Download 5. Lecture. Compact Spaces.

5. Lecture. Compact Spaces.
In the last lecture we have seen that the Cantor set is
perfect, totally disconnected, homogeneous, not homeomorphic to [0, 1] and homeomorphic to the product
with itself. Any infinite discrete space has the same
properties. Nevertheless, the Cantor set is not homeomorphic to any discrete space. The distinguishing
feature is compactness. The Cantor set is compact
and an infinite discrete space is not. In this lecture
we introduce the notion of compactness and discuss
some of its properties. Compact spaces are the spaces
of choice. If one has a non-compact space one often
tries to compactify it. Here we describe two such compactifications - the one-point compactifiction and the
compactifications of trees.
Basics.
We say that a collection of open subspaces of a topological space X is an (open) covering of X if its
union equals X. A subcovering of a covering is a
subset of the covering.
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5.1. Definition. A topological space X is compact
if
(1) X is Hausdorff, and if
(2) every open cover of X has a finite subcovering.
A topological space that has only property (2) is also
called quasi-compact. A subset A ⊂ X is relative
compact if its closure Ā is compact.
Remark. In the literature a space with (1) and (2)
is called compact. Some other authors say a space is
compact if it has only property (2). In order to avoid
ambiguity we also use the term ”compact Hausdorff”
when we want to remind ourselves that the space is
quasicompact and Hausdorff.
5.2. Example. Many topological spaces are not compact. For instance, the Euclidean line R is not compact. Indeed, the open covering given by {(n, n +
2) | n ∈ Z} has no finite sub-covering. An infinite
tree such as the Cayley graph of a free group is not
compact.
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Here is the non-trivial standard example of a compact
space.
5.3. Theorem. Let a, b ∈ R be real numbers with
a < b. Then the closed interval [a, b] is compact.
Proof. Assume the converse. Then there is an open
cover O of [a, b] which has no finite subcover. We
say a closed interval J in [a, b] is bad if no finite subcollection of O covers J. In this terminology J0 =
[a0 , b0 ] := [a, b] is a bad interval. If Jn := [an , bn ] ⊂
[a0 , b0 ] is a bad sub-interval, let xn = 12 (an + bn )
be its mid-point. and observe that either [an , xn ] or
[xn , bn ] must be a bad sub-interval. This is immediate
from the fact that the union of two compact spaces is
compact again.
Thus, starting with J0 and selecting bad sub-intervals
through splitting along mid-points, one constructs a
nested strictly decreasing sequence Jn = [an , bn ] of
bad sub-intervals. The end-points an and bn define
Cauchy sequence. Therefore they converge an → x
and bn → y, say. Moreover, x = y since an −bn → 0.
Let U ∈ O be a covering element that contains x.
Then there is an open interval (c, d) with x ∈ (c, d) ⊂
U . In particular, there is an index m with Jm =
[am , bm ] ⊂ (c, d) since x lies in all [an , bn ]. It follows
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that Jm is not bad. But this contradicts our choice
of Jm . This contradiction proves the theorem. ♦
Remark. It follows that the open interval (0, 1) is
relative compact but not (0, ∞).
The compactness of the closed interval is the basis for
compactness properties of many other spaces. For instance the next result implies the compactness of the
unit cube [0, 1]n .
5.4. Theorem. The product of two compact spaces
is compact.
Proof. Let X, Y be two compact spaces. We are
asked to show that X × Y is compact in the product
topology. First we are going to show that X × Y is
Hausdorff and after that we show that every covering
of X × Y has a finite subcovering.
Let p1 = (x1 , y1 ) and p2 = (x2 , y2 ) be two different
points of X × Y . Then x1 6= x2 or y1 6= y2 .
Thus w.l.o.g. there are two disjoint neighborhoods
U (x1 ), U (x2 ) for x1 , x2 in X since X is Hausdorff.
Hence U (x1 ) × Y and U (x2 ) × Y are two disjoint
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neighborhoods of p1 , p2 in X × Y . This is what we
needed in order to show that X × Y is Hausdorff.
