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.

]]>In the category of sets, the two morphisms ( and ) in a direct product are monomorphisms and the two morphisms in a direct sum are epimorphisms. Is this true in every category?

There are two mistakes in this wording.

- The author has exchanged “monomorphisms” and “epimorphisms”. In a direct product of sets, the morphisms are generally not monomorphisms (injective), and in a direct sum of sets, they are generally not epimorphisms (surjective).
- In a direct product of sets, the morphisms are
*almost*always epimorphisms. If one of the sets is empty, the product is empty, and the morphism associated with the other set (an empty mapping) is not an epimorphism; unless this other set is empty too.

The general theorem I will prove is this:

Let and be objects of a category , and a -direct product of and . If is not empty then is an epimorphism.

We consider the above conditions satisfied. There exists a morphism .

Let be the identity morphism and the morphism . Applying the universal property of the direct product to the triplet , we obtain that there exists a unique such that both and .

The first of these two relations says that is the identity on . If we have an object and two morphisms and from to such that , then, composing this relation on the right with and eliminating the resulting identity morphisms, we get . Hence is an epimorphism.

We obtain, of course, a similar result exchanging the roles of and .

Reversing the arrows, we get the corresponding result for direct sums:

Let and be objects of a category , and a -direct sum of and . If is not empty then is a monomorphism.

Plus, of course, the similar result obtained by exchanging and .

It so happens that in the category of sets, in the case of a direct product, both and are *always* monomorphisms, even if , respectively , are empty.

See also: the equivalent page on direct sums.

In the following, we will always have and topological spaces and a direct product thereof.

In the construction, this is just the trivial fact that if you know , you know both and .

Theorem:

For every , if and , then .

Proof:

Let us suppose that we have such that and .

Let us take a singleton topological space, that is that is a singleton (singletons exist!) and we equip it with the only possible topology, which is both the indiscrete and the discrete one.

We define and as the constant mappings, and respectively, that map the only element of to , which is also , and to , which is also , respectively.

We define and as the constant mappings, both , that map the only element of to and to respectively.

Since these four mappings are constant, or, alternatively, since the topology on is discrete, they are all continuous.

Furthermore, for both and we have:

Since is a direct product of and , there can be only one morphism satisfying these two relations. Hence necessarily , which implies .

If we have and , does there exist a such that and ? Again, this is trivial in the representation: is the answer. Sticking to the universal definition of the direct product, we have:

For any and , there exists such that and .

Proof:

We take as a singleton topological space as above, and and as the mappings and respectively that map the unique element of to and to respectively. These two mappings again are continuous, since has the discrete topology.

The universal definition of as a direct product implies that there exists such that and . Taking as the image by of the unique element of , we thus have and . .

This result and the preceding one come together as:

For any and , there exists a unique such that and .

At least, usually.

If or , then is surjective. If or , then is surjective.

This is just a consequence of the general result proved in exercise 7 (corrected): in any category, if is a direct product of and , and if , then is an epimorphism (and similarly, exchanging the roles of and , and and , respectively); plus the fact that in the category of topological spaces, epimorphisms are surjective mappings.

That the topology on the the direct product space is the one generated by the collection of all and for open in and open in is a stipulation of the explicit construction of as . It can however be proven without reference to this construction.

The topology of is the topology generated in by the collection of all and for open in and open in . A subset of is open if and only if it is the union of some collection of subsets of the form with open in and open in .

Proof:

Let be the direct product topology on and the topology on generated in by the collection of all and for open in and open in .

Since and are defined as continuous, necessarily contains all the and for open in and open in . Hence is finer than . We wish to show that is also finer than .

The mapping , considered from to , remains continuous, since all the for open in are in . Similarly, is continuous . Hence we have a topological space and two continuous mappings and , from to and respectively. Since is a direct product of and , there exists a unique continuous mapping such that and .

This implies that any element has the same image as by both and . The first theorem above implies that . Thus .

