If and are such that is separately coprime with each , then is coprime with their product .

Proof: Through Bézout. being coprime with each , we can write, for each :

If we take the product of each of these expressions, we get:

Among the terms of this product, all but one contain as a factor. The one that doesn’t have as a factor is . Hence we obtain an expression of the form:

Hence and are coprime.

The same result holds with polynomials: if and are polynomials over a field , such that is coprime with each , then is coprime with their product .

The proof is identical.

]]>The degree of a polynomial P, written , is an element of . The symbol is taken to have the usual properties of order and of addition relative to natural numbers.

The family is a -vector space basis of . For any , there exists an unique , only finitely many of which are nonzero, such that . If all the are zero, that is, if is the zero element of , then is defined as . If not all the are zero, since only finitely many are nonzero, there is a greatest value of such that . Then is defined as this greatest value.

The following are easily checked to hold for any polynomials and (even when one or both are zero):

if , then

The notion of the degree of a polynomial allows us to formulate the following property of Euclidean division:

For all polynomials and , with nonzero, there exists exactly one pair of polynomials such that with .

For the proof, we fix the value of nonzero polynomial , and recurse over the degree of .

More specifically, being given, we consider the following proposition dependent on :

iff for all polynomials such that , there exist polynomials , with , such that .

is trivially true (for means that is zero, and satisfies with ).

Let us suppose true for some , and consider a polynomial the degree of which is less or equal to the successor of , that is to if , or otherwise.

If the degree of is less or equal to , then applies and there exist the wanted and .

If the degree of is less than that of , then we can directly write with and .

Remains the case in which is the successor of .

]]>itself, as a vector space over itself and the usual multiplication in as the third law, is a -aua,

Morphisms between two -auas are linear mappings that conserve the third law and map the unit of the first algebra onto the unit of the second one.

All that will be said of the polynomials over will be based on the universal property given below. The polynomials will thus be defined uniquely up to an isomorphism.

The polynomials over is any -aua together with a distinguished element of such that the following universal property holds:

For any -aua and any , there exists exactly one morphism such that .

Let and be two sets of polynomials over . The universal property implies the existence of a morphism such that , and of a morphism such that .

We then have . Hence is a morphism that maps to . Now is another such morphism. By virtue of the universal property of , it follows that , that is, is the -aua category identity on .

The same reasoning shows that is the -aua category identity on .

Hence is an isomorphism that maps the formal variable of to that of .

Let be an associative unital algebra over , and an element of . Then represents the polynomials over if and only if the family , with and , is a family-base of the -vector space .

Let us first suppose that the latter condition holds on , and prove the universal property.

Let be a -aua and an element of .

Let us suppose that is a morphism such that . Then for all , we have . Since as a -aua morphism, it is in particular a linear mapping. Since a linear mapping is defined by the image of a basis, and is a basis, necessarily is the unique linear mapping such that . Now it is easy to check that this linear mapping does preserve the multiplication in , and is hence a -aua morphism.

Thus the universal property is proven for .

Let us now suppose the universal property for , and show that is a family-base of the -vector space .

Let us first show that is linearly independent.

Let be a free -vector space on set . This entails that is a family-base of . To make into an associative unital algebra, we must define a multiplication law that is bilinear, and also associative.

A bilinear law can be defined by the images of all the ordered pairs of base elements. Hence we may define the multiplication in by:

We then have .

Hence the multiplication is associative with regard to the elements of the base; it is easy to check that this follow through to arbitrary elements, which makes an associative algebra.

is a unit; hence is an associative unital algebra.

Applying the universal property of , we find that there exists exactly one -aua morphism such that .

We then have ; ; and, for any , . Hence for all , .

Let us consider a linear combination of the family that vanishes; that is, a finite sequence of values of , , such that . Taking the image of this linear combination by , we obtain . Since the form a basis, all the must be zero.

Thus the family is linearly independent.

Lastly, we must show that this family generates the -vector space .

Let be the sub-aua of generated by .

There exists a unique aua morphism such that .