What about the second property? So let us be given an
open covering C of X ×Y . Without loss of generality
we may suppose that all the covering elements of C
are actually boxes U × V since, by definition, the
boxes form a basis of the product topology (i.e., all
open sets are unions of boxes).
Now, we need a procedure for selecting a finite subcovering. Let us proceed as follows. First observe that
for every x in X the subspace x × Y is compact
in X × Y (why?). Thus already finitely many of the
boxes from C cover x × Y .
Let us denote those finitely many boxes by U1 (x) ×
V1 , . . . , Un (x) × Vn . The intersection
U (x) := U1 (x) ∩ . . . ∩ Un (x)
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is an open set (a topology is closed under taking finite
intersections). But notice that the strip U (x) × Y is
still covered by the previous covering elements. Thus
it remains to show that X × Y is covered by only
finitely many of all the strips U (x) × Y .
The collection {U (x) | x ∈ X} is a covering of the
compact space X and so has a finite subcovering. In
other words there are finitely many points x1 , . . . , xm
such that X = U (x1 ) ∪ . . . ∪ U (xm ). Thus X ×
Y is covered by the finitely many strips U (x1 ) ×
Y, . . . , U (xm ) × Y . This is what we needed. ♦
Properties of Compact Spaces.
A closed subset need not be compact as e.g. the subspace [0, ∞) ⊂ R shows. However, we have:
5.5. Theorem. Every closed subset of a compact
space is compact. Every compact subspace of a Hausdorff space is closed.
Proof. Let A be a closed subspace of a compact
space X. LetS O be a collection of open subsets of
X with A ⊂ O. Then O ∪ {X − A} is an open
cover of X. Then compactness of X implies that
also O has a finite sub-cover that covers A.
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Let Y ⊂ X be a compact subspace of a Hausdorff
space X. To show that Y is closed, it suffices to
prove that X−Y is open. So let x ∈ X−Y . For every
point y ∈ Y , there are open neighborhoods U (y) of
y and U (x, y) of x with U (y) ∩ U (x, y) = ∅. The
collection {U (y) | y ∈ Y covers Y . Hence it has a
finite subcover. In other words there are finitely many
points y1 , . . . , yn ∈ Y with Y ⊂ U (y1 ) ∪ . . . ∪ U (yn ).
But then the finite intersection V (x) = U (x, y1 )∩. . .∩
U (x, yn ) is an open neighborhood of x that does not
meet U (y1 ) ∪ . . . ∪ U (yn ) and so it does not meet Y .
Since x was arbitrary, we have shown that X − Y
is open. ♦
5.6. Theorem. Every compact space is regular.
Proof. Same as the previous proof. ♦
5.7. Theorem. The Cantor set is compact. Every
compact discrete space is finite.
Proof. The Cantor set is a closed subset of the unit
interval. Hence it is compact. Every point in a discrete
space is open. Thus the set of all points of a discrete
space is an open covering of the space. It can only have
a finite subcovering when the space itself is finite. ♦
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Expanding on the idea of the proof of the previous
theorem somewhat, we show that all compact spaces
are normal. Recall that a space is normal if disjoint
closed subsets can be separated.
5.8. Theorem. Every compact space is normal.
Proof. In order to prove the lemma let us be given
two disjoint and closed subsets A, B ⊂ X.
We already know that X is regular. Thus, for every x ∈ B, there are disjoint, open neighborhoods
U (x), Vx (A) of x and A, respectively. The collection {U (x)}x∈B is an open covering of B. But
B is compact as a closed subset of a compact space.
Hence there are finitely many points x1 , . . . , xn ∈ B
such that B ⊂ U (x1 ) ∪ . . . ∪ U (xn ). In other words,
the union U (B) := U (x1 ) ∪ . . . ∪ U (xn ) is an open
neighborhood of B. Moreover, the finite intersection
U (A) := Vx1 (A)∩. . .∩Vxn (A) is an open neighbrhood
of A. Finally, U (A) ∩ U (B) = ∅, by construction.
