MA3421 (Functional Analysis 1) Tutorial sheet 4
[October 23, 2014]
Name: Solutions
1. Let X be a first countable topological space. Show that for each x0 ∈ X there is a neighbourhood base Bx0 = {B1 , B2 , . . .} at x0 such that B1 ⊇ B2 ⊇ B3 ⊇ · · · .
Solution: Fix a point x0 ∈ X. Since X is first countable, there is a countable neighbourhood base Bx0 at x0 . We can write the sets in Bx0 in a list (finite or infinite) Bx0 =
{B1 , B2 , . . .}.
The list can’t be empty as X is a neighbourhood of x0 and so there must be B ∈ Bx0 with
x ∈ B ⊆ X. If the list is finite, we can treat it as infinite by repeating the last set infinitely
often. So, if Bx0 = {B1 , B2 , . . . , Bn } let Bj = Bn for j = n + 1, n + 2, . . ..
Now define B10 = B1 , B20 = B1 ∩ B2 and in general
Bn0
=
n
\
Bi = B1 ∩ B2 ∩ · · · ∩ Bn
i=1
for n ≥ 1. Then we can see that each Bn0 is a neighbourhood of x0 because the intersection of two neighbourhoods (of x0 ) is again a neighbourhood. (So that extends to finite
intersections by induction, or we can use that Bn0 = b0n−1 ∩ Bn for n > 1.)
Let Bx0 0 = {Bn0 : n ∈ N}. We claim it is a neighbourhood base at x0 . Indeed we have
already noted that each Bn0 is a neighbourhood of x0 . If N is any neighbourhood of x0 ,
then we know that there is n so that Bn ⊆ N (because Bx0 is a neighbourhood base). It
follows that Bn0 ⊆ Bn ⊆ N .
Clearly B10 ⊇ B20 ⊇ B30 ⊇ · · · .
2. Let Xn be first countable
Q topological spaces for n = 1, 2, . . . and let x = (x1 , x2 , . . .) be a
point in the product ∞
n=1 Xn and let Bn = {Bn,1 , Bn,2 , . . .} be a neighbourhood base at
xn ∈ Xn for each n. Assume that Bn,1 ⊇ Bn,2 ⊇ Bn,3 ⊇ · · · for each n.
Show that the sets of the form
B1,k × B2,k × · · · × Bk,k × Xk+1 × Xk+2 × · · ·
make a countable neighbourhood base at x (in the product topology).
[So it follows form these two exercises that countable products of first countable spaces
are first countable.]
Solution: Let N be a neighbourhood of x. Then x ∈ N ◦ with N ◦ open in the product
topology. So there is a basic open set B for the product topology with x ∈ B ⊆ N ◦ ⊆ N .
Basic open sets for the product topology are finite intersections
B = πj−1
(U1 ) ∩ πj−1
(U2 ) ∩ πj−1
(U` )
1
2
`
where ` ≥ 0, j1 , j2 , . . . , j` are distinct elements indices 1 ≤ ji < ∞ and Ui ⊆ Xji is open
for 1 ≤ i ≤ `. As usual, we use πj for the coordinate projection of the product space onto
Xj . We can alternative write
∞
Y
B=
Vn
n=1
where
(
Ui
Vn =
Xn
if n = ji for some i, 1 ≤ i ≤ `
otherwise.
Alternatively we could write
B = V1 × V2 × · · · × Vr × Xr+1 × Xr+2 × · · ·
where r = max{ji : 1 ≤ i ≤ `}.
From x ∈ B we have xm ∈ Vm for m = 1, 2, . . . , r.
Q
Note that if y = (y1 , y2 , . . .) ∈ ∞
n=1 Xn then y ∈ B is equivalent to ym ∈ Vm for
1 ≤ m ≤ r.
Since xm ∈ Vm , Vm is open and Bm = {Bm,1 , Bm,2 , . . .} is a neighbourhood base at
xm ∈ Xm , there must be km ≥ 1 with xm ∈ Bm,km ⊆ Vm . Since the sets in Bm are in
decreasing order, we then also have Bm,k ⊆ Vm for k ≥ km .
Let k0 = max{k1 , k2 , . . . , kr }. Then
B1,k × B2,k × · · · × Br,k × Xr+1 × Xr+2 × · · · ⊆ B
for k > k0 . So, if we also ensure that k ≥ r we have
B1,k ×B2,k ×· · ·×Bk,k ×Xk+1 ×Xk+2 ×· · · ⊆ B1,k ×B2,k ×· · ·×Br,k ×Xr+1 ×Xr+2 ×· · · ⊆ B.
So we have the neighbourhood base property, but finally we need to check that the sets
B1,k × B2,k × · · · × Bm,k × Xk+1 × Xk+2 × · · ·
◦
are neighbourhoods of x. That is quite easy since xm ∈ Bm,k
(for all m) and
◦
◦
B1,k
× B2,k
× · · · × Bm,k ◦ ×Xk+1 × Xk+2 × · · ·
is an open set in the interior containing x.
Richard M. Timoney
2