Let then be , but with all of as codomain; in other words, . It is clear that too is an aua morphism, and it is such that .

But there is exactly one aua morphism that maps to ; and is such.

Hence .

The image of is that of , and is a subset of . But the image of is itself. Hence .

Thus the sub-aua of generated by is itself. This sub-aua is the set of linear combinations of all the powers of ; hence the sub-vector space generated by is .

Thus is a family-base of the -vector space .

]]>Let be a commutative field. The polynomials form a commutative unital algebra over containing an element such that the following universal property holds:

For any unital algebra (commutative or otherwise) and any element , there exists exactly one unital algebra morphism such that .

If and are elements of , let be the unique morphism such that . Then the composition of and is defined by .

In terms of polynomial functions, the composition of polynomials is the equivalent to the composition (via “”) of the associated functions.

Let us first define the function that is associated with a polynomial.

The set of functions is made into a commutative and unital associative algebra by the usual operations of summing (, multiplication by a scalar ( and internal multiplication ().

Hence, following the universal property of the algebra of polynomials, there exists a unique morphism such that .

By definition, the function associated with a polynomial is simply .

We wish to show that if and are polynomials and and the respective associated functions, the function associated with is .

Let be a polynomial and . Let us consider the mapping .

It is easy to check that is a morphism.

Furthermore, .

On the other hand, is also a morphism from to , and .

Thus both and are morphisms and they agree on the value . The universal property of polynomials entails that they are equal.

Hence for any , .

That is, .

If and , the evaluation of at is, by definition, the value at of the unique morphism such that .

Given polynomials and and , we wish to evaluate at , and, specifically, show that .

Since , we have .

Both and are morphisms . Both evaluate at to . Hence they are equal:

which entails

hence: , as announced.

In particular: if and , then .

If , and are polynomials in , we wish to show that:

Using the definition of composition, this translates into:

That is:

We need to show that for any and in , the two mappings and are identical.

Both are endomorphisms of .

We have:

- by definition of .
- by definition of .

Hence both morphisms agree on the image of . By virtue of the universal property of polynomials, they are equal.

Hence for any , we indeed have , that is, .

todo

todo

]]>Let be a set.

Traditional definition of a filter on : A collection of subsets of is a filter on iff ; and ; and .

Alternative definition: A collection of subsets of is a filter on iff ; and .

The two definitions are equivalent.

Let and be sets, a filter on and .

Traditional definition: The filter on image of by is .

Alternative definition: The filter on image of by is .

Again, the two definitions are equivalent.

]]>For any and real , let be the open ball centered on and with radius . We know that an open ball is an open subset.

The lemma says:

Let be a metric space, an open subspace of and a compact subspace of . Then there exists such that for all in , .

**Proof**:

For each , since , there exists an such that .

The collection of all with is an open cover of . Hence there exists a finite such that covers .

Since is finite, there exists a real such that for all , .

Let be an eventual point of .

Since covers , there exists such that , that is, .

Let be an eventual point of .

We have and , which leads to , that is, . But is in and was taken such that . Hence .

Hence .

We thus have such that .

]]>In this post I discuss the concept of open tubes, particular subsets of the Cartesian product with a set and a metric space.

Let be a set and a metric space, with distance .

For any and any , we note the open ball of of radius and centred on , that is set of all such that . Open balls are notoriously open subsets.

Let be any mapping , and a real .

Then the open tube of radius and centred on is the following subset of :

Equivalently:

The “open” in “open tube” refers to its being based on open balls (second expression above), but does not imply that it is itself necessarily open; indeed, we have not given any topology to , so we don’t have any particular one on and cannot speak of being open or not.

However, if is a topological space and is continuous , then is open, as I will show.

If is a topological space and is a continuous mapping , then any open tube centred on is an open subset of .

Proof:

I will show that in the above circumstances is a neighbourhood of each of its points.

Let be an eventual point of .

Then .

Lettuce define . We have .

Let . Since is continuous and is an open subset of , is an open subset of . Since , .

Let , open subset of . Clearly, , so is a neighbourhood of .