Thus disjoint closed sets can be separated by open
neighborhoods. ♦
Recall from the previous lecture that normal spaces
have a sandwich property. The next theorem gives
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a stronger version for components of compact Hausdorff spaces. At this point recall that a component of
a topological space is always closed but not necessarily
open.
5.9. Theorem. Let X be a compact Hausdorff
space. Then for every component C of X and every
open neighborhood U (C), there is an open subset D
with
C ⊂ D = D̄ ⊂ U (C).
Proof. Exercise. ♦
Remark. Observe that D must be a union of components.
Compact spaces have other nice properties. One of the
most basic one is
5.10. Theorem. Every infinite sequence in a compact metric space has a convergent subsequence.
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Proof. Assume for a moment the converse, i.e., assume there is an infinite sequence xn with no convergent subsequence. We get a contradiction in two
steps.
Step 1. In this step we form the set A := {xn | n ∈ N }
of all sequence elements. W.l.o.g. this set A is infinite
(otherwise we find a constant subsequence which is
convergent) and has no limit point x (otherwise we
find a convergent subsequence through choices xm ∈
1
A ∩ U (x, m
)). The first property will be important in
a moment. The second property means that for every
point x not in A there is a neighborhood which does
not meet A. In other words X − A is open and so
A is closed and so it is compact.
Step 2. No element a ∈ A is a limit point either. Thus
for every such A there is a neighborhood U (a) which
intersects A only in the point a. In other words
A has a covering by sets each containing exactly one
point of A.
Now, the covering above has a finite subcovering since
A is compact. It follows that A is finite. Contradiction. ♦
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For us the above property is one of the characterizing
properties for compact spaces. In fact, it is hard to
think of any space that does have this property and
is not compact. Amazingly, such examples exist but
they are pretty weird (see e.g. [Munkres]).
Another characterizing property is the finite intersection property. We say a collection C of subsets of a
space X satisfies the finite intersection property
if
\
Ci 6= ∅,
1≤i≤n
for every finite subcollection {C1 , . . . , Cn } ⊂ C.
Example. A nested sequence C1 ⊃ C2 ⊃ C3 ⊃ . . .
of non-empty sets is a typical example of a collection
with the finite
T intersection property. Note that the
intersection i Ci may be empty. Here is an example.
Let X be the open unit-interval X = (0, 1) (with
the subspace topology). This is a non-compact space.
The sets Ci :=
T (0, 1/i) are closed in (0, 1) and their
intersection i (0, 1/i) is empty.
5.11. Theorem. Let X be a topological space. Then
X is compact if and only if
\
C 6= ∅,
C∈C
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for all collections C of closed subsets of X with the
finite intersection property.
Proof. X is compact
⇔
no collection O of open sets can cover X, if no finite
subcollection of O covers X.
⇔
S
if C is a collection of closed sets, then, if 1≤i≤n (X −
Ci ) 6= S
X, for all finite subcollections C1 , . . . , Cn ∈ C,
then C∈C (X − C) 6= X.
⇔
T
if C is a collection of closed sets, then, if 1≤i≤n Ci 6=
∅, for all finite subcollections C1 , . . . , Cn ∈ C, then
T
C∈C C 6= ∅. ♦
Compact Spaces and Continuous Maps.
Compact spaces are well behaved with respect to continuous maps.
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5.12. Theorem. Let f : X → Y be a continuous map between topological spaces. Suppose X is
compact and suppose f is surjective. Then Y is
compact.
Proof. Let O be an open cover of Y . Then
f −1 (O) := { f −1 (U ) | U ∈ O }
is an open cover of X since f is continuous. Since
X is compact, there is a finite subcover.
particular,
S In
there are U1 , . . . , Un ∈ O with X ⊂S i f −1 Ui . Since
f is surjective, it follows that Y = i Ui . Hence O
has a finite sub-cover. ♦
5.13. Theorem. Let X be a topological space and
let f : X → R be a continuous map. Suppose X is
compact. Then there are real numbers a, b ∈ R with
a, b ∈ f (X) and
a ≤ f (x) ≤ b, for all x ∈ X.
