Let be any set. A topology on is a subset of the power set of such that the following three axioms hold:
Definition (topological space):
A topological space is a set together with a topology on it.
Example (Euclidean topology):
Let and consider the set . Then we define
is a topology on , called the Euclidean topology.
Definition (open set):
Let be a topological space and its topology. An open set of is any set contained in .
Definition (closed set):
Let be a topological space. A subset is called closed if and only if is open, ie. contained in the topology of .
Definition (clopen set):
Let be a topological space. A subset is called clopen if and only if it is both open and closed at the same time.
Proposition (definition of topologies from closed sets):
Let be a set, and let be a set of subsets of such that the following axioms hold:
Then there exists a unique topology on such that contains precisely the closed sets of that topology, and conversely the closed sets of any topology satisfy 1. - 3.
Proof: We claim that we have a bijection between sets that satisfy 1.-3. in the proposition (call this set ) and the topologies on , which we shall denote by , so that if this bijection is denoted by , then the closed sets of a topology are given precisely by . Indeed, this map is given by
- , and the inverse is .
These are inverse to each other by a straightforward computation, and the closed sets of are precisely . We must show that is well-defined, that is, maps to and the inverse is also well-defined in that it maps to . Let first be a topology. Since and , 1. of the above is satisfied. Suppose now that , so that for we have for a certain . Since is a topology, , and further, by de Morgan
Finally, suppose that is a family of elements of , and for write again . Since is a topology, , and by de Morgan
and satisfies 1. - 3. The reverse direction is proved by analogous computations. Thus, the unique topology whose closed sets are is , and if is a topology, , its closed sets, satisfy 1.-3.
Proposition (topologies are ordered by inclusion):
If is any set and are topologies on , then defines an order on the topologies of .
Proof: This is a special case of the general fact that subsets of the power set are ordered by inclusion.
Definition (comparison of topologies):
Let be any set and topologies on . Whenever , we say that is coarser than and is finer than .
Proposition (intersection of topologies is a topology):
Let be a family of topologies on a set . Then
is a topology on .
Proof: This follows from intersections preserving closure properties, by noting that finite intersection and arbitrary union are operations on (as well as are the whole and the empty set, namely 0-ary operations), and that each topology is closed under these.
Definition (open neighbourhood):
Let be a topological space, and let . A neighbourhood of is an open set (ie. ) such that .
Let be a topological space, and let A neighbourhood of is a set such that there exists an open set such that and . The set of all neighbourhoods of is denoted .
Theorem (characterisation of topologies by neighbourhood systems):
Suppose that is a set, and that for each we are given a set such that for each . Then the sets form the neighbourhoods of a topology on if and only if the following conditions are satisfied:
- (closedness under supersets)
In this case, the topology of which the are the neighbourhoods is uniquely determined by the neighbourhoods and equals
Proof: Let's first show that whenever the sets are the neighbourhoods of a topology on , then they satisfy 1. - 4. Indeed, is an (open) neighbourhood of every of its points. Then, whenever for some , we find open sets such that and , note that and that is open, as finite intersections of open sets are open. If , open and arbitrary, then so that . Finally, if , choose an open s.t. , then is open, hence a neighbourhood of all of its points. Now suppose that for each we are given and these neighbourhoods satisfy the conditions 1.-4. Then we define
and claim that is a topology. Indeed, , since then the condition is trivially satisfied, since there are no . Furthermore, , since whenever , and is a neighbourhood of . Then, suppose that is a family of sets contained in , and set . We claim that . Indeed, if , pick so that , then such that since , and finally note that . Then, let and pick so that and . Then , so that , since .
Now we claim that for each , the neighbourhoods of with respect to are precisely . Indeed, let first a neighbourhood of be given. Choose open. Then by definition of , there exists such that , and since is closed under supersets, . Conversely, suppose . Define , and claim that and is open in . Indeed, is clear, since is necessary for being a neighbourhood of . Let now . By 4., choose so that and is in for all . Thus for all , so that , and since was arbitrary, is open, and is a neighbourhood in of .
Finally let be any topology which has the sets as neighbourhoods, and claim that as defined above. Indeed, if , then is itself a neighbourhood of each of its points and hence in . Conversely, if , then for each choose a neighourhood of that is contained within (in order to avoid the axiom of choice, choose them to be the union of all such sets) and note that
- Let be a set.
- Prove that , the power set, is a topology on (it's called the discrete topology) and that when is equipped with this topology and is any function where is a topological space, then is automatically continuous.
- Prove that is a topology on (called the trivial topology), and that when is equipped with this topology, is any topological space and is any function, then is continuous.