Download Product spaces

CHAPTER 5
Product spaces
5.1. Finite products
Definition 5.1. Let (X1 , T1 ),...,(Xn , Tn ) be a collection of n ≥ 2 topological spaces.
Then their Cartesian product X1 × ... × Xn = {(x1 , ..., xn ) | xi ∈ Xi , i = 1, ..., n}
becomes a topological space with the product topology TX1 ×...×Xn defined as the topology
generated by the basis
BX1 ×...×Xn = {U1 × ... × Un ⊂ X1 × ... × Xn | Ui ∈ Ti , i = 1, ..., n}
The space (X1 × ... × Xn , TX1 ×...×Xn ) is called the product of (X1 , T1 ), ..., (Xn , Tn ) or
simply a product space. We shall refer to the spaces (Xi , Ti ), i = 1, ..., n as the factors
of (X1 × ... × Xn , TX1 ×...×Xn ). The functions
πi : X1 × ... × Xn → Xi
πi (x1 , ..., xn ) = xi
shall be referred to as the projection maps or simply as the projections.
To see that the set BX1 ×...×Xn from definition 5.1 is in fact the basis of some topology,
we invoke lemma 2.31: Since Xi ∈ Ti for each i = 1, ..., n, we see that X1 × ... × Xn ∈
BX1 ×...×Xn so that the first condition from lemma 2.31 is met. As for the second, pick
arbitrary elements Ui , Vi ∈ Ti for each i = 1, ..., n and let (x1 , ..., xn ) ∈ (U1 × ... × Un ) ∩
(V1 × ... × Vn ) be any point. Since
(U1 × ... × Un ) ∩ (V1 × ... × Vn ) = (U1 ∩ V1 ) × .. × (Un ∩ Vn )
and because Ui ∩ Vi ∈ Ti , we see that (U1 × ... × Un ) ∩ (V1 × ... × Vn ) ∈ BX1 ×...×Xn and
therefore indeed, BX1 ×...×Xn is a basis of a topology.
Most of the time we shall apply definition 5.1 to the case of n = 2. This is no less
general than the case of arbitrary n ≥ 2 since the n-fold product space (X1 × ... ×
Xn , TX1 ×...×Xn ) is homeomorphic to the space gotten by taking n − 1 successive twofold
products (in arbitrary order) of the spaces (X1 , T1 ), ..., (Xn , Tn ) (exercise ??).
Remark 5.2. It may be tempting, but incorrect, to assume that every open set in
(X × Y, TX×Y ) is a product of open sets from X and Y . Rather, according to definition
5.1, every element W ∈ TX×Y is a union of such products:
W = ∪i∈I Ui × Vi
Ui ∈ TX , Vi ∈ TY , i ∈ I
with
for some indexing set I.
71
72
5. PRODUCT SPACES
Example 5.3. Consider the topological spaces (X, TX ) = (R, Tp ) and (Y, TY ) =
(R, Tq ) where Tp and Tq are the included point topologies associated to the points
p, q ∈ R. An open set W ⊂ X × Y is a union of open set U × V with U ∈ Tp and
V ∈ Tq . In particular, if U and V are non-empty, then p ∈ U and q ∈ V so that
(p, q) ∈ W .
The converse of this is false, i.e. a subset W of X × Y that contains (p, q) may not
be open. An example is any set W = {(p, q), (x, y)} with x 6= p and y 6= q. Thus, the
product topology of Tp and Tq on R2 is not equal to the included point topology T(p,q)
on R2 .
The result of the following lemma is certainly expected though not completely
immediate from definition 5.1.
Lemma 5.4. Let X and Y be two topological spaces and let A ⊂ X and B ⊂ Y be
two non-empty subsets.
(a) A × B is an open subset of X × Y with respect to the product topology TX×Y if
and only if A and B are both open.
(b) A × B is a closed subset of X × Y with respect to the product topology TX×Y if
and only if A and B are both closed.
Proof. (a) If A and B are both open then A × B is open simply by definition of
TX×Y . Suppose conversely that A × B is open. Then we can write A × B = ∪i∈I Ui × Vi
with Ui ⊂ X and Vi ⊂ Y each open. But then A = ∪i∈I Ui and B = ∪i∈I Vi showing
that they are both unions of open sets and thus open themselves.
