and roots” is unfamiliar to you, you can simply calculate:

    \[ a+b=\frac{x-\sqrt{x^{2}-4y}}{2}+\frac{x+\sqrt{x^{2}-4y}}{2}=x\]

as well as :

    \[ ab=\frac{x-\sqrt{x^{2}-4y}}{2}\;\frac{x+\sqrt{x^{2}-4y}}{2}=\frac{x^{2}-\left(x^{2}-4y\right)}{4}=y\]

The reverse inclusion is established and finally:

    \[ \boxed{\varphi\left\langle \mathbb{R}^{2}\right\rangle =V}\]

By the way, V is the union of the equation parable y=\frac{x^{2}}{4} and its “outside”. In the figure below, V is colored in blue:


Example 6.

We have known for almost 23 centuries that there is an infinity of prime numbers. The evidence proposed by Euclid is based on the following observation: Direct Image

If p_{1},\cdots,p_{n} are prime numbers all distinct, so the integer 1+\prod_{i=1}^{n}p_{i} is not divisible by any of the p_{i} for 1\leqslant i\leqslant n. Its prime factors are therefore all out of the box \left\{ p_{1},\cdots,p_{n}\right\}. Euclid proves that the whole \left\{ p_{1},\cdots,p_{n}\right\} can not contain all prime numbers, from which he infers that the whole \mathbb{P} prime numbers are infinite.

Now consider the application \psi:\mathbb{N}^{\star}\rightarrow\mathbb{N} defined by recurrence, as follows:

  •  \psi\left(1\right)=2
  •  For any integer n\geqslant1,\psi\left(n+1\right) denotes the smallest prime factor of Q_{n}={\displaystyle 1+\prod_{i=1}^{n}\psi\left(i\right)}

By construction and according to Euclid’s argument, the numbers \psi\left(n\right) are all first and are two to two distinct. compute \psi\left(n\right) for small values of n :

  •  Q_{1}=1+2=3, so \psi\left(2\right)=3
  •  Q_{2}=1+2\times3=7, so \psi\left(3\right)=7
  •  Q_{3}=1+2\times3\times7=43, so \psi\left(4\right)=43
  •  Q_{4}=1+2\times3\times7\times43=1807=13\times139, so \psi\left(5\right)=13
  •  Q_{5}=1+2\times3\times7\times43\times13=23479=53\times443, so \psi\left(6\right)=53
  •  Q_{6}=?\ldots It’s yours !! Reply in the APPENDIX , at the end of the article .

The following \left(\psi\left(n\right)\right)_{n\geqslant1} is known as the Euclid-Mullin suite.

It is obvious that \psi\left\langle \mathbb{N}^{\star}\right\rangle \subset\mathbb{P}, but the question of whether any prime number is (or not) a term of this sequence remains, for more than half a century, an open problem! In other words, we do not know to date the direct image of \mathbb{N}^{\star} by \psi.

4 – Reciprocal image: definition and first examples

Consider an application f:X\rightarrow Y and a part (= a subset) B of Y . We call reciprocal image of B by f all :

    \[ \boxed{f^{-1}\left\langle B\right\rangle =\left\{ x\in X;\thinspace f\left(x\right)\in B\right\} }\]

This reads: Direct Image

f^{-1}\left\langle B\right\rangle is the set of elements x of X whose image by f belongs to B “.

In other words : f^{-1}\left\langle B\right\rangle is the set of antecedents by f elements of B .

Example 7.

We start again with the “potatoes” …


Noting again f the application shown opposite and X,Y his start and finish sets:

f^{-1}\left\langle \left\{ A\right\} \right\rangle =\left\{ a,b\right\}

f^{-1}\left\langle \left\{ B,C\right\} \right\rangle =\left\{ c,d\right\}

f^{-1}\left\langle \left\{ C\right\} \right\rangle =\emptyset

Example 8.

