Lecture Notes Topology (2301631) Phichet Chaoha Department of ...

CHAPTER 2
Countability and Separation Axioms
1. The Countability Axioms
Definition 2.1. Let X be a space and x ∈ X. A collection B x of open subsets
of X is called a (local) basis at x if for each neighborhood U of x, there exists
B ∈ B x such that x ∈ B ⊆ U.
Definition 2.2. A space X is called
(1) first-countable if it has a countable basis at each point.
(2) second-countable if it has a countable basis.
(3) separable if it has a countable dense subset.
Theorem 2.3. If X is second-countable, then it is first-countable and separable.
Proof. Let X be second-countable. Then it is clear from the definition that
X is also first-countable. Now, let B be a countable basis for X. By the axiom of
choice, we can form a subset D of X by choosing one element from each set in B.
It follows that D is the desired countable dense subset of X.
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Example 2.4. R n is clearly second-countable by taking the collection all products
of intervals with rational endpoints as a basis. Hence, it is first-countable by
the previous theorem. Moreover, it is separable since Q n is its countable dense
subset. Hence, R n satisfies all countability axioms.
Example 2.5. R l is first-countable and separable, but not second-countable.
It is easy to see that the intervals [x, x + 1 n ) form the basis at x ∈ R l, and clearly
Q is dense in R l . To see that R l is not second-countable, let B be a basis for R l .
Then, for x, x ′ ∈ R l , there exists B x , B x ′ ∈ B such that x ∈ B x ⊆ [x, x + 1) and
x ′ ∈ B x ′ ⊆ [x ′ , x ′ + 1). Clearly, inf(B x ) = x because {x} ⊆ B x ⊆ [x, x + 1), and
similarly, inf(B x) ′ = x ′ . Hence, if x ≠ x ′ , we must have B x ≠ B x ′. So, B contains
an uncountable subcollection {B x | x ∈ R l }, and hence B is uncountable.
Exercise 2.6. Let X be a second-countable space. Prove that
(1) Every collection of disjoint open sets in X is countable.
(2) The subspace topology on an uncountable subset of X cannot be discrete.
Theorem 2.7. A subspace of a first-countable [second-countable] space is firstcountable
[second-countable]. A countable product of first-countable [second-countable]
spaces is first-countable [second-countable].
Proof. The first statement is clear since we can construct a basis element
of a subspace simply by intersecting a basis element of the whole space with the
subspace. For a countable product, we use the fact that a countable union of
countable sets and a finite product of countable sets are countable. For example,
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