(b) =⇒ Suppose that A ⊂ X and B ⊂ Y are closed subsets and note that
(X × Y ) − (A × B) = X × (Y − B) ∪ (X − A) × Y
Both of the sets on the right hand side of the above are open subsets of X × Y showing
that (X × Y ) − (A × B) is open and thus that A × B is closed.
⇐= Suppose that A × B ⊂ X × Y is closed and let x ∈ X − A be any point. Pick
y ∈ B arbitrarily and observe that then (x, y) lies in (X × Y ) − (A × B) which is an
open set. Thus there must be neighborhoods Ux ⊂ X of x and Vy ⊂ Y of y such that
Ux × Vy ⊂ (X × Y ) − (A × B). We claim that Ux is disjoint from A. If it were not,
we could pick a point z ∈ Ux ∩ A giving us the relation (z, y) ∈ (Ux × Vy ) ∩ (A × B),
a contradiction since (Ux × Vy ) ∩ (A × B) = ∅. Therefore Ux ∩ A = ∅ showing that
X − A = ∪x∈X−A Ux making X − A an open and hence A a closed set. The proof that
B is closed is analogous.
Definition 5.5. Let fi : Xi → Yi , i = 1, ..., n be continuous functions and define
the function
f1 × ... × fn : X1 × ... × Xn → Y1 × ... × Yn
by f1 × ... × fn (x1 , ..., xn ) = (f1 (x1 ), ..., fn (xn )). We shall refer to f1 × ... × fn as the
product of the functions f1 , ..., fn and shall call each fi a factor function.
Theorem 5.6. Let (X, TX ) and (Y, TY ) be two topological spaces.
5.1. FINITE PRODUCTS
73
(a) The projection maps πX : X × Y → X and πY : X × Y → Y are continuous
and surjective functions. Moreover, both maps πX and πY are open maps.
(b) For any two points x0 ∈ X and y0 ∈ Y , the two inclusion maps ιy0 : X → X ×Y
and ιx0 : Y → X × Y given by ιy0 (x) = (x, y0 ) and ιx0 (y) = (x0 , y), are
continuous functions.
(c) If (Z, TZ ) is a topological space and f : Z → X × Y is function with coordinate
functions f = (f1 , f2 ), then f is continuous if and only if each of f1 : Z → X
and f2 : Z → Y is continuous.
(d) A product of finitely many functions is continuous if and only if each of the
factors is continuous.
Proof. (a) Surjectivity of πX and πY are obvious. To prove their continuity, let
U ⊂ X and V ⊂ Y be open sets. Then
−1
πX
(U ) = U × Y
and
πY−1 (V ) = X × V
which are both open sets showing that πX and πY are continuous.
Let W ⊂ X × Y be an open set and consider the set WX = πX (W ). We’d like
to show that WX is open. Towards that goal, pick an arbitrary point x ∈ Wx and
let (x, y) ∈ W for some y ∈ Y . Since W is open and (x, y) ∈ W , there must exist
neighborhoods Ux and Vy of x and y respectively, such that Ux × Vy ⊂ W . But then
clearly Ux = πX (Ux × Yy ) ⊂ πX (W ) = WX and so WX = ∪x∈WX Ux showing that WX is
open. In a similar vein one shows that WY = πY (W ) is also open demonstrating that
πX and πY are open maps.
(b) To show that ιy0 : X → X × Y is continuous, let W ⊂ X × Y be any open set
and define Wy0 as Wy0 = ι−1
y0 (W ). We’d like to show that Wy0 is an open set. Given
any x ∈ Wy0 , we know that (x, y0 ) lies in W . Then there must be neighborhoods Ux
and Vy0 of x and y0 respectively such that Ux × Vy0 ⊂ W . But then Ux ⊂ Wy0 showing
that Wy0 = ∪x∈Wy0 Ux which is an open set. Consequently ιy0 is continuous, a very
similar argument show that ιx0 is also continuous.
(c) =⇒ Suppose first that f = (f1 , f2 ) : Z → X × Y is continuous. To see that
f1 : Z → X is also continuous, let U ⊂ X be an open set and consider f1−1 (U ). Note
that f1−1 (U ) = f −1 (U × Y ) and, since f is continuous and U × Y open in X × Y , we see
that f1−1 (U ) must be open. Therefore f1 is continuous and by symmetry of argument,
f2 is also continuous.
