# General Topology/Continuity

**Definition (continuity)**:

Let be topological spaces and let be a function. is called **continuous** if and only if for every open , the set is open.

**Proposition (characterisation of continuity via subbasis)**:

Let be a function between topological spaces , and let be a subbasis of the topology of . If is open in for all , then is continuous.

**Proof:** Let be an arbitrary open set; we are to show that is open. Now a basis of the topology on is given by

- .

Since intersections commute with preimages, is open for . But all open sets of are unions of elements of , and preimages commute with unions, so that is continuous.

**Example (identity is a continuous function)**:

Let be a set with a topology , and let denote the identity function on . Then is continuous.

**Proposition (category of topological spaces)**:

The class of all topological spaces, together with the continuous functions between them as morphisms, forms a category.

**Proof:** Indeed, the composition of continuous functions is again continuous, and further, the identity (which is unique, by composing any other identity with the above identity) is well-defined.

**Proposition (characterisation of continuity of functions to a space with initial topology)**:

Let be a set equipped with the initial topology of certain functions , where is a family of topological spaces. Let be another topological space. A function is continuous if and only if all the functions () are continuous.

**Proof:** First suppose that all the functions are continuous, and let be open, so that we may write

- ,

since we saw that the sets thus defined form a basis of the initial topology. Then is open in since taking preimages commutes with taking unions and intersections. Further, observe that all functions are continuous, so that the function are continuous when is.

**Proposition (product of topological spaces is categorical product)**:

Let be topological spaces, and let be their product with product topology. Then , together with the canonical projections , is a product of in the category of topological spaces and continuous functions.

**Proof:** Let another cone over the diagram be given that contains as points the and no morphism, and suppose that the object of that cone is called and the arrows of that cone are called , so that is a topological space and the are continuous functions. To ease notation, set , and define

- .

Note that is the unique map that has the property that , since the projection is simply given by taking the -th component. Furthermore, note that is continuous, since the are continuous, has the initial topology of the , and we conclude by the characterisation of continuity of functions to a space with initial topology. Thus, is a product.

Continuity is a local property, in that it may be characterized by a property that a function might have at every point.

**Definition (continuity at a point)**:

Let be topological spaces, a point, and a function. is called **continuous at ** iff for every open neighbourhood of , there exists an open neighbourhood of such that .

**Proposition (continuity is equivalent to continuity at each point)**:

Let be topological spaces and be a function. is continuous if and only if it is continuous at all .

**Proof:** Suppose first that is continuous, and let . Let be an open neighbourhood of , then by continuity is an open neighbourhood of and by definition of the preimage . Suppose now that is continuous at each point , and let be any open set. We are to show that is open. Indeed, let be arbitrary, so that . By continuity at , we find that the sets of neighbourhoods of such that is nonempty. (Note: We are inserting here an additional step of choosing a canonical neighbourhood for each in order to avoid the axiom of choice.) Define

- ,

which is an open neighbourhood of and has the property that , that is, . Then we obtain

- , ie. ,

which is open as the union of open sets.

**Proposition (characterisation of continuity by closed sets)**:

Let be a function between topological spaces.

is continuous if and only if is closed in for all closed subsets .

**Proof:** is continuous if and only if for all open subsets , the set is open in . The open subsets of are in bijective correspondence to the closed subsets of via the map , and the same holds for . Now note that , so that the latter is closed precisely when the former is. In particular, the latter is always closed whenever is always open, and if the latter is always closed, then is always open.

**Proposition (continuity on closed union)**:

Let be topological spaces, let be closed subsets of such that and let be a function such that and are both continuous. Then is continuous.

**Proof:** We show that is closed for all closed subsets , thereby proving continuity by the characterisation of continuous functions by the preimages of closed sets. Indeed, let be closed. Note that

- ,

that is, is the union of two closed sets and thus itself closed.

**Proposition (restriction of continuous function is continuous)**:

Let be topological spaces, a subset and a continuous function. Then is a continuous function from (with the subspace topology) to .

**Proof:** Let be open. Then is open in , so that is open in the subspace topology on .

**Definition (homeomorphism)**:

Let be topological spaces, and let be a function which is invertible, ie. bijective. is called a **homeomorphism** if and only if both and are continuous.

**Proposition (restriction of homeomorphism is homeomorphism)**:

Let be topological spaces and a homeomorphism. Let further be a subset. Then is a homeomorphism, where and bear the subspace topologies induced by resp. .

**Proof:** is continuous since the restriction of a continuous function is continuous. Further, , which is likewise continuous as the restriction of a continuous map. Therefore, is a homeomorphism.

**Definition (equicontinuity)**:

Let be topological spaces. A subset (where denotes the set of all functions from to ) is called equicontinuous at a point iff for each and each open neighbourhood of , there exist open neighbourhoods of and of so that for all whenever , then . is called **equicontinuous** iff it is equicontinuous at each point.

**Proposition (function in equicontinuous set is continuous)**:

Let be topological spaces, let be equicontinuous at and let . Then is continuous at .

**Proof:** Set . By definition of equicontinuity, whenever is a neighbourhood of , we find neighbourhoods of and of so that whenever for arbitrary, then . But since , we have , so that and is continuous at .

When is a uniform space, the definition of equicontinuity simplifies, and furthermore in this situation equicontinuous subsets are related to compact subsets of . This we will see in the chapter on uniform structures.

**Definition (local homeomorphism)**:

Let be topological spaces. A **local homeomorphism** is a function such that for all , we find an open neighbourhood of so that is a homeomorphism and is an open subset of .

In other words, a function is a local homeomorphism if and only if for all , there exists an open neighbourhood of and an open neighbourhood of so that is a homeomorphism from to .

**Proposition (local homeomorphisms are open)**:

Let be a local homeomorphism. Then is an open map.

**Proof:** Indeed, let be open, and let be arbitrary. Pick arbitrary. Then since is a local homeomorphism, we find open with such that is open. Further, . Since was arbitrary, is open.

**Definition (embedding)**:

An **embedding** is a continuous function between topological spaces so that is a homeomorphism between and .

**Proposition (continuity criterion for the greatest lower bound topology)**:

Let be a set, let be a family of topologies on , and suppose that is the greatest lower bound topology of on . Let be a function, where is a topological space. Then is continuous if and only if is continuous with respect to all topologies () on .

**Proof:** is continuous iff for all open , the set is open in . This in turn is equivalent to being open in all topologies () for arbitrary open . And this condition translates that is continuous with respect to all topologies () on .

**Proposition (continuity criterion for the least upper bound topology)**:

Let be a topological space, a set and a family of topologies on . Let be a function. Then is continuous with respect to the least upper bound topology of the if and only if is continuous for all topologies .

**Proof:** is continuous iff for all open , the set is open. If that is the case, then all the sets are open, where ( arbitrary), so that is continuous with respect to all topologies . Suppose now that is continuous with respect to all these topologies. Note that since the least upper bound topology on with respect to the is the topology generated by , the latter forms a subbasis of the least upper bound topology. Hence, we may apply the characterisation of continuity via subbasis.