Download Let (X, τ) be a topological space, a base B is a

Let (X, τ ) be a topological space, a base B is a subset of τ such that each element
of τ could be written as an union of element from B, every base i) covers X and ii)
has the property that the intersection of any two base sets contains another base set
(and conversely every subset of 2X which satisfies these properties determines a unique
topology, but for different sets which satisfiy i) and ii) the topologies they induce could
be different, for example with X = R the topologies generated by {(a, b) : a, b ∈ R} and
the topoloy generated by {[a, b) : a, b ∈ R} are much different). Note that a base for
a topology is not unique, even not when looking for a base with as least members as
possible. (a class of topological space which has this property are the Alexandrov and
the finite spaces, for them there exists canonical ”minimal” bases). A family of subsets
B is a base for a given topology τ if and only if for any open set U and x ∈ U there
exists a B ∈ B such that x ∈ B ⊆ U . By this, an arbibtrary subset U ⊆ X is open if and
only for each x ∈ U there exists a base set B ∈ B such that x ∈ B ⊆ U . Furthermore for
two bases B1 , B2 and their induced topologies τ1 , τ2 it is τ1 ⊆ τ2 if and only if for each
base set B of B1 and x ∈ B there exists a base set B 0 from B2 such that x ∈ B 0 ⊆ B,
this is equivalent to saying that each basis element from B1 could be written as a union
of base elements from B2 . Bases are like the ”building blocks” for topologies, and most
properties expressible in terms of open sets could be equivalently expressed in terms of
bases, which most often are more convenient to work with.
A family of subsets Bx ⊆ τ is called a local base at x if for every open set U containing
x there exists some B ∈ Bx such that x ∈ B ⊆ U .
Now call a topological space (X, τ ) first countable if every point has a countable local
base, call it second countable if it has a countable base, and call it separable if it contains
a countable dense subset.
Lemma: Every second countable space is first countable and separable.
Proof: Let (X, τ ) be second countable with base {Bn }. Then for each x the set Bx :=
{Bn : x ∈ Bn } is a countable local base. Furthermore from each base set Bn select some
point xn ∈ Bn , then the set {xn } is countable and dense in X. For this let x ∈
/ {xn }.
Then each open set U with x ∈ U contains some Bn ⊆ U with x ∈ Bn , and because
x∈
/ {xn } it must be x 6= xn . But not every first countable space is also second countable. Consider an uncountable
set X with the discrete topology, then point is isolated, hence this singleton set form
a local basis, but there does not exists a countable base because the singletons sets as
open sets have to be part of such a base and there are uncountable many. Furthermore,
because this space has no limit points because each point is isolated, it could not be
separable.
An example of a separable, but not second countable space is the topology induced on
R by the basis sets
{[a, b) : a, b ∈ R}
the so called Sorgenfrey line, the rational Q with respect to this toplogy are dense,
because each non-empty basis set contains the open interval and therefore a rational
number. But this topological space is not second countable.
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Example: The co-finite topology. Let X be some set and define
τ = {A ⊆ X : A = ∅ or AC is finite }.
Then each finite set is closed by definition and has no limit points, because let A be
finite and x ∈ A, then {x} ∪ AC is an open set containing x but no other point of A, i.e.
{x} is open in the subspace A. Each infinite set is dense in X, i.e. every point of X is
a limit point of this set. For this let A be infinite and let x ∈ X be arbitrary. Let U be
an open set with x ∈ U , then U C is finite, so for infinite A the set A ∩ U C is at most
finite too, therefore there must be (infinite many) points of A inside of U
Now let i) X be finite, then by definition each set is open, so we get the discrete topology.
Because each point is isolated in X, i.e. {x} is open, this singleton set forms a finite
(and therefore countable) local base at x, so the space is first countable. Because the set
of all open subsets form a (trivial) base, it is second countable too. Furthermore since X
is finite (and hence countable) and it’s closure is trivially X itself the space is separable.
Let ii) X be countable. Then because every infinite subset is dense the space is separable.
Now we show that the set of all open sets is countable. For this notice that the families
S
of subsets Sn = {A ⊆ X : |A| = n} are all countable for fixed n, now S = n Sn is the
family of all finite subsets, and as a countable union of countable sets is itself countable.
Now the family of all open sets form a basis, and so the space is second countable, and
by this also first countable.
Lastly let iii) X be uncountable. Then any infinite subset is dense, in particular the
countable infinite ones, and so the space is separable. We show that the space is not
first countable, which implies that it could also not be second countable. Suppose it is
first countable and fix some x ∈ X, then this has a countable local base {B1 , B2 , B3 , . . .}.
Now
[
Y := BnC
n
is a countable union of finite sets, and so is countable. Therefore there exists some y 6= x
which lies in all Bn . Now consider X − {y}, this set is open and contains x, but there
could be no Bn such that Bn ⊆ X − {y} for otherwise there would be y ∈ X − {y},
which is not possible.
To summarize, we have:
separable first countable second countable
finite
x
x
x
countable infinite x
x
x
uncountable
x
Remark: A nice proof using the properties of N as numbers to show that the set of all
finite sets of N are countable. Let A denote the set of all finite subsets of N. We need
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to find an injection f : A → N. Let p1 < p2 < p3 < . . . denote the prime numbers in
increasing order. Given S = {x1 , . . . , xn } ⊆ N such that x1 < x2 < . . . < xn , then
f (S) := px1 1 px2 2 · · · pxnn
is an injection, because of the uniqueness of the prime decomposition. Another proof that the uncountable finite complement space is not first countable I
found here
http://www.proofwiki.org/wiki/Uncountable_Finite_Complement_Space_is_
not_First-Countable
Here the fact that
\
Bx = {x}
holds because the finite complement topology is T1 , and so if there would be two different
x, y then we have an open set U with x ∈ U and y ∈
/ U , and therefore x ∈ B ⊆ U for
T
some B ∈ Bx and by this y ∈
/ Bx .
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