⇐= Assume that f1 : Z → X and f2 : Z → Y are both continuous and let
W ⊂ X × Y be an arbitrary open subset. Find open subsets Ui ⊂ X and Vi ⊂ Y ,
i ∈ I, such that W = ∪i∈I Ui × Vi . Then
f −1 (W ) = f −1 (∪i∈I Ui × Vi ) = ∪i∈I f −1 (Ui × Vi ) = ∪i∈I (f1−1 (Ui ) × f2−1 (Vi ))
showing that f −1 (W ) is a union of open sets and hence open. Thus, f must be continuous as claimed.
(d) The coordinate functions of the product function f = f1 ×f2 : X1 ×X2 → Y1 ×Y2 ,
are the two functions f1 ◦ πX1 and f2 ◦ πX2 . If f1 and f2 are continuous, then parts (a)
and (c) of the theorem imply that f is also continuous. Conversely, if f is continuous,
74
5. PRODUCT SPACES
then so are f1 ◦πX1 and f2 ◦πX2 (by part (c) of the theorem). But then, for any open set
U ⊂ Y1 , the preimage f1−1 (U ) equals πX1 ((f1 ◦ πX1 )−1 (U )) which is open since f1 ◦ πX1
is continuous and πX1 is open (by part (a)). This implies continuity of f1 and by a
similar argument one obtains continuity of f2 as well.
The next example demonstrates that projection maps πX , πY don’t need to be
closed maps.
Example 5.7. Consider the topological spaces (X, TX ) = (Y, TY ) = (R, TEu ) and
let Γf ⊂ R × R be the graph of the function f : R → R defined as f (x) = sin(x)(1 −
2
e−x ). We leave it as an exercise to show that Γf is a closed subset of R × R. On the
other hand, πY (Γf ) = h−1, 1i which is not a closed subset of (R, TEu ). Ergo, πY in this
example is not a closed map.
Inclusion maps on product spaces need neither be open maps nor closed maps:
Example 5.8. Let (X, TX ) and (Y, TY ) be two topological spaces and consider the
inclusion map ιy0 : X → X × Y (remember that ιy0 (x) = (x, y0 )). For any subset
A ⊂ X we obtain ιy0 (A) = A × {y0 }. If A is an open set then, according to lemma 5.4,
A × {y0 } is an open set if and only if {y0 } is open. On the other hand, if A is closed
then, again using lemma 5.4, the set A × {y0 } is closed if and only if {y0 } is closed.
Examples of spaces (Y, TY ) in which {y0 } fails to be both open and closed, abound.
An easy example is to take (Y, TY ) = (R, Tindis ), the indiscrete topology on R. For this
example, the inclusion map ιy0 is neither open nor closed for any choice of y0 ∈ R.
Corollary 5.9. Let X and Y be two topological spaces. Then, for any choice of
y0 ∈ Y , the inclusion map ιy0 : X → X × Y is a homeomorphism onto its image. This
allows us to view X as a subspace of X × Y . If additionally Y is T1 , then X is a closed
subspace of X × Y (for any choice of y0 ∈ Y ). Identical conclusions hold for Y .
Proof. Clearly ιy0 is injective and it is continuous by theorem 5.6, part (b). Its
inverse function ι−1
y0 = πX |X×{y0 } is also continuous since the restriction and composition
of continuous functions is continuous (theorem 3.12).
If Y is T1 then {y0 } is a closed subspace of Y (lemma 4.3) and so X × {y0 } is a
closed subspace of X × Y (lemma 5.4).
We next explore how properties of topological spaces, such as second countability
and the Hausdorff property, transfer from factors to the product space and vice versa.
Theorem 5.10. Let X, Y be two topological spaces and let X × Y be equipped with
the product topology. Let A and B be chosen from the sets of properties of topological
spaces


Separabiltiy,






Second countability,
Normality ( T1 & T4 ),
A∈
B∈
First countability, 
Complete normality ( T1 & T5 ).





T0 , T1 , T2 , , T3 .
Then
5.1. FINITE PRODUCTS
75
(a) X × Y possesses property A if and only if both X and Y possess property A.
(b) If X × Y possesses property B, then so do X and Y (the converse is generally
not true, see example 5.12 below).
Proof. Concerning the properties A, one direction of the claim is immediate.
Namely, according to corollary 5.9, we can view X and Y as subspaces of X × Y . But
according to theorems 2.45 and 4.4, each of the A properties is inherited by subspaces.
Therefore, if X × Y has property A then so do X and Y .
The same conclusion holds for the two B properties if one observes that in a T1
space, any singleton is a closed set (lemma 4.3). With this, according to corollary 5.9,
X and Y become closed subsets of X × Y and so the conclusion of the theorem for the
B properties then follows from theorem 4.4, part (b).
