An area in mathematics concerning a special kind of topological spaces called Baire spaces is an example of a notion that intersects three branches of mathematics, namely topology, analysis and set theory – this intersection in particular is called descriptive set theory. We start off by defining the space, then exploring its properties and ending up looking at its applications throughout analysis and topology. Let be the set of functions and the set of functions for .

**Definition 1.** Let and . Define

to be the basic open subsets of .

Examples of these basic open sets are for instance if , then the elements of are all the infinite sequences starting with . For example, is in . This then leads up to the construction of the open sets on :

**Definition 2. **Define to be *open* if there exist a class of basic open sets of satisfying .

From this definition we can then define the Baire space , where is the open subsets of . The only thing we’re missing now is actually to *prove* that our definition of the Baire space is correct, that the defined open sets really satisfies the properties of a topology. This shows in the following theorem:

**Theorem 3.** The Baire space is a topological space.

*Proof. *We have to check that the three properties of topologies are satisfied. First of all that and . For , just choose to be empty in Definition 2. For notice that . for being the empty sequence.

Next, we show that is closed under arbitrary unions. Let be a class of open sets. Each of these open sets is a union of basic open sets by Definition 2, so would then be a union of unions of basic open sets, making it itself a union of basic open sets, thus making it open.

Lastly, as the hardest step, we show that is closed under finite intersections. We show that . Let and , where and are classes of basic open sets. Define to be the class of basic open sets such that there exist satisfying

- and
- or
- .

The last two clauses 2 and 3 implies that and are *comparable* – that is, either or . Thus would be the longest of the two, due to clause 3. We will have to show that

.

““: Let . Then there is , such that and . and would then be comparable, because they’re restrictions of the same infinite sequence . Let . Then , since it satisfies all above three clauses by construction of and , meaning .

““: Let , meaning there is a satisfying . Then by clause 1 above, there is such that and , satisfying and . Thus and , meaning .

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