Proof. The image f (X) is compact since X is
compact. It follows that f (X) ⊂ R must be bounded
from above and from below. Moreover, f (X) is closed
as a compact subset of the Hausdorff space R. Hence
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it contains its least upper bound, say b, as well as its
greatest lower bound, say a. In particular, f (X) ⊂
[a, b]. ♦
5.14. Theorem. Any continuous bijection f : X →
Y between compact spaces X, Y is a homeomorphism.
Proof. We have to show that the inverse map f −1
is continuous. Since f is bijective, this is the same
as saying that f is an open map in the sense that
it maps open subsets to open subsets. Of course, a
continuous bijection is an open map if and only if it is
a closed map in the sense that it maps closed sets to
closed sets.
To continue the proof, let A ⊂ X be any closed subset
of X. Then A is compact as a closed subset of a
compact space. Hence the image f (A) is compact and
so closed as a compact subset of a Hausdorff space. ♦
Remark. The proof allows the somewhat stronger
statement that a continuous bijection f : X → Y is
a homeomorphism if X is quasi-compact and Y is
Hausdorff.
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We next establish an important property of compact
spaces, namely the Urysohn lemma. Actually, the
property is valid for normal spaces which are more
general than compact spaces.
5.15. Theorem. Let X be a normal space (e.g., a
compact space) and let A, B ⊂ X be two disjoint and
closed subspaces of X. Then there is a continuous
map
h : X → [0, 1] with h(A) = 0 and h(B) = 1.
Proof. We view h as a height function. The idea
is to approximate this height function by a sequence
of step functions. We visualize the process in X × R,
i.e., we approximate the graph of the height function.
The steps of the step function are given by
Up × p ⊂ X × R,
where Up ⊂ X are open subsets in X, indexed by
rational numbers p, with
Ūq ⊂ Up , iff p < q.
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To any such collection of steps we associate a step
function
hn (x) := max { p | x ∈ Up },
where the maximum is taken over all the finitely many
steps constructed so far. Here is an illustration of such
a step function:
h(B)
1
q
q
r
p
h(A)
p
0
Up
Uq
Up
Ur Uq
Given a step function as above, we refine it as follows.
First, we look for two adjacent steps Up × p, Uq × q
and Uq ⊂ Up whose distance p − q is maximal among
the distances of all pairs of adjacent steps.
Since Ūq ⊂ Up there is an open subset U with
Ūq ⊂ U ⊂ Ū ⊂ Up . For this subset U we choose the
index r := 12 (p + q) and set Ur := U . Then the step
Ur × r lies between Uq × q and Up × p.
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This finishes the construction. The step functions converge to the map
h(x) = sup { p | x ∈ Up },
where this time the supremum is taken over an infinite
collection of indices.
Clearly h(A) = 0 and h(B) = 1. It therefore
remains to show that h is a continuous map. So let
x ∈ X be any point and let (c, d) ⊂ R be any open
interval containing h(x). We are asked to find an
open neighborhood U of x with h(U (x)) ⊂ (c, d).
For this choose indices
c < p < h(x) < q < d.
They clearly exist. We claim U := Up − Ūq is one of
the desired neighborhoods.
First, U is open since Up is open and Ūq is closed.
Moreover, x ∈ Up , for otherwise h(x) ≤ p. Similarly,
x 6∈ Ūq , for otherwise h(x) ≥ q. Hence x ∈ U .
Finally, h(U ) ⊂ (c, d). To see this let h(z) ∈ f (U ).
Then z ∈ U = Up − Ūq . Since z ∈ Up , we have
h(x) ≥ p. Since z 6∈ Ūq , implies z 6∈ Uq and so
h(z) ≤ q. Thus h(z) ∈ [p, q] ⊂ (c, d).
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This finishes the proof. ♦
Appendix: Compactifications.