For the converse of the A properties, we examine each value of A separately.
• A is “Separability”
Assume that both X and Y are separable and let A ⊂ X and B ⊂ Y be countable
dense subsets. Then A × B is a countable subset of X × Y and A × B = Ā × B̄ (see
exercise ??). Thus X × Y is separable.
• A is “Second Countability”
Assume that X and Y are second countable and let BX = {Ui | i = 1, 2, 3, ...} and
BY = {Vj | j = 1, 2, 3, ...} be two countable bases for X and Y respectively. Define
BX×Y as BX×Y = {Ui × Vj | i, j = 1, 2, 3, ...}. This is a countable set of open subsets
of X × Y . To see that it is a basis, let W ⊂ X × Y be any open set. By definition
5.1 of the product topology, we can find open sets Ai ⊂ X and Bi ⊂ Y , i ∈ I, such
that W = ∪i∈I Ai × Bi . For each i ∈ I, let Ji and Ki be indexing sets such that
Ai = ∪j∈Ji Uj and Bi = ∪k∈Ki Vk , then
W = ∪i∈I Ai × Bi = ∪i∈I [(∪j∈Ji Aj ) × (∪k∈Ki Bk )] = ∪i∈I,j∈Ji ,k∈Ki Aj × Bk
Thus W is a union of elements from BX×Y and so BX×Y is a basis.
• A is “First Countability”
This proof mirrors the one just given for second countability and the details are left as
an exercise (see exercise ??).
• A is the “ T0 property”, the “ T1 property”or the “ T2 property”.
We address the T0 − T2 properties together since their proofs are very similar. Let
(x1 , y1 ) and (x2 , y2 ) be two distinct points in X × Y .
T0 : If x1 6= x2 , find either a neighborhood U1 ⊂ X of x1 that doesn’t contain x2 or
else a neighborhood U2 ⊂ X of x2 that doesn’t contain x1 . Then either U1 × Y
is a neighborhood of (x1 , y1 ) not containing (x2 , y2 ) or U2 × Y is a neighborhood
of (x2 , y2 ) not containing (x1 , y1 ). If x1 = x2 then y1 6= y2 and one proceeds in
a similar fashion.
T1 : If x1 6= x2 , let U1 U2 ⊂ X be neighborhoods of x1 and x2 respectively with
x2 ∈
/ U1 and x1 ∈
/ U2 . Then U1 × Y and U2 × Y are neighborhoods of (x1 , y1 )
and (x2 , y2 ) respectively with (x2 , y2 ) ∈
/ U1 × Y and (x1 , y1 ) ∈
/ U2 × Y . The case
of y1 6= y2 goes similarly.
76
5. PRODUCT SPACES
T2 : If x1 6= x2 , let U1 U2 ⊂ X be disjoint neighborhoods of x1 and x2 . Then U1 × Y
and U2 × Y are disjoint neighborhoods of (x1 , y1 ) and (x2 , y2 ) respectively. The
case of y1 6= y2 goes accordingly.
• A is “T3 property”
Assume that X and Y are T3 . Let C ⊂ X × Y be a closed subset and pick a point
(x, y) ∈ X × Y − C. Consider X × Y − C as a neighborhood of (x, y) and find
neighborhoods U of x and V of y such that U × V ⊂ X × Y − C. By regularity of X
we can separate x and the closed set X − U by neighborhoods Wx and WX−U . Note
that
Wx ⊂ X − WX−U =⇒ W x ⊂ X − WX−U ⊂ U
(since X − WX−U is a closed set). In the same vein we find neighborhoods Wy and
WY −V of y and Y − V with W y ⊂ V . Then Wx × Wy is a neighborhood of (x, y) with
W x × W y ⊂ U × V ⊂ X × Y − C so that X × Y − W x × W y becomes a neighborhood
of C. Clearly Wx × Wy and X × Y − W x × W y are disjoint.
Corollary 5.11. The product X × Y is a regular space ( T0 and T3 ) if and only
if each of the factors X, Y is regular.
The following example shows even when X and Y are normal spaces, X × Y need
not be (so that part (b) of theorem 5.10 cannot be an “if and only if”statement).