Let’s take the application again g example 3 :

    \[ g:\mathbb{R}\rightarrow\mathbb{R},\thinspace x\mapsto x^{2}\]

and determine g^{-1}\left\langle \left[1,4\right]\right\rangle. By definition :

    \[ g^{-1}\left\langle \left[1,4\right]\right\rangle =\left\{ x\in\mathbb{R};\thinspace1\leqslant x^{2}\leqslant4\right\} =\left\{ x\in\mathbb{R};\thinspace x^{2}\geqslant1\right\} \cap\left\{ x\in\mathbb{R};\thinspace x^{2}\leqslant4\right\} \]

On the one hand,

    \[ \left\{ x\in\mathbb{R};\thinspace x^{2}\geqslant1\right\} =\left]-\infty,-1\right]\cup\left[1,+\infty\right[\]

And on the other hand :

    \[ \left\{ x\in\mathbb{R};\thinspace x^{2}\leqslant4\right\} =\left[-2,2\right]\]

So :

    \[ \boxed{g^{-1}\left\langle \left[1,4\right]\right\rangle =\left[-2,-1\right]\cup\left[1,2\right]}\]

This relationship is visualized with the following illustration:


Example 9.

In the first year algebra class, it is shown that if E,F are two \mathbb{K}– vector spaces and if u:E\rightarrow F is linear, then for any vector subspace (sev) F' of F, the reciprocal image of F' by u is a vector subspace of E . Let’s check this property together. Direct Image

We will note 0_{E} and 0_{F} the respective null vectors of E and of F .

This is to prove on the one hand that 0_{E}\in u^{-1}\left\langle F'\right\rangle and on the other hand that u^{-1}\left\langle F'\right\rangle is stable by linear combination.

First of all, as u is linear, we know that u\left(0_{E}\right)=0_{F}. Otherwise, F' being a sev from F we also know that 0_{F}\in F'. It follows that :

    \[ \boxed{0_{E}\in u^{-1}\left\langle F'\right\rangle }\]

Then do we give vectors x,y in u^{-1}\left\langle F'\right\rangle and a scalar \lambda. In order to prove that \lambda x+y\in u^{-1}\left\langle F'\right\rangle we calculate:

    \[ u\left(\lambda x+y\right)=\lambda\thinspace\underbrace{u\left(x\right)}_{\in F'}+\underbrace{u\left(y\right)}_{\in F'}\]

and we invoke stability by linear combination for F'. We see that u\left(\lambda x+y\right)\in F', which settles the question.

It should be noted in passing that a particular fundamental case is one where F'=\left\{ 0_{F}\right\}; in that case, u^{-1}\left\langle F'\right\rangle is none other than the core of u . This is how the core of any linear application is a sev of its starting space. Direct Image

5 – An ambiguity that is not an ambiguity

Given a bijection u:X\rightarrow Y, the notation u^{-1} is used to designate its reciprocal bijection. Therefore, if B\subset Y, a potential ambiguity arises when writing u^{-1}\left\langle B\right\rangle.

Is it:

  •  of the reciprocal image of B by u?
  •  or the direct image of B by u^{-1}?

Rest assured, this ambiguity is only apparent because these two sets are confused. Let’s prove that by carefully checking for double-inclusion.

note A the reciprocal image of B by u : the elements of A are, by definition, the elements x of X checking u\left(x\right)\in B.

note A' the direct image of B by u^{-1} : the elements of A' are those of the form u^{-1}\left(y\right), for y in B.

If x\in A, then x=u^{-1}\left(u\left(x\right)\right)\in A'. This proves that A\subset A'

Conversely, if x\in A' then there is y\in B such as x=u^{-1}\left(y\right), then u\left(x\right)=u\left(u^{-1}\left(y\right)\right)=y, which proves that u\left(x\right)\in B in other words, x\in A. This proves that A'\subset A.

Double-inclusion is established. Direct Image

Finally, let’s add that if u:X\rightarrow Y is not bijective, then the symbol u^{-1} is not defined ! But of course, we can write u^{-1}\left\langle B\right\rangle for B\subset Y.

