This section covers measure and category on \mathbb{R}.
For any sequence \{a_n\} of real numbers and for any interval I there exists a point p in I such that p \not = a_n for every n
Let I_1 be a closed subinterval of I such that a_1 \not \in I_1. Let I_2 be a closed subinterval of I_1 such that a_2 \not \in I_2. We can continue this pattern inductively, where I_n is a closed interval of I_{n-1} such that a_n \not \in I_n. The nested sequence of closed intervals I_n has a non-empty intersection. If p \in \bigcap I_n, then p \in I and p \not = a_n for every n.
A set A is nowhere dense if it is dense in no interval on \mathbb{R}. This is equivalent to saying that every interval on \mathbb{R} has a subinterval entirely contained in the compliment of A. An equivalent definition is that A is nowhere dense if and only if Int(Clos(A)) = \varnothing
Nowhere dense sets are closed under certain options, namely:
Any subset of nowhere dense sets is nowhere dense.
The union of finitely many nowhere dense sets is nowhere dense.
The closure of a nowhere dense set is nowhere dense.
- Any subset of nowhere dense sets is nowhere dense.
Let A be a nowhere dense set. If B \subseteq A is dense in some interval on \mathbb{R}, than A must be too, which contradicts our assumption A is nowhere dense. Therefore B is also nowhere dense.
- The union of finitely many nowhere dense sets is nowhere dense.
If A_1 and A_2 are nowhere dense, than for each interval I there exist subintervals I_1 such that I_1 \cap A_1 = \varnothing and a subinterval I_2 \subseteq I_1 such that I_2 \cap A_2 = \varnothing. Hence I_2 \cap (A_1 \cap A_2) = \varnothing. Thus A_1 \cup A_2 is nowhere dense.
- The closure of a nowhere dense set is nowhere dense.
For this proof we use the second definition of a nowhere dense set, A is nowhere dense if and only iff Int(Clos(A)) = \varnothing. Note that Clos(A) = Clos(Clos(A)).
A set A is meagre if it is a countable union of nowhere dense sets. A set is comeagre if its complement is meagre.
\mathbb{Q} is a meagre set, because it is composed of a countable union of singletons, i.e.
\mathbb{Q} = \bigcup_{q \in \mathbb{Q}} q
All comeagre sets are dense.
Let A be a meagre set and A = \bigcup A_n be the representation of A as a countable union of nowhere dense sets. Let I be any interval on \mathbb{R}. Let I_1 = I - A_1, I_2 = I_1 - A_2, I_n = I_{n-1} - A_n. The interval \bigcup I_n is a non-empty subset of I - A. So any interval on \mathbb{R} contains points in A^c, therefore A^c is dense in \mathbb{R}.
No interval in \mathbb{R} is meagre
Let A be a meagre set. Because every interval contains points in A^c, there is no interval fully contained in A.
Any subset of a meagre set is meagre. The union of any countable family of meagre sets is meagre.
The properties above are obvious for meagre sets. However, the closure of a set A is meagre if and only if A is nowhere dense. Therefore the closure of a meagre set may not be meagre.
A class (informal collection) of sets that contains countable unions and arbitrary subsets of its members is called a \pmb\sigma-ideal.
The class of meagre sets is a \sigma-ideal, along with the class of countable sets. The class of nullsets (sets of measure zero) also forms a \sigma-ideal
Let the length of an interval I be given by |I|. A set A \subseteq R is called a nullset if for all \varepsilon > 0, there exist a sequence of intervals I_n such that A \subset \bigcup I_n and \Sigma |I_n| < \varepsilon.
If a finite or infinite sequence of intervals I_n covers an interval I, \Sigma |I_n| \geq |I|
The proof is too long for me to want to write down. The theorem implies that no interval is a nullset.
Every countable set is both meagre and of measure zero. However, some uncountable sets also have these two properties.
The Cantor set C is both meagre and a nullset.
Meagre
The Cantor set is nowhere dense, because for any C_n, the largest interval is of length \frac{1}{3^n}. For any interval I \subset R, there exist a sufficiently large n such that a hole can be found in I.
Nullset
The sum of the lengths of all closed intervals for any step C_n is exactly \frac{2}{3^n}. There exist a sufficiently large n such that for any \varepsilon > 0, the sum of the interval length is below \varepsilon.