Example 5.12. Consider the topological space (X, TX ) = (R, Tll ) where Tll is the
lower limit topology from example 2.17 generated by the subbasis Sll = {[a, bi | a, b ∈
R, a < b}. It was already shown in example 4.11 that X is T0 −T5 and thus in particular
also T4 and regular. To see that X × X is not T4 with the product topology consider
the subdiagonal ∆0 = {(x, x) | x ∈ X}. Since ha, bi = ∪∞
n=1 [a + 1/n, bi is an open subset
0
of X, it is easy to see that ∆ is a closed subset of X × X. The relative topology on
∆0 is the discrete one because ∆0 ∩ ([a, a + 1i × [−a, −a + 1i) = {(a, −a)}. Therefore,
the two disjoint sets
AQ = {(a, −a) ∈ ∆0 | a ∈ Q}
and
AI = {(a, −a) ∈ ∆0 | a ∈ I = R − Q}
are both closed in ∆0 and thus also in X × X.
If AQ and AI were separated by disjoint open sets UQ , UI ⊂ X × X, then for every
point (a, −a) ∈ AQ there would have to a basis element [b, ci × [d, ei with (a, −a) ∈
[b, ci × [d, ei ⊂ UQ , and similarly for points (a, −a) ∈ AI . But the only basis elements
[b, ci × [d, ei that intersect one but not both of AQ and AI are those for which a = −c.
Thus we arrive at the following requirement:
For all a ∈ R there exist ba , ca > a such that [a, ba i × [−a, −ca i ⊂
UQ ; if a ∈ Q
UI ; if a ∈ I
and such that (∪a∈Q [a, ba i × [−a, −ca i) ∩ (∪a∈I [a, ba i × [−a, −ca i) = ∅.
This is clearly impossible and so AQ and AI cannot be separated by disjoint neighborhoods. Consequently, X × X is not T4 and so neither T5 .
5.2. INFINITE PRODUCTS
77
Since X is regular, according to corollary 5.11, X × X is also regular. This shows
that there are regular spaces that aren’t normal.
5.2. Infinite products
In this section we briefly turn to the investigation of infinite product spaces. While
in the case of finite products, the product topology is a natural choice, in the case of
infinite products there are two such natural choices. Depending on context, one may
prove more advantageous than the other (see for example Tychonoff’s compactness
theorem in section 9.4).
Let (X, TX ) and (Y, TY ) be two topological space. The product topology TX×Y on
X × Y , generated by the basis {U × V | U ∈ TX , V ∈ TY }, can be characterized in two
ways:
1. TX×Y is the smallest topology which contains all of the products of open sets
from X and Y .
2. TX×Y is the smallest topology on X × Y for which the two projection maps
πX : X × Y → X
and
πY : X × Y → Y
are continuous. Indeed, if we assume their continuity, then for any U ∈ TX and
any V ∈ TY , the sets π −1 (U ) = U × Y and πY−1 (V ) = X × V would have to be
open in X × Y . But then their intersection (U × Y ) ∩ (X × V ) = U × V is also
forced to be open.
A similar characterization holds for the product topology of finitely many factors. It
is of course easy to see that these two descriptions lead to the same topology. This is
no longer the case when infinitely many factors are involved.
Let I be an indexing set, possibly infinite, and for each i ∈ I let (Xi , Ti ) be a
topological space. Let X = ×i∈I Xi denote their Cartesian product, i.e. let X be
X = {x : I → ∪i∈I Xi | x(i) ∈ Xi }
As customary, we shall denote x(i) simply by xi . For example, if Xi = Y for all i ∈ I,
then X is simply the set of functions from I to Y .
Definition 5.13. Let (Xi , Ti ), i ∈ I be a collection of topological space and set
X = ×i∈I Xi .
(a) The box topology Tbox on X is the topology generated by the basis
Bbox = {×i∈I Ui | Ui ∈ Ti }
(b) The product topology Tprod on X is the topology generated by the basis
Bprod = {×i∈I Ui | Ui ∈ Ti and Ui = Xi for all but finitely many i ∈ I}
The verification that the sets Bbox and Bprod from definition 5.13 are indeed basis
of a topology, is easy and is left as an exercise (exercise ??). It should be noted that
Tbox is strictly finer than Tprod when I is infinite. When I is finite, the two topologies
agree.
78
5. PRODUCT SPACES
Our main use of the box topology will be in section 9.4 where it shall play a
prominent role in the proof the Tychonoff’s compactness theorem. For now, we contend
ourselves with studying a few examples.