6 – Four legs \textrightarrow{} A sheep

In combinatorics, one sometimes uses the following result, known under the name of “lemma of the shepherds”.

Given two sets E,F, we suppose that F is finished and there is an application u:E\rightarrow F whose fibers all have the same cardinal k\geqslant1.

So E is finished and:

    \[ \boxed{\text{card}\left(E\right)=k\thinspace\text{card}\left(F\right)}\]

As for the “fibers” of u they are (by definition) the reciprocal images of singletons. Here is a formalized version of the same statement:

    \[ \left[\exists k\in\mathbb{N}^{\star};\thinspace\forall y\in F,\thinspace\text{card}\left(u^{-1}\left\langle \left\{ y\right\} \right\rangle \right)=k\right]\Rightarrow\left[\text{card}\left(E\right)=k\thinspace\text{card}\left(F\right)\right]\]

To prove it, just see that E is the union of the fibers of u. As a finite union of finite sets, E is finished. Moreover, this union being disjointed, the cardinal of E is the sum of the cardinals of the fibers, whence the announced formula. Direct Image

The name “lemma of shepherds” probably comes from a joke: we imagine a shepherd wanting to know the number of sheep in his herd … he just has to count the legs and divide by 4 the whole obtained!

The illustration below represents the application that goes from the set of legs to that of the sheep and which, with each leg, associates the sheep to which it belongs

For this application, the lemma applies with k=4. It is obviously assumed that the sheep are in good health and that they have not been genetically manipulated …


An example of how to use the lemma of shepherds:

Proposal. If E,F are finished with respective Cardinals p and n with 1\leqslant p\leqslant n, then there is n\left(n-1\right)\cdots\left(n-p+1\right) injections of E towards F.

This proposition can be proved by reasoning by recurrence on the cardinal of E (that of F being fixed). For p=1 , the n possible applications are obviously all injective, which initiates recurrence. Suppose the property established at the rank p , for a certain p<n, and we give two sets E,F of respective cardinals p+1 and n. Fix an element x of E and consider the application

    \[ u:\text{Inj}\left(E,F\right)\rightarrow\text{Inj}\left(E-\left\{ x\right\} ,F\right)\]

who at each injection of E towards F associates its restriction with E-\left\{ x\right\}. By hypothesis of recurrence, \text{card}\left(\text{Inj}\left(E-\left\{ x\right\} ,F\right)\right)=n\left(n-1\right)\cdots\left(n-p+1\right). Like every injection of E-\left\{ x\right\} towards F can be extended n-p ways in an injection of E towards F (by associating with x any of the n-p elements still available), we see that the lemma of the shepherds applies to u and :

    \[ \text{card}\left(\text{Inj}\left(E,F\right)\right)=\left(n-p\right)\:\text{card}\left(\text{Inj}\left(E-\left\{ x\right\} ,F\right)\right)=n\left(n-1\right)\cdots\left(n-p\right)\]

as wished.

7 – Reciprocal image of a direct image

If f:X\rightarrow Y and A\subset X, then f^{-1}\left\langle f\left\langle A\right\rangle \right\rangle \supset A. Indeed, if a\in A, then f\left(a\right)\in f\left\langle A\right\rangle(by definition of f\left\langle A\right\rangle) and so a\in f^{-1}\left\langle f\left\langle A\right\rangle \right\rangle (by definition of f^{-1}\left\langle B\right\rangle withB=f\left\langle A\right\rangle).

But this inclusion is usually strict. Consider the only application f:\left\{ 0,1\right\} \rightarrow\left\{ 2\right\} and ask A=\left\{ 0\right\}. So :

    \[ f^{-1}\left\langle f\left\langle A\right\rangle \right\rangle =f^{-1}\left\langle \left\{ 2\right\} \right\rangle =\left\{ 0,1\right\} \neq A\]


If f:X\rightarrow Y is injective, then f^{-1}\left\langle f\left\langle A\right\rangle \right\rangle =A, for any part A of X.

