MA3421 (Functional Analysis 1) Tutorial sheet 3
[October 16, 2014]
Name: Solutions
1. Let (X, dX ) and (Y, dY ) be metric spaces and Z = X × Y . Verify that d1 : Z × Z → R
given by
d1 ((x1 , y1 ), (x2 , y2 )) = dX (x1 , x2 ) + dY (y1 , y2 )
defines a metric on Z.
Solution: We just need to use the metric properties of dX and dY to show that d1 is a metric.
Certainly d1 ((x1 , y1 ), (x2 , y2 )) = dX (x1 , x2 ) + dY (y1 , y2 ) ≥ 0 always (that is for all
(x1 , y1 ), (x2 , y2 ) ∈ X × Y ) and if equality holds then dX (x1 , x2 ) ⇒ x1 = x2 and
dY (y1 , y2 ) ⇒ y1 = y2 , so that (x1 , y1 ) = (x2 , y2 ). Also d1 ((x1 , y1 ), (x1 , y1 )) = dX (x1 , x1 )+
dY (y1 , y1 ) = 0.
d1 is symmetric since dX and dY are:
d1 ((x1 , y1 ), (x2 , y2 )) = dX (x1 , x2 ) + dY (y1 , y2 )
= dX (x2 , x1 ) + dY (y2 , y1 )
= d1 ((x2 , y2 ), (x1 , y1 ))
Finally d1 satisfies the triangle inequality since if (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) ∈ X × Y , then
d1 ((x1 , y1 ), (x3 , y3 )) =
≤
=
=
dX (x1 , x3 ) + dY (y1 , y3 )
dX (x1 , x2 ) + dX (x2 , x3 ) + dY (y1 , y2 ) + dY (y2 , y3 )
dX (x1 , x2 ) + dY (y1 , y2 ) + dX (x2 , x3 ) + dY (y2 , y3 )
d1 ((x1 , y1 ), (x2 , y2 )) + d1 ((x2 , y2 ), (x3 , y3 ))
2. Let X denote the Sorgenfrey line (so X = R as a set, but the topology has as a base for its
open sets the intervals [a, b) with a < b). Let Y = X × X with the product topology and
Z = {(x, −x) : x ∈ R} with the relative topology for Z ⊆ Y .
Show that
(a) Q2 is dense in Y
Solution: Let E denote the closure of Q2 in Y . If Q2 is not dense, that means E 6= Y
and so here is a point (x, y) ∈ Y \ E. But then (since Y \ E is open in Y ) there is a
basic open set U × V for Y (with U ⊂ X and V ⊂ X open) with (x, y) ∈ U × V ⊆
Y \ E.
So x ∈ U and there must be a basic open set [a, b) for X with x ∈ [a, b) ⊆ U .
Similarly y ∈ V and there must be a basic open set [c, d) for X with y ∈ [c, d) ⊆ V .
But inside [a, b) there is a rational q ∈ [a, b) ∩ Q and there is another rational r ∈
Q ∩ [c, d). Then we get a contradiction from
(q, r) ∈ [a, b) × [c, d) ⊆ U × V ⊆ Y \ E ⊆ Y \ Q2
(which holds since Q2 ⊆ E). Thus E = R2 must hold and Q2 is dense as required.
(b) Y is separable
Solution: To say Y is separable means Y has a countable dense subset and we have
just shown that Q2 (which is countable) is dense.
(c) Z has the discrete topology and is not separable.
[A conclusion from this is that subspaces of separable topological spaces need not be
separable.]
Solution: The open sets of Z in the relative topology (also known as the subspace
topology or the topology induced form the topology of Y ) are those of the form
O ∩ Z with O ⊆ Y open.
But amongst such open O we have O = [x, x + 1) × [−x, −x + 1) and for that O,
O ∩ Y = {(x, −x)}. (The reason is that points (z1 , z2 ) ∈ Z satisfy z1 + z2 = 0 while
if (z1 , z2 ) ∈ O = [x, x + 1) × [−x, −x + 1) then
x + (−x) ≤ z1 + z2 < x + 1 + (−x + 1) = 2
but we can only get 0 = x + (−x) when x = z1 and −x = z2 . The figure is means to
make this more clear.)
Figure 1: O ∩ Y = {(x, −x)}
Once every singleton subset of Z is open, them every subset of Z is open (union of
the singletons it contains) and so Z has the discrete topology.
Now, in the discrete topology each subset is open, each subset is closed and so every
subset is its own closure. There are no dense subsets of Z except Z itself. Since Z is
uncountable (same number of points as R), Z has not countable dense subset — that
is Z is not separable.
Richard M. Timoney
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