Example 5.14. Choose I = N and let (Xi , Ti ) = (R, TEu ). To tell copies of R apart
we shall label them as Ri , i ∈ I. Denote the product ×i∈N R by RN and note that RN
is simply the set of sequences in R. We shall consider the subset Y ⊂ X of bounded
sequences in R. On Y , consider the metric d : Y × Y → [0, ∞i given by
∞
X
|xi − yi |
d(x, y) =
2i
i=1
The associated metric topology is the relative topology on Y induced by Tbox . On the
other hand, the relative topology on Y induced by Tprod is not a metric topology since
it is not Hausdorff. For example, there are no disjoint neighborhoods of xi = 1/i and
yi = −1/i in Tprod .
Example 5.15. Pick I = R and let (Xλ , Tλ ) equal the Euclidean line for each
λ ∈ I. To distinguish the copy of R corresponding to λ ∈ I, we shall denote it by Rλ .
We’ll denote the Cartesian product ×λ∈R Rλ by RR . As in the previous example, we
can interpret RR as a set of functions, namely functions from R to R. We will examine
when a sequence of functions fi ∈ RR converges to a function f ∈ RR with respect to
the two topologies Tprod and Tbox .
• Tprod Consider first RR with the product topology and suppose that limi→∞ fi = f .
Let λ ∈ R be arbitrary and let U ⊂ R be a neighborhood of f (λ). Consider the
open set in the product topology
U = (×µ<λ Rµ ) × U × (×µ>λ Rµ )
Then U is a neighborhood of f and so there must exist an integer iλ such that
fi ∈ U for all i ≥ iλ . This last statement is equivalent to saying that fi (λ) ∈ U
for all i ≥ iλ which is simply saying that the sequence fi (λ) converges to f (λ)
in the Euclidean topology on R.
Conversely, suppose that for each λ ∈ R, the equation limi→∞ fi (λ) = f (λ)
holds with respect to the Euclidean topology on R. Pick any neighborhood of
V of f and suppose that
V = ×µ∈R−{λ1 ,...,λn } Rµ × (Vλ1 × ... × Vλn )
where λ1 , ..., λn ∈ R are chosen arbitrarily and where Vλ1 , ..., Vλn are arbitrary
neighborhoods of f (λ1 ), ..., f (λn ) respectively. Then there exists an integer i0
such that i ≥ i0 implies that fi (λj ) ∈ Vλj for all i ≥ i0 and for any choice
j ∈ {1, ..., n}. This of course is equivalent to saying that fi ∈ V showing that
limi→∞ fi = f with respect to the product topology. In conclusion, we have
shown that convergence of function in RR with respect to the product topology,
is equivalent to pointwise convergence with respect to the Euclidean topology
limi→∞ fi = f in Tprod
⇔
limi→∞ fi (λ) = f (λ) in TEu for every λ ∈ R.
5.3. EXERCISES
79
• Tbox Now let us consider RR with the box topology and suppose again that limi→∞ fi =
f . For each λ ∈ R, let Uλ be a neighborhood of f (λ) with respect to the Euclidean topology on Rλ and let U be the neighborhood of f (with respect to the
box topology) defined as
U = ×λ∈R Uλ
By assumption, there exists an integer i0 such that i ≥ i0 implies that fi ∈ U.
The latter inclusion is equivalent to fi (λ) ∈ Uλ for every λ ∈ R which in turn is
simply saying that fi converges uniformly to f (with respect to the Euclidean
topology) on all of R.
Conversely, if fi converges uniformly to f , and if V = ×λ∈R Vλ is any neighborhood of f , then Vλ is a neighborhood of f (λ) for every λ ∈ R, and so there
is again some i0 with i ≥ i0 giving fi (λ) ∈ Vλ for all λ ∈ R. Thus fi converges
to f with respect to the box topology on RR . In summary
limi→∞ fi = f in Tbox
⇔
limi→∞ fi = f uniformly in TEu .
For instance, define fi : R → R as fi (t) = et−i . Then, for every fixed but arbitrary t ∈
R, the relation limi→∞ fi (t) = 0 holds in (R, TEu ) so that limi→∞ fi = 0 in (RR , Tprod ).
However, the convergence limi→∞ fi (t) = 0 in (R, TEu ) is not uniform (since for every
fixed i, the quantity |fi (t)| can be arbitrarily large as we vary t) so that limi→∞ fi 6= 0
in (RR , Tbox ).
5.3. Exercises
5.16. Show that d : X × X → [0, ∞i is continuous.