Math 6131 Fall 2011 Jan Mandel
Homework 1 solutions
1. Prove that when (X, dX ) and (Y, dY ) are metric spaces, then the distance function
dX×Y ((x1 , y1 ) , (x2 , y2 )) = dX (x1 , x2 ) + dY (y1 , y2 )
(1)
induces the product topology on X × Y .
Solution.
Notation. Since there is no possibility of confusion (the kind of distance function is
determined unambiguously by where its arguments are from), we omit the subscripts
X Y and X×Y . Br (w) means the open ball {z : d (z, w) < r}, in the appropriate space
(X, Y , or X × Y ) that w is from.
Let Z ⊂ X × Y be open in the product topology and (u, v) ∈ Z. By definition of the
product topology, there exist open sets U ⊂ X and V ⊂ Y such that (u, v) ∈ U × V ⊂
Z. By definition of open set, there exist ε1 , ε2 such that the open balls Bε1 (u) ⊂ U
and Bε2 (v) ⊂ V , that is
∀x ∈ X : d (x, u) < ε1 =⇒ x ∈ U
∀y ∈ Y : d (y, v) < ε2 =⇒ v ∈ V
Define ε = min {ε1 , ε2 } .Then ε > 0. Let (x, y) ∈ X × Y and
d ((x, y) , (u, v)) = dX (x, u) + dY (y, v) < ε.
Then
d (x, u) < ε ≤ ε1 and d (y, v) < ε ≤ ε2 .
Thus (u, v) ∈ U × V ⊂ Z. We have proved that Bε ((u, v)) ⊂ Z. Thus Z is open in
the metric space (X × Y, dX×Y ).
Conversely, assume that Z is open in (X × Y, dX×Y ). Let (u, v) ∈ Z. We need to show
that there exist open sets U ⊂ X and V ⊂ Y such that (u, v) ∈ U × V ⊂ Z. There is
ε > 0 such that Bε ((u, v)) ⊂ Z. Let d (x, u) < 2ε and d (y, v) < 2ε . Then
d ((x, y) , (u, v)) = dX (x, u) + dY (y, v) < ε
and so (x, y) ∈ Z. So, we have proved that B 2ε (u) × B 2ε (v) ⊂ Z. So we can set
U = B 2ε (u) and V = B 2ε (v), which are open balls and thus open sets.
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2. Prove that if (X, dX ) and (Y, dY ) areSseparable metric spaces, then every open set
Z ⊂ X × Y is a countable union Z = ∞
k=1 Uk × Vk , where Uk are open in (X, dX ) and
Vk are open in (Y, dY ).
Solution. We continue in the same notation. Since the product of separable metric
spaces is separable and subspace of separable metric space is separable, Z is separable,
i.e., there exists a countable dense set {(uk , vk ) : k ∈ N} in Z. Define
δk = sup {r > 0 : Br ((uk , vk )) ⊂ Z} .
(2)
If any δk = ∞, then Z = X × Y , we put U1 = X, V1 = Y , the rest empty, and we are
done. So consider the remaining case when δk ∈ (0, ∞) for all k only. Since, using the
definition of supremum,
[
Bδk ((uk , vk )) =
Br ((uk , vk )) ,
0<r<δk
we have
Bδk ((uk , vk )) ⊂ Z.
(3)
Since Z is open, the set in (2) is nonempty and thus all δk > 0. Define
Uk = Bδk /2 (uk ) ,
Vk = Bδk /2 (vk ) .
(4)
Recall that the distance function on X × Y is defined as in (1). Let k ∈ N. We will
prove that Uk × Vk ⊂ Z. So, let x ∈ Uk and y ∈ Vk . Then
d ((x, uk ) , (y, vk )) = d (x, uk ) + d (y, vk ) <
δk δk
+
= δk ,
2
2
thus (x, y) ∈ Bδk ((uk , vk )) ⊂ Z from (3). We have proved that Uk × Vk ⊂ Z for all k,
and, consequently,
∞
[
Uk × Vk ⊂ Z.
k=1
It remains to prove the opposite inclusion,
Z⊂
∞
[
Uk × Vk .
k=1
Let (s, t) ∈ Z; if we prove that there exists k such that (s, t) ∈ Uk × Vk we are done.
Since Z is open, there exists δ > 0 such that
Bδ ((s, t)) ⊂ Z.
(5)
By density, there exists a point (uk , vk ) such that
δ
d ((s, uk ) , (t, vk )) < .
4
2
(6)
We will need to find a lower bound on δk in order to establish that (s, t) ∈ Uk × Vk .
Now first prove that
Bδ/2 ((uk , vk )) ⊂ Bδ ((s, t)) .
(7)
Let (x, y) ∈ Bδ/2 ((uk , vk )), that is,
δ
d ((x, uk ) , (y, vk )) < .
2
By the triangle inequality and using (6),
d ((s, x) , (t, y)) ≤ d ((s, uk ) , (t, vk )) + d ((x, uk ) , (y, vk )) <
δ δ
+ < δ,
4 2
which proves (7). Now (7) and (5) yield
Bδ/2 ((uk , vk )) ⊂ Bδ ((s, t)) ⊂ Z
and from the definition of δk in (2), we have the desired lower bound
δ
δk ≥ .
2
Now from (6),
d ((s, uk ) , (t, vk )) <
δk
δ
≤ ,
4
2
which implies
δk
δk
, d (t, vk ) < ,
2
2
which, by the definition of Uk and Vk in (4), is the same as (s, t) ∈ Uk × Vk .
d (s, uk ) <
Remark 1 Obviously, I had to draw some pictures to see where I am going. Do that yourself
to understand the proof ! I have formulated the proof in terms of the open balls as much as
possible so that you can just draw circles. This has cost me few extra lines (I had to poass
from inclusion of balls to inequalities) but I think it was worth it.
Remark 2 I have decided not to introduce new symbols for points in X × Y and I just wrote
everything in terms of the ordered pairs. This way I have avoided definining new symbols
that the reader would have to remember, but the writing was a bit longer. This is a matter
of taste. I could have also used notation such as x = (x1 , x2 ) ∈ X1 × X2 which in hindsight
might have been the best.
Remark 3 Just as obviously, after writing the first version I had to go back few times and
juggle the various fractions of δ to make it work. That’s why this proof is time consuming to
derive in class.
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Remark 4 You can verify the proof in detail step by step, without too much thinking and
understanding the big picture. There are crossreferences to previous formulas as needed,
nothing is left to the imagination. This is the kind of writing you must get used to so that
you can verify what you write, and I will expect this in your homework solutions. Otherwise
it will be impossible for me to verify your solutions; I will not attempt to invent missing
arguments and bridge the gaps; I will simply mark the first point that does not make sense
to me, and stop there.
Remark 5 Even if it takes few pages when done in detail, this is a straightforward exercise
that the author of our book considers self-evident and just mentions as a fact in passing. You
need to learn how to work out such exercises whenever needed and give it whatever time and
effort it takes.. This is an indispensable part of reading math books at this level and it will
get easier as you get more experience, though it will still take lots of time.
Remark 6 It is also possible there is some simpler and shorter solution that I just did not
see.
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