Indeed, given A\subset X, if x\in f^{-1}\left\langle f\left\langle A\right\rangle \right\rangle, then f\left(x\right)\in f\left\langle A\right\rangle which means that it exists a\in A such as f\left(x\right)=f\left(a\right). As f is injective, then x=a and so f^{-1}\left\langle f\left\langle A\right\rangle \right\rangle \subset A. As for the other inclusion, it has been established in general terms.

The reciprocal of the previous implication is true:

If f:X\rightarrow Y is such that f^{-1}\left\langle f\left\langle A\right\rangle \right\rangle =A for any part A of X, so f is injective. Direct Image

Indeed, be x,x'\in X such as f\left(x\right)=f\left(x'\right). By choosing A=\left\{ x\right\} in the event, we find that \left\{ x\right\} =f^{-1}\left\langle f\left\langle \left\{ x\right\} \right\rangle \right\rangle =f^{-1}\left\langle \left\{ f\left(x\right)\right\} \right\rangle. Or x'\in f^{-1}\left\langle \left\{ f\left(x\right)\right\} \right\rangle and so x'=x.

8 – Direct image of a reciprocal image

If f:X\rightarrow Y and B\subset Y, then f\left\langle f^{-1}\left\langle B\right\rangle \right\rangle \subset B. Indeed, if b\in f\left\langle f^{-1}\left\langle B\right\rangle \right\rangle, then (by definition of f\left\langle A\right\rangle with A=f^{-1}\left\langle B\right\rangle) it exists a\in f^{-1}\left\langle B\right\rangle such as b=f\left(a\right) and so b\in B.

Reverse inclusion is not true in general, as seen by considering, for example f:\left\{ 0\right\} \rightarrow\left\{ 1,2\right\} defined by f\left(0\right)=1 and B=\left\{ 1,2\right\}. We observe that :

    \[ f\left\langle f^{-1}\left\langle B\right\rangle \right\rangle =f\left\langle \left\{ 0\right\} \right\rangle =\left\{ 1\right\} \neq B\]


If f:X\rightarrow Y is surjective, then f\left\langle f^{-1}\left\langle B\right\rangle \right\rangle =B for any part B of Y.

Indeed, given B\subset Y, if b\in B, then it exists (by surjectivity of f) an element x\in X such as f\left(x\right)=b. obviously x\in f^{-1}\left\langle B\right\rangle and so b\in f\left\langle f^{-1}\left\langle B\right\rangle \right\rangle.

Reciprocally \ldots{}

If f:X\rightarrow Y is such that f\left\langle f^{-1}\left\langle B\right\rangle \right\rangle =B for any part B of Y , then f is surjective.

Indeed, either y\in Y . By choosing B=\left\{ y\right\} in the hypothesis, we see that \left\{ y\right\} =f\left\langle f^{-1}\left\langle \left\{ y\right\} \right\rangle \right\rangle \subset f\left\langle X\right\rangle. So y\in f\left\langle X\right\rangle; in other words, y admits an antecedent by f.

9 – Reciprocal image and topology

Given a metric space E (eg a standardized vector space or part of such a space) as well as a part \Omega of E, we sometimes need to prove that \Omega is an open of E. For this, several methods exist. We can\ldots{}

  •  apply the definition of an open (that is, to verify, for each x\in\Omega, the existence of an open center ball x included in \Omega)
  •  prove that the complementary in E of \Omega is a closed, by applying for example the sequential characterization of the closed ones (that is to say by checking that for any convergent sequence \left(x_{n}\right) at term in the complementary B of \Omega, the limit of this sequel still belongs to B)
  •  find a metric space F, an open W of F and a continuous application u:E\rightarrow F such as \Omega=u^{-1}\left\langle W\right\rangle.

This third method is based on the fact that “the reciprocal image of an open by a continuous application is an open”. Direct Image 

By simple passage to the complementary, one has the twin statement: “the reciprocal image of a closed by a continuous application is a closed”. Direct Image

Both statements are very useful in practice. Let’s give three examples of use.