A non-compact space can often be compactified.
Sometimes the compactification is unique but often it
is not. In the last case it is a challenge to find a compactification (if there is one at all) that really reflects
the topological properties of the space at infinity. We
will give examples to illustrate the problem. First, the
definition.
5.16. Definition. Let Y be a compact Hausdorff
space and let X ⊂ Y be a subspace. We say Y is a
compactification of X if Y = X̄.
As examples we consider the one-point compactification and two compactifications for infinite trees.
One-Point Compactification.
Every locally-compact space can be compactified
through adding a single point. Here a space X is
called locally-compact if every point x ∈ X has an
open neighborhood which is contained in a compact
subset. Equivalently, every point x ∈ X has an open
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neighborhood U whose closure Ū is compact. Of
course, every compact space is locally compact. We
need local-compactness in the following theorem to
make sure that the compactification is Hausdorff.
5.17. Theorem. Let X be a locally-compact Hausdorff space. Then X has a one-point compactification in the sense that there is a topological space
X + with X ⊂ X + and X + − X a single point.
Moreover, any two compact spaces Y, Z are homeomorphic if X ⊂ Y, Z and if Y − X, Z − X is a single
point.
Proof. (Idea) Let ∞ be any symbol (not contained
in X). Then define
X + := X ∪ {∞}.
Define a topology on X + by declaring a subset U ⊂
X + to be open if
(1) U ⊂ X and U is open in X, or
(2) X − U is a closed compact subset of X.
It is easy to verify that the set of open sets forms a
topology of X + which turns X + into a compact
space. Furthermore, use local-compactness to show
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that X + is a Hausdorff space. To show uniqueness
verify that the identity id : X → X extends to a
homeomorphism Y → Z. ♦
Remark. We did not spend much time proving the
above theorem because the one-point compactification
is not really that useful. It is too insensitive. It destroys all the delicate features that a non-compact
space often has at infinity.
Compactifications of Trees.
Let T be a tree which is infinite but locally finite,
i.e., every vertex is end-point of only finitely many
edges. Then T is locally compact and so has a onepoint compactification. But we will see that for applications this particular compactification is not very
useful. Better compactifications can be obtained by
looking at embeddings into the unit disk. There are
two essentially different compactifications of T that
one can obtain in this way - the connected compactification and the totally disconnected compactification.
Here is the first construction.
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The embedding of T is described through a recursive
process. Start with 6 equally spaced points on the
boundary ∂D2 of the unit disk. Let T1 be the
cone over three of those points with the other points
in between its edges. Next, divide the intervals in the
boundary ∂D2 by their midpoints and replace the
tree T1 by the tree T2 on the right, Continue the
process. The closure T̄ of this subspace in R2 is the
union of T with the boundary ∂D of the unit disk
D2 . Thus the boundary
∂T = T̄ − T = ∂D2
is connected. The boundary ∂T as well as the closure T̄ is a compact Hausdorff space. Thus T̄ is a
compactification. To distinguish this compactification
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from the next we say that T is compactificed by a
connected boundary.
Here is the second construction.
The embedding of the tree T is again described
through a recursive process. This time start by dividing the boundary circle into six equal intervals, three
black intervals and three white intervals. Let T1 be
the cone over the mid-points of the black intervals.
Next, take out the middle third of every black interval
and replace T1 by the tree T2 on the right with endpoints in the middle of the black intervals. Continue
the process ad infinitum. In the end the tree can be
viewed as a subspace of the unit disk. The closure T̄
of this subspace is the union of T with the Cantor
set. Thus the boundary
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∂T = T̄ − T = C
is totally disconnected. The boundary ∂T as well as
the closure T̄ is a compact Hausdorff space. Thus T̄
is a compactification. We say that T is compactificed by a totally disconnected boundary.
Remark. We have carried out the construction for
a homogeneous infinite tree of valence three. An easy
modification yields compactifications for other trees as
well.
Here are two examples that indicate how both constructions come up in concrete situations.