Hence is continous ; which means that is finer than .

Hence , that is the topology of the direct product is the topology of generated by the collection of all and for open in and open in .

The open sets of the topology generated by a collection of sets are the unions of any number of finite intersections of elements of .

In this case, . A finite intersection of elements of will be of the form with the ‘s open subsets of and the ‘s open subsets of . Since intersections are well-behaved respective to inverse images, and since finite intersections of open subsets are open subsets, this can be written with open subset of and open subset of . Hence an open subset of is a union of subsets of of the form with open subset of and open subset of .

If is a topological space and a mapping , then is continuous if and only if both and are continuous.

If is continuous, since both and are continuous, by composition of continuous mappings we conclude that and are continuous.

Conversely, suppose that we know that and are continuous. Let be an eventual open subset of . We wish to show that is an open subset of .

Let be an eventual point of . Then . Since is open in and (see above) “a subset of is openif and only if it is the union of some collection of subsets of the form with open in and open in ”, there exist open in and open in such that . Taking of this expression, we obtain . But . Since we know that both and are continuous, this last expression is an open subset of , that contains and is included in . Thus is a neighbourhood of . This being the case for any , is an open subset of .

]]>Prove that both the direct product and the direct sum of two Hausdorff topological spaces is Hausdorff.

We will consider Hausdorff topological spaces and and their direct sum (in the first part) or product (in the second part) .

We will proceed without reference to the explicit constructions of direct topological sums and products, using instead the results in “Theorems about the direct product and sum of two topological spaces”.

Let and be two eventual points of , with . The first result about direct sums in the above page implies that . The second result implies that . Two cases are possible:

- One of and is in and the other in .
- Both and are in , or both are in .

We will consider successively the case where and , and the case where both are in . The other possible cases are similar and will lead to similar results.

First case ( and ):

We have and , with and .

itself is an open neighbourhood of , and hence (last result about direct sums on the above page) is an open neighbourhood of in . Similarly, is an open neighbourhood of in . Since and are disjoint, we have found disjoint neighbourhoods respectively of and .

Second case ():

We have and , with . Since , we have . Since is Hausdorff, there exist and disjoint open neighbourhoods in respectively of and .

//TODO (we need injective)

Let and be two eventual points of , with .

The first result about direct products in the above page implies that or .

Let us suppose that .

Since is Hausdorff and and are distinct points of , there exist open sets and of such that , and .

Then and are disjoint (the image by of a common element would be both in and , which are disjoint). They are open (since and are open and is continuous). Since , we have ; similarly, . Thus there exist two disjoint open neighbourhoods respectively of and in .

The same would result if we had .

This being the case for all with , the topological space is Hausdorff.

]]>See also: the equivalent page on direct products.

In the following, we will always have and topological spaces and a direct sum thereof.

and are disjoint.

Let and be eventual elements respectively of and .

Let and be any two distinct objects (such exist), and , equipped with any topology (the discrete one will do). Then the constant mappings and are continuous.

Hence there exists a continuous mapping such that and .

This implies in particular that and , that is, and . Since , it follows that .

Thus and cannot have a common point, which would both be the image by of a point of and by of a point of .

Let with the topology induced on by that of .

Let be the mapping and the mapping .

Since an open set on is the intersection of with an open set on and since certainly contains no points of , its preimage by is equal to the preimage of by and hence is open in . Thus is continuous. Similarly, is continuous.

We can thus apply the universal property of the direct sum , which implies that there exists a continuous mapping such that and .

Let be the mapping . The preimage by of an open set of is again the same as its preimage by , hence is an open set of . Hence is continuous.

For any , ; hence . Similarly, .

But we also have and , with continuous . The uniqueness requirement of the universal property of the direct sum thus implies that .

But the image of is that of , and is included in . Thus the image of , which is , is included in . It follows that , that is, that .

The topology on the direct sum is the topology generated by the collection of the and for all open in and all open in .

//TODO

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