Example 10-a.

In \mathbb{R}^{2}, the hyperbole of equation xy=1 is a closed since it is the reciprocal image of the closed \left\{ 1\right\} of \mathbb{R} by the application \mathbb{R}^{2}\rightarrow\mathbb{R},\thinspace\left(x,y\right)\mapsto xy which is continuous (since polynomial). More generally, any equation curve f\left(x,y\right)=0 or f:\mathbb{R}^{2}\rightarrow\mathbb{R} is continuous, is a closed part of \mathbb{R}^{2}. This shows that the curves \Gamma_{a} of polar equation: Direct Image

    \[ \rho=e^{a\theta}\qquad\left(a>0\right)\]

(known as logarithmic spirals ) do not have an equation of form f\left(x,y\right)=0 with f keep on going. Indeed, \Gamma_{a} is not closed, since the origin is an asymptotic point that does not belong to the curve.


Example 10-b.

In \mathcal{M}_{n}\left(\mathbb{C}\right), the group GL_{n}\left(\mathbb{C}\right) Inverted square matrices is an open since it is the reciprocal image of the open \mathbb{C}-\left\{ 0\right\} of \mathbb{C} by the application \mathcal{M}_{n}\left(\mathbb{C}\right)\rightarrow\mathbb{C},\thinspace M\mapsto\det\left(M\right) which is continuous (since polynomial, too). We can moreover show, more precisely, that GL_{n}\left(\mathbb{C}\right) is a dense open : any matrix A\in\mathcal{M}_{n}\left(\mathbb{C}\right) is the limit of a sequence of invertible matrices (just consider the general term matricial suite A_{k}=A+\frac{1}{k}I_{n},fork\ge1).

Example 10-c.

In \mathbb{R}^{2}, consider two parts A and B, not empty and disjointed (we could, without changing anything essential, replace in what follows \mathbb{R}^{2} by any metric space). A classic question is whether we can separate them, that is, whether there are open U and V disjointed such as A\subset U and B\subset V. The answer is usually negative, but if A and B are closed, it becomes affirmative!


Indeed, note:

  •  \left\Vert v\right\Vert the Euclidean norm of a vector v\in\mathbb{R}^{2}.
  •  d\left(v,X\right) the distance of a vector v to a party X, defined by:

    \[ d\left(v,X\right)=\inf\left\{ \left\Vert v-x\right\Vert ;\thinspace x\in X\right\} \]

We know that, whatever X\subset\mathbb{R}^{2}, the application v\mapsto d\left(v,X\right) is continuous because 1– Lipschitz. We also know that d\left(v,X\right)=0 if and only if, x\in\overline{X} (the adhesion of X). In particular, if X is closed in \mathbb{R}^{2}, then : d\left(v,X\right)=0\Leftrightarrow v\in X. Would you know how to establish these points? Solution in APPENDIX !

  •  f the application set on \mathbb{R}^{2} by :

    \[ \forall v\in\mathbb{R}^{2},\: f\left(v\right)=d\left(v,A\right)-d\left(v,B\right)\]

As f is continuous (difference of two continuous applications), then the sets

    \[ \boxed{U=f^{-1}\left\langle \left]-\infty,0\right[\right\rangle }\qquad\text{and}\qquad\boxed{V=f^{-1}\left\langle \left]0,+\infty\right[\right\rangle }\]

are open from \mathbb{R}^{2} because everyone is the reciprocal image of an open \mathbb{R}. They are obviously disjoint and moreover:

    \[ A\subset U\qquad\text{and}\qquad B\subset V\]

Indeed: if v\in A, then on the one hand d\left(v,A\right)=0 And on the other hand, v\notin B from where d\left(v,B\right)>0 since B is closed; so : f\left(v\right)<0. This shows that A\subset U. We also see that B\subset V. So we can always separate, in \mathbb{R}^{2} as in any metric space, two closed non-empty and disjointed.