Let be an eventual point of .

, hence , that is .

, hence .

Since and , this leads to:

With:

Thus we have: .

Which implies: .

Hence .

Thus we have found a neighbourhood of that is a subset of , which implies that itself is a neighbourhood of .

Thus is an open subset of .

]]>This may well be “clear” intuitively, but is not so easy to prove. Here I will prove that the third (that the author does not name, but that I call the open set topology) is finer than the second, which is the uniform convergence topology.

I will only need to specify that is a topological space and a metric space with distance .

I define the set of continuous mappings from to .

I note the open ball in centred on and with radius . It is the set of all such that .

Furthermore, for any subset of , I note the set of all continuous mappings the graph of which is a subset of ; in other words, .

The uniform convergence topology on is such that a subset of is open if and only if for any , there exists such that for any , if , then . We can express this condition in terms of subsets of that I have named *open tubes*: for any mapping (not necessarily continuous) and for any , the open tube is . Then is open for the uniform convergence topology if and only if for any , there exists such that .

The third topology ‑ the “open set” topology ‑ is the topology on generated by all the for all open subsets of . (The book says the topology is this collection itself, but I have not managed to prove that it directly forms a topology ‑ as is more explicitly requested in exercise 215.)

To show that the open set topology is finer than the uniform convergence one, we consider an open set for the uniform convergence topology and endeavour to show that it is open in the open set one.

Let be an open set in the uniform convergence topology.

Let be an eventual member of .

There exists an such that .

I prove here that, being continuous, is an open subset of . It includes the graph of . Hence contains , and is by definition an open set for the open set topology on . Since it is a subset of , is a neighbourhood of for the open set topology.

Hence for any , is a neighbourhood of for the open set topology. This implies that is itself an open set in the open set topology.

]]>Let and be connected subsets of topological space such that intersects . Prove that is connected. Find an example to show that it is not enough to assume that intersects .

Let us assume that and are open subsets of such that is empty and that . To show that is connected, we need to show that either or does not intersect .

Since is connected, and is empty and , necessarily is entirely in or in . Similarly, is entirely in or entirely in .

Let us suppose that they are not both in or both in ; for instance, that and .

If , then with open in . Since does not intersect , cannot be in . Hence cannot intersect . This contradicts our assumption.

Hence it is impossible that and . Similarly, it is impossible that and . Thus, and are either both in or both in , which implies that either or does not intersect .

If we only assume that and intersect, it does not always follow that is connected. For instance, let be the topological plane and and two open disks that “almost touch”, that is such that the corresponding closed disks ‑ which are their closures ‑ intersect in one point. The union of and is then not connected.

]]>Prove the following generalization of theorem 40. Let ( in ) be a collection of connected subsets of topological space X, and let the equivalence relation on this collection of sets generated by if have just one equivalence class. Then is connected.

For clarity, rather than a family I will consider a set of connected subsets of . The condition is that the equivalence relation in generated by the above relation has only one class.

Let .

Let and be eventual open subsets of such that and . To show that is connected, we must show that either or is empty.

First, let be an eventual element of . Since , the two subsets of , and are disjoint, and their union is . Furthermore, they are open in . Since is connected, this implies that one or the other is empty. If is empty, then ; if instead is empty, then . Hence in all cases an element of is entirely in or entirely in .

Let us consider the relation on defined by ; in other words, if and only if both are subsets of , or both are subsets of . Clearly, this is an equivalence relation.

Furthermore, if , that is, , we cannot have and , for and are disjoint. Hence and are either both in or both in ; that is, . Hence is an equivalence relation greater than the equivalence relation generated by .

If had two or more classes, there would be some element of in one class and another in another; these two elements would not be equivalent following . Since is greater than the equivalence relation generated by , they would not be equivalent either following this latter relation; which contradicts the fact that this has only one class. Hence too can have only one class, which means that for all and in , either both are in or both are in . But this implies that all are in or all are in , which in turn implies that their union is a subset of either or , hence that either or is empty.

]]>