5.18. Example. Recall the group SL2 Z of all 2 × 2matrices with entries in Z and determinant 1. We
know that it acts on the simplicial complex ∆(Q),
where Q = {(x, y) ∈ Z2 | gcd(x, y) = ±1 }. Next we
take the quotient under the antipodal map (a, b) 7→
(−a, −b) and we get the quotient complex on the right:
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2
1
3
_13
_10 7
3 _ 1
2
_1
4 _
3
_8
_
-_3 3
1
7
5 -_
-8 -_ 3
2
-_2 4
1
-_8
-_5 5
-_7 3
1
1_2
1
4
1
1 -_
-2 -_ 5
4
_6 5
5 _ _9
4 7
_4
3
_11 _7 _10
8 5 7
_3
2
13
11
_ 8_ _
7 5 8
5_
3
_
_
_ -1 7
-3
7 -2 _-5 -3 -4
_
_ -3
_
_
_ -2
_
-1
8 5 7 _
7 5 8
3
3
2
_-4 5 _3 -5
4
9_
1_2 7_ 5
7 4
2_
-
9_
_5
4 _
-1
1
_-4
7_
3
_-7
3
2
-_3
5
5
5_
5
_1 =-1
_
0 0
_1
4
3 _
1
1_
3_
4
_4 3
11 _ _5
8 1
3
_2
5
_5 _3 _4
12 7 9
_1
2
3_
5
5_ 4_ 7_
9 7 12
2_
7_
8_ 5_ 11
13 8
7_
8_
5_ 11
7
0
1
5
6
_5
9
3
_1
9
2_
5
_7
1_
6
_4
1_
1
_1
0_
_3
11 _2
7 _3
10
1
94
4
3
The complex on the right is obtained from the identification (±a, ±b) ↔ ab of the lattice points from
Q/{±1} with the rational numbers. The rational
numbers Q ⊂ R are then mapped under the stereographic projection into S 1 and two rational numbers
a c
b , d are joined by some circular arc iff (a, b), (c, d)
are joined by some 1-simplex in ∆(Q). Now, let
lim(Q) = {x ∈ S 1 | there is a sequence xn ∈ Q with xn → x
be the so called limit set of the action of PSL2 Z :=
SL2 Z/{±id}. Then Q ⊂ S 1 as indicated in the picture on the right. The edges of the dual complex of
∆(Q) form an infinite tree, say T . The closure of
this tree is a compact space. It is the compactification
of T with a connected boundary. The group PSL2 Z
acts on T . Moreover, the action extends continuously
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to an action on the boundary ∂T . The fixpoints on
∂T of a matrix from PSL2 (Z) correspond to the
eigenlines of the matrix. Thus the induced action of
PSL2 Z on the boundary contains important information about matrices.
5.19. Example. Consider the Cayley tree of the free
group F2 on two letters a, b, say. This is an infinite, homogeneous tree of valence four. A modification
of the above construction yields the compactifications
with connected and disconnected boundaries, respectively. The action of the free group F2 on its Cayley
tree extends continuously on both boundaries. However, this is different for the action of the entire automorphism group Aut(F2 ). To see this consider the
compactification with connected boundary:
B
a
b
B
b
A
a
a
A
b
B
x
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For simplicity, we write A = a−1 , B = b−1 for the
inverses. Notice that the point x is approached by two
edge paths. One starting with A and the other with
b. It follows that the automorphism ϕ : F2 → F2
defined by
na →
7 a
b 7→ B
maps those two edge paths to edge paths with different end-points. This means that ϕ does not extend
to a continuous map on the connected boundary since
it maps pairs of points that are close together to pairs
of points that are far apart (in fact, ϕ fixes one of
the two halfs given by the broken line and rotates the
other half). However, one can verify that ϕ extends to
the totally disconnected boundary given by the second
construction. It therefore would appear that adding
the totally disconnected boundary to the tree is a better idea when it comes to studying the automorphisms
of the free group. We will return to this problem later.
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