Let’s finish by noting that the direct image of an open by a continuous application is not , in general, an open space of arrival. For example, the application \mathbb{R}\rightarrow\mathbb{R},x\mapsto\sin\left(x\right) transforms the open \left]0,\pi\right[in\left]0,1\right], which is not open.


Here are detailed answers to the questions raised in the text: Direct Image

Section 1 – Example 3

 This is a consequence of the intermediate value theorem. Consider a real y\geqslant0. As {\displaystyle \lim_{x\rightarrow+\infty}g\left(x\right)=+\infty}, it exists a\in\mathbb{R} such as g\left(a\right)>y. As 0 and g\left(a\right) are reached by g, then everything real between 0 and g\left(a\right) is also achieved: this is particularly the case of y. This proves that \left[0,+\infty\right[\subset g\left\langle \mathbb{R}\right\rangle. Reciprocal inclusion being obvious (the square of any real number is positive or zero), we have equality. Variations of g (decrease on \left]-\infty,0\right] and growth on \left[0,+\infty\right[ ) and its continuity show that

    \[ g\left\langle \left[-1,0\right]\right\rangle =\left[g\left(0\right),g\left(-1\right)\right]=\left[0,1\right]\]


    \[ g\left\langle \left[0,2\right]\right\rangle =\left[g\left(0\right),g\left(2\right)\right]=\left[0,4\right]\]

Therefore :

    \[ g\left\langle \left[-1,2\right]\right\rangle =g\left\langle \left[-1,0\right]\cup\left[0,2\right]\right\rangle =g\left\langle \left[-1,0\right]\right\rangle \cup g\left\langle \left[0,2\right]\right\rangle =\left[0,4\right]\]

Section 3 – Example 6

After 2,3,7,43,13 and 53 the next term is 5. Indeed :

    \[ 1+2\times3\times7\times43\times13\times53=1\thinspace244\thinspace335\]

and the smallest prime factor of this integer is obvious 5. As for the next, it is a little bigger … We calculate:

    \[ 1+2\times3\times7\times43\times13\times53\times5=6\thinspace221\thinspace671\]

who is first.

Section 9 – Number 3

Two properties of distance to a party

– Let v,w\in\mathbb{R}^{2}. For everything x\in X :

    \[ d\left(v,X\right)\leqslant\left\Vert v-x\right\Vert =\left\Vert v-w+w-x\right\Vert \leqslant\left\Vert v-w\right\Vert +\left\Vert w-x\right\Vert \]

and so :

    \[ d\left(v,X\right)-\left\Vert v-w\right\Vert \leqslant\left\Vert w-x\right\Vert \]

Passing to the lower limit:

    \[ d\left(v,X\right)-\left\Vert v-w\right\Vert \leqslant d\left(w,X\right)\]

In other words :

    \[ d\left(v,X\right)-d\left(w,X\right)\leqslant\left\Vert v-w\right\Vert \]

But v and w here play symmetrical roles! So we also have:

    \[ d\left(w,X\right)-d\left(v,X\right)\leqslant\left\Vert v-w\right\Vert \]

from where finally:

    \[ \boxed{\left|d\left(v,X\right)-d\left(w,X\right)\right|\leqslant\left\Vert v-w\right\Vert }\]

This shows that the application \mathbb{R}^{2}\rightarrow\mathbb{R},\thinspace v\mapsto d\left(v,X\right) East 1– Lipschitzian, so continue.

– If d\left(v,X\right)=0 then, according to the sequential characterization of a lower bound, there is a continuation \left(x_{n}\right)_{n\in\mathbb{N}} of elements of X as {\displaystyle \lim_{n\rightarrow\infty}\left\Vert v-x_{n}\right\Vert =0}. We see that v is adherent to X. Conversely, if v is a member of X, then such a continuation exists and passing to the limit in the frame 0\leqslant d\left(v,X\right)\leqslant\left\Vert v-x_{n}\right\Vert, we obtain d\left(v,X\right)=0Direct Image

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