General Topology/Nets

Definition (net):

Let be a topological space, and let be a directed set. A net is a family .

Definition (net convergence):

Let be a topological space, let be a net (where then of course is a directed set) and let . The sequence is said to converge to if and only if for every neighbourhood there exists so that for all we have . For this we write .

As are filters, nets are analogues of sequences, which are used to adapt theorems which otherwise would only hold for "nice" spaces to the setting of general topological spaces. The downside is, as with filters, that theorems involving nets often use the axiom of choice. Whether you use nets or filters is a matter of taste, and a matter of selecting that tool which uses the least amount of choice in your given situation.

Proposition (closedness is equivalent to containing all net limits):

Let be a topological space. Then a set is closed if and only if for all nets so that which converge to a point , we have .

(On the condition of the axiom of choice.)

Proof: Suppose first that is closed, and that is a convergent net in to a limit . Suppose . Then for all , we find so that at least , which implies that , and since , . Since a set is closed if and only if it contains its boundary, we conclude that , a contradiction. Suppose then that contains the limit of all its convergent nets. Then let , and note that is a directed set, ordered by inclusion. For each , choose by definition of , and observe that forms a net convergent to . Thus, , and since was arbitrary, and is closed.

Proposition (characterisation of continuity by net convergence):

Let be topological spaces and a function. Then is continuous if and only if for all nets in that converge to a certain point , the net converges to .

(On the condition of the axiom of choice.)

Proof: Suppose first that is continuous, and suppose that converges to . Then let be any open neighbourhood of , and set , which is open by the continuity of . Then pick so that for all we have ; for such we will then have by definition of the preimage.

Suppose now that has the property that net convergence to implies net convergence of to . Then suppose that is closed. If we prove that is closed, we've proven the theorem. It suffices to prove that contains all limits of its nets. So suppose that is a net in that converges to some . Then and thus , so that is indeed closed.

Definition (subnet):

Let be a topological space and be a net in it. A subnet of is a net together with a final order homomorphism so that .

Proposition (compactness is equivalent to the existence of convergent subnets):

Let be a topological space, and let be a subset. is compact if and only if for all nets in , there exists and a subnet of so that .

(On the condition of the axiom of choice.)

Proof: Suppose first that is compact, and let be a net in . Suppose that does not have a convergent subnet. Then for each , we find an open neighbourhood of and a such that for all , we have ; for otherwise, if is a point so that for all open neighbourhoods and there exists so that , then we note that the set of all (that is, the sets of all open neighbourhoods of ) forms a directed set when ordered inversely by inclusion (which shall henceforth be denoted by ). Note that we may take the product order on , and then we may define an order morphism by sending . Consider the subset so that . We define a subnet of as thus: Define by , and the order morphism induced by , which is final since for any , we find so that , when is arbitrary. Now converges to , a contradiction. But then we apply compactness to the , so that we gain points so that covers . Then we choose an upper bound of by successively choosing an upper bound of , then an upper bound of that upper bound and and so on. Then for , is not contained in any of the , so that in particular , a contradiction. Conversely, if every net contains a convergent subnet, let be an open cover of . Suppose that it doesn't admit any finite subcover, and consider the set of all finite subfamilies of ordered by inclusion, which is a directed set. Then define a net indexed over that set by choosing, for each finite index set , the element . By assumption, upon defining to be the set of all finite subfamilies of , we find that has a convergent subnet . Let be the point to which this subnet converges, and since is a cover, pick so that . Note that is a finite subfamily of , so that by finality of the order homomorphism we find so that contains . On the other hand, since converges to , we find so that for we have . Finally, since is directed, let be an upper bound for and , then contains , but then is not contained in , a contradiction.

Definition (sequence):

Let be a topological space. A sequence is a net indexed over the directed set of the natural numbers.

In order to denote that a sequence converges to a point , we shall resort to the useful notation .

Definition (sequentially closed):

Let be a topological space. A subset is called sequentially closed iff for all sequences in that converge in , the limits are all contained within .

Proposition (characterisation of closedness by sequences in first-countable spaces):

Let be a first-countable topological space. Then a subset is closed iff it is sequentially closed.

(On the condition of the countable choice.)

Proof: Suppose first that is closed. Then certainly every limit of a convergent sequence in is contained within , because a sequence is a net, and the corresponding statement for nets is true without the axiom of choice. Suppose now that is sequentially closed. Then let ; we are to show that . Choose a countable basis for the neighbourhood filter . By the axiom of countable choice, pick for each . Then the sequence converges to , and hence .

Definition (sequential continuity):

A function between topological spaces is called sequentially continuous iff for all sequences in that converge to some , we have .

Proposition (sequence criterion for continuity in first countable spaces):

Let be a first-countable space, a topological space and a function. Then is continuous if and only if it is sequentially continuous.

(On the condition of the countable choice.)

Proof: If is continuous, certainly whenever , since a neighbourhood of gives rise to a neighbourhood of , into which will be for sufficiently large, and then by def. of the preimage.

If is sequentially continuous, let be closed, and set ; we are to show that is closed. Let be a sequence in . By the characterisation of closedness by sequences in first-countable spaces, it will suffice to prove that is sequentially closed. Hence, let be a sequence in that converges to some . Then , so that and hence .