Reference: Introductory Real Analysis, Kolmogorov and Fomin. Dover Publications. Translated and edited by Richard A. Silverman:
Problem 1:
Let X be an uncountable set, and let be the ring consisting of all finite subsets of X and their complements. Is
a
-ring?
Solution 1:
Recall the definition: A ring of sets is called -ring if it contains the union
whenever it contains the sets
.
There are countably many finite subsets of X (given uncountable set) and their complements (also especially) so that the union of all these is the whole set X. Hence, X is a -ring as it is closed under countably many set unions. QED.
Problem 2:
Are open intervals Borel sets?
Solution 2:
Recall the following:
(1) Borel sets or B-sets are the subsets of the real line belonging to the minimal B-algebra generated by the set of all closed intervals .
Problem 3:But a B-algebra is also a ring and is closed under differences/complements. Open intervals are complements of closed intervals. And, they belong hence to B-algebra. Note we still have to answer the question whether open sets are Borel sets and so we have to check if they are contained in a minimal B-algebra. As per the following theorem: Given any non-empty system of sets , there is a unique irreducible (minimal) B-algebra
generated by the system
or the Borel closure of
. So, yes indeed open sets are Borel sets. QED.
Problem 3A:
Let be a function defined on a set M and taking values in a set N. Let
be a system of subsets of M, and let
denote the system of all images
of sets
. Moreover, let
be a system of subsets of N, and let
denote the system of all preimages
of sets
. Prove that:
i) If is a ring, so is
.
ii) If is an algebra, so is
iii) If is a B-algebra, so is
iv) where
stands for a ring.
v) where
is irreducible or minimum Borel algebra generated by a non-empty system of sets
.
Problem 3B:
Prove the following:
i) If is a ring, so is
.
ii) If is an algebra, so is
.
iii) If is Borel-algebra, so is
.
iv)
v) where
is irreducible minimum Borel algebra generated by a non-empty system of sets
.
Solution 3A:
Part 1: Given mapping , that is,
, Let
be subsets of N and let
be subsets of M. Then, as it is given that
is a ring then :
The above implies:
, that is,
, that is,
.
Also, so that
, that is,
, that is, we conclude
.
Also, as
is a ring; so that
so that for some
, which is
So, we have shown that the system of sets is closed under set union, set intersection, set difference, and hence, also set symmetric difference. Hence, it is is a ring. QED.
Part ii:
We have already shown that is a ring, now
is also an algebra meaning a ring of sets with a unit element. Let it contain a unit E
such that if
is any set belonging to
, then
. so that
, hence,
, which in turn implies that
is a unit of
. Hence,
is a ring of sets with a unit element, or in other words, an algebra of sets.
Part iii:
To prove that: if is a Borel algebra, so is
. From parts i and ii above,
is a ring with a unit or in other words,
is an algebra. Moreover, if it contains
whenever it contains the sets
all of which are members of
.
But for an arbitrary (finite or infinite) collection of sets:
Now,
hence
Consequently, , which in turn implies that
is a Borel algebra also. QED.
Part iv and part v:
Using parts i and iii above, these can be very easily proved.
Solutions 3b:
i) Prove: If is a ring, so is
. Note that
,
. Proof (i): As
is a system of subsets of M. and it is a ring, so if
, and
, it implies that
;
;
. We know that
and
. And, quite obviously,
so that it is true that if
is a ring, then
is also a ring.
ii) Check if it is true or false: if is an algebra, so is
. Argument: As
is an algebra, it is already a ring. But additionally,
contains an element E such that
(whenever
but
but it need not be true that
is a unit element of
. So, we can say that in case
is an algebra, then
may or may not be an algebra. QED.
iii) Check if it is true or false: If is a Borel algebra, then so also
is also a Borel algebra. Argument: As in previous case,
may not have a unit element. So,
need not be a Borel algebra. QED.
iv) . Check if this is true or false. Argument: Let
,
. By definition of a ring,
,
,
, hence we get the following:
;
and
. This in turn means that
,
and
. Hence,
. Similarly, we can show that
. QED.
v) . Check if this is true or false. (We will discuss this later; meanwhile, if you wish, please try as homework and let me know).
Cheers,
Nalin Pithwa.
Comments are most welcome.