Baby Rudin Problems
Last update: 19 March 2014
Prove that the empty set is a subset of every set.
A complex number is said to be algebraic if there are integers
not all zero, such that
Prove that the set of all algebraic numbers is countable. Hint: For every positive integer there are only finitely many equations with
Prove that there exist real numbers which are not algebraic.
Is the set of all irrational real numbers countable?
Construct a bounded set of real numbers with exactly three limit points.
Let be the set of all limit points of a set Prove that
is closed. Prove that and
have the same limit points. (Recall that
Do and always have the same limit points?
be subsets of a metric space.
Show, by an example, that this inclusion can be proper.
Is every point of every open set a limit point of
Answer the same question for closed sets in
Let denote the set of all interior points of a set
[See Definition 2.18(e); is called the interior of
Prove that is always open.
Prove that is open if and only if
If and is open, prove that
Prove that the complement of is the closure of the complement of
Do and always have the same interiors?
Do and always have the same closures?
Let be an infinite set. For and define
Prove that this is a metric. Which subsets of the resulting metric space are open? Which are closed? Which are compact?
For and define
Determine, for each of these, whether it is a metric or not.
Let consist of and the numbers
Prove that is compact directly from the definition (without using the Heine-Borel theorem).
Construct a compact set of real numbers whose limit points form a countable set.
Give an example of an open cover of the segment which has no finite subcover.
Show that Theorem 2.36 and its Corollary become false (in for example) if the word
"compact" is replaced by "closed" or by "bounded."
Regard the set of all rational numbers, as a metric space, with
Let be the set of all such that
Show that is closed and bounded in but that is not compact. Is
Let be the set of all whose decimal
expansion contains only the digits and Is countable?
Is dense in Is
compact? Is perfect?
Is there a nonempty perfect set in which contains no rational number?
If and are disjoint closed sets in some metric space
prove that they are separated.
Prove the same for disjoint open sets.
define to be the set of all for which
similarly, with in place of Prove that
and are separated.
Prove that every connected metric space with at least two points is uncountable. Hint: Use (c).
Are closures and interiors of connected sets always connected? (Look at subsets of
Let and be separated subsets of suppose
[Thus if and only if
Prove that and are separated subsets of
Prove that there exists such that
Prove that every convex subset of is connected.
A metric space is called separable if it contains a countable dense subset. Show that is separable.
Hint: Consider the set of points which have only rational coordinates.
A collection of open subsets of is said to be a
base for if the following is true: For every and every open set
such that we have
In other words, every open set in is the union of a subcollection of
Prove that every separable metric space has a countable base. Hint: Take all neighborhoods with rational radius and center in some countable
dense subset of
Let be a metric space in which every infinite subset has a limit point. Prove that is separable.
Hint: Fix and pick
choose if possible, so that
for Show that this
process must stop after a finite number of steps, and that can therefore be covered by finitely many neighborhoods of radius
and consider the centers of the corresponding neighborhoods.
Prove that every compact metric space has a countable base, and that is therefore separable.
Hint: For every positive integer there are finitely many neighborhoods of radius
whose union covers
Let be a metric space in which every infinite subset has a limit point. Prove that is compact.
Hint: By Exercises 23 and 24, has a countable base. It follows that every open cover of has a
If no finite subcollection of covers then
the complement of
is nonempty for each but is empty. If
is a set which contains a point from each consider a limit point of
and obtain a contradiction.
Define a point in a metric space to be a condensation point of a set
if every neighborhood of contains uncountable many points of
Suppose is uncountable, and let
be the set of all condensation points of Prove that is perfect and that at most countable
many points of are not in In other words, show that
is at most countable. Hint: Let
be a countable base of
let be the union of those for which
is at most countable, and show that
Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable.
(Corollary: Every countable closed set in has isolated points.) Hint: Use Exercise 27.
Prove that every open set in is the union of an at most countable collection of disjoint segments.
Hint: Use Exercise 22.
Imitate the proof of Theorem 2.43 to obtain the following result:
where each is a closed subset of then at least one
has a nonempty interior.
Equivalent statement: If is a dense open subset of
is not empty (in fact, it is dense in
(This is a special case of Baire's theorem; see Exercise 22, Chap. 3, for the general case.)