- Let d =
x,y ∈ ℝ. Show that this is a metric on ℝ.
- Now consider ℝn. Let
Show that this is a metric on ℝn. In the case of n = 2, describe the ball B.
Hint: First show that
- Let C denote the space of functions which are continuous on
Verify the following.
. Then use to show that
≡ is a metric and that with this metric,
is a metric
- Recall that is compact. This was done in single variable advanced
calculus. That is, every sequence has a convergent subsequence. (We will go
over it in here as well.) Also recall that a sequence of numbers
Cauchy sequence means that for every
ε > 0 there exists N such that if
m,n > N, then
< ε. First show that every Cauchy sequence is
bounded. Next, using the compactness of closed intervals, show that every
Cauchy sequence has a convergent subsequence. It is shown later that if
this is true, the original Cauchy sequence converges. Thus ℝ with the
usual metric just described is complete because every Cauchy sequence
- Using the result of the above problem, show that is a complete metric
space. That is, every Cauchy sequence converges. Here
- Suppose you had is a metric space. Now consider the product
with d = max
. Would this be a metric space? If so,
prove that this is the case.
Does triangle inequality hold? Hint: For each i,
Now take max of the two ends.
- In the above example, if each is complete, explain why
- Show that C is a complete metric space. That is, show that if
is a Cauchy sequence, then there exists
f ∈ C such that
n→∞d = lim
n→∞ = 0.
Hint: First, you know that is a
Cauchy sequence for each
t. Why? Now let f be the name of the thing to which
fn converges. Recall why the uniform convergence implies
t → f is
continuous. Give the proof. It was done in single variable advanced calculus. Review
and write down proof. Also show that
- Let X be a nonempty set of points. Say it has infinitely many points. Define
d = 1 if
x≠y and d = 0 if
x = y. Show that this is a metric. Show that in
every point is open and closed. In fact, show that every set is open and every
set is closed. Is this a complete metric space? Explain why. Describe the open
- Show that the union of any set of open sets is an open set. Show the intersection of
any set of closed sets is closed. Let A be a nonempty subset of a metric space
. Then the closure of
A, written as A is defined to be the intersection
of all closed sets which contain A. Show that A = A ∪ A′. That is, to
find the closure, you just take the set and include all limit points of the
- Let A′ denote the set of limit points of A, a nonempty subset of a metric space
. Show that A′ is closed.
- A theorem was proved which gave three equivalent descriptions of compactness of a
metric space. One of them said the following: A metric space is compact if and only
if it is complete and totally bounded. Suppose is a complete metric space
K ⊆ X. Then is also clearly a metric space having the same metric as
X. Show that is compact if and only if it is
closed and totally bounded.
Note the similarity with the Heine Borel theorem on ℝ. Show that on ℝ, every
bounded set is also totally bounded. Thus the earlier Heine Borel theorem for ℝ is
- Suppose is a compact metric space. Then the Cartesian product is also a
metric space. That is
is a metric space where
is compact. Recall the Heine Borel theorem for
is compact in ℝn with the distance given by d = max
suffices to show that is sequentially compact. Let
m=1∞ be a
m=1∞ is a sequence in Xi. Therefore, it has a subsequence
k1=1∞ which converges to a point x1 ∈ X1. Now consider
the second components. It has a subsequence denoted as k2 such that
k2=1∞ converges to a point x2 in X2. Explain why limk2→∞x1k2 = x1.
Continue doing this n times. Explain why limkn→∞xlkn = xl ∈ Xl for
each l. Then explain why this is the same as saying limkn→∞xkn = x in
- If you have a metric space and a compact subset of
K, suppose that
L is a closed subset of K. Explain why L must also be compact. Hint: Go right to
the definition. Take an open covering of L and consider this along with the open set
LC to obtain an open covering of K. Now use compactness of K. Use this to
explain why every closed and bounded set in ℝn is compact. Here the distance is
given by d
- Show that compactness is a topological property in the following sense.
are both metric spaces and f : X → Y has the property that f is one to one, onto,
and continuous, and also f−1 is one to one onto and continuous, then the two
metric spaces are compact or not compact together. That is one is compact if and
only if the other is.
- Consider ℝ the real numbers. Define a distance in the following way.
Show this is a good enough distance and that the open sets which come from this
distance are the same as the open sets which come from the usual distance
. Explain why this yields that the identity mapping
continuous with continuous inverse as a map from to
. To do
this, you show that an open ball taken with respect to one of these is
also open with respect to the other. However,
is not a complete
metric space while
is. Thus, unlike compactness. Completeness is
not a topological property.
Hint: To show the lack of completeness of
, consider xn = n. Show it is a Cauchy sequence with respect to
- A very useful idea in metric space is the following distance function. Let be a
metric space and
S ⊆ X,S≠∅. Then dist
. Show that
this always satisfies
This is a really neat result.
- If K is a compact subset of and
K, show that there always exists x ∈ K
such that d = dist
. Give an example in
ℝ to show that this is might not
be so if K is not compact.
- You know that if f : X → X for X a complete metric space, then if
< rd it follows that
f has a unique fixed point theorem. Let
f : ℝ → ℝ be given by
, but f has no fixed point.
- If is a metric space, show that there is a bounded metric
ρ such that the
open sets for are the same as those for
- Let be a metric space where
d is a bounded metric. Let C denote the
collection of closed subsets of X. For A,B ∈C, define
where for a set S,
Show x → dist is continuous and that therefore,
Sδ is a closed set containing
S. Also show that ρ is a metric on C. This is called the Hausdorff metric.
- ↑Suppose is a compact metric space. Show
is a complete
Hint: Show first that if Wn ↓ W where Wn is closed, then
→ 0. Now let be a Cauchy sequence in
C. Then if ε > 0 there
exists N such that when m,n ≥ N, then ρ
< ε. Therefore, for each
n ≥ N,
Let A ≡∩n=1∞∪k=n∞Ak. By the first part, there exists N1 > N such that for
n ≥ N1,
Therefore, for such n, Aε ⊇ Wn ⊇ An and
ε ⊇ A because
- ↑ Let X be a compact metric space. Show is compact.
Hint: Let Dn be a
2−n net for X. Let Kn denote finite unions of sets of the form B
p ∈Dn. Show Kn is a 2−