![]() | (P-1.1) |
Let U be the set ⋃ x∈AUx where Ux is a particular, yet arbitrary, set for which the property in P-1.1 holds with respect to x. As U is the union of subsets of A, clearly U ⊂ A. Also for each a ∈ A, a ∈ Ua ⊂U, so A ⊂U. Thus A equals U, a union of open sets, i.e. A is open.
Now let {Uα} be an arbitrary collection of sets in . The complement of the union of these sets, X \⋃
αUα, is equal to
⋂
α(X \ Uα) by DeMorgan, which has a cardinality that is no larger than that of its largest set. As each X \ Uα is
countable, then ⋂
α(X \ Uα), and therefore X \⋃
αUα, is countable. So
is closed under arbitrary union of its
elements.
Let {Ui} be a finite collection of sets in . The complement of the intersection of these sets, X \⋂
iUiA, by DeMorgan,
is also given by ⋃
i(X \Ui). This set, however, is a countable union of countable sets and thus has countable cardinality. So
is closed under finite union of its elements.
With the above two properties of and, as we saw, the fact that it contains ∅ and X, then
is a topology on
X.
Now let x ∈⋂
A∈(X \ A). Thus for all A ∈
, x is in X \ A. Then there is no A for which x ∈ A, so x is in
X \⋃
A∈
A. Hence ⋂
A∈
(X \ A) ⊂ X \⋃
A∈
A.
With these two results, we have that
which we can use to prove that
Using a little complement-trickery, we have that X \⋂
A∈A is equal to X \⋂
A∈
(X \ (X \ A)), which from our above
proof, we know to be equal to X \ (X \⋃
A∈
(X \ A)). However, this is simply ⋃
A∈
(X \ A)). Thus we
have
as desired.
So let {Uα} be a subset of elements of of arbitrary size. Allow x to be an element of ⋃
αUα. Then x is in some Uα
and since Uα ∈
, we can find an interval (a,b) such that x ∈ (a,b) ⊂ Uα. Therefore (a,b) is also contained in ⋃
αUα and
so it’s an element of
. Thus
is closed under arbitrary union.
Now let {Ui} be a finite collection of elements of . If ⋂
iUi is the empty set, then its in
as per above, so let it be
nonempty. Let x ∈⋂
iUi. Then there is a collection {(ab)i} of intervals such that for each i, x ∈ (a,b)i ⊂ Ui. This then
implies that x ∈⋂
i(a,b)i, but ⋂
i(a,b)i ⊂ Ui for each i. Hence x ∈⋂
i(a,b)i ⊂⋂
iUi, and since ⋂
i(a,b)i is an open
interval, then ⋂
iUi ∈
; i.e. it is closed under finite union.