Base

 

Def 1

$(X, \mathscr{T})$ : topological space

A family $\mathscr{B}$ of open sets of $X$ ( i.e. $\mathscr{B} \subseteq \mathscr{T}$ ) is a base for $\mathscr{T}$ if

every open set in $X$ is a union of sets in $\mathscr{B}$.

 

 

 

Thm 1

Let $(X, \mathscr{T})$ be a topological space  and  $\mathscr{C} \subseteq \mathscr{T}$ .

T.F.A.E.

(1)  $\mathscr{C}$ is base for $\mathscr{T}$  

(2)  $\forall p \in X$  and  for every open set $U$ containing $p$,  $\exists O \in \mathscr{C}$  s.t. 

$p \in O \subseteq U$

 

pf)

(1) $\Rightarrow$ (2)

  $\forall p \in X$

  Let $U$ be an open set containing $p$

 

  Since $\mathscr{C}$ is a base of $\mathscr{T}$, there exists $O_{\alpha} \in \mathscr{C}$  for $\alpha \in I$  s.t.

$\displaystyle U = \bigcup_{\alpha \in I} O_{\alpha}$

 

  Since $p \in U$, there exists $\alpha_{0} \in I$  s.t.

$\displaystyle p \in O_{\alpha_{0}} \in \mathscr{C}$

 

  Since $U = \displaystyle \bigcup_{\alpha \in I} O_{\alpha} $,

$O_{\alpha_{0}} \subseteq U$

 

(2) $\Rightarrow$ (1)

  Let $U$ be an open set.

  $\forall x \in U \subseteq X$

 

  By assumption (2), there exists $O_{x} \in \mathscr{C}$  s.t. 

  $x \in O_{x} \subseteq U$

 

  Therefore

$U = \displaystyle \bigcup_{x \in U} \left\{x \right\} \subseteq \bigcup_{x \in U} O_{x} \subseteq U$

 

  $\therefore$  $U = \displaystyle \bigcup_{x \in U} O_{x}$

  $\therefore$  $\mathscr{C}$ is base for $\mathscr{T}$

 

 

 

Thm 2

$X$ : set

A family $\mathscr{B}$ of subset of a set $X$ is a base for some topology of $X$ iff the following conditions holds.

$(\mathrm{i})$   The union of the members of $\mathscr{B}$ is $X$.

$(\mathrm{ii})$  For any $B_{1}, B_{2}$ in $\mathscr{B}$  and  $x \in B_{1} \cap B_{2}$,  $\exists B_{x} \in \mathscr{B}$  s.t. 

$x \in B_{x} \subseteq B_{1} \cap B_{2}$

 

# So we can define topology from base

 

pf)

$\Rightarrow)$

  Suppose $\mathscr{B}$ is a base for a topology $\mathscr{T}$ for $X$.

 

 

  Show 1 : $(\mathrm{i})$

  Since $X$ is open set, there exists $B_{\alpha} \in \mathscr{B}$  for $\alpha \in I$ s.t.

  $X = \displaystyle \bigcup_{\alpha \in I} B_{\alpha}$

 

  Therefore

  $X = \displaystyle \bigcup_{\alpha \in I} B_{\alpha} \subseteq \bigcup_{B_{\alpha} \in \mathscr{B}} B_{\alpha} \subseteq X$

  $\therefore$  $\displaystyle \bigcup_{B_{\alpha} \in \mathscr{B}} B_{\alpha} = X$

 

  end

 

 

  Show 2 : $(\mathrm{ii})$

  Let $B_{1}, B_{2} \in \mathscr{B} \subseteq \mathscr{T}$

  $\forall x \in B_{1} \cap B_{2}$

 

  Since $B_{1} \cap B_{2}$ is open set, by Thm 1 there exists $B_{x} \in \mathscr{B}$  s.t.

$x \in B_{x} \subseteq B_{1} \cap B_{2}$

 

  end

 

$\Leftarrow)$

  Let $\mathscr{B} = \left\{B_{\alpha} \; | \; \alpha \in J \right\}$.

  We define

  $\mathscr{T} =$ the collections of all unions of element of $\mathscr{B}$

        $= \left\{\displaystyle \bigcup_{\alpha \in D} B_{\alpha} \; | \; D \subseteq J \right\}$

 

 

  Show 1 : $\varnothing \in \mathscr{T}$,  $X \in \mathscr{T}$

  $\varnothing = \displaystyle \bigcup_{\alpha \in \varnothing} B_{\alpha} \in \mathscr{T}$

  By assumption $(\mathrm{i})$,

  $X = \displaystyle \bigcup_{\alpha \in J} B_{\alpha} \in \mathscr{T}$

 

  end

 

 

  Show 2 : $U_{\alpha} \in \mathscr{T}$  for $\alpha \in I$  $\Rightarrow$  $\displaystyle \bigcup_{\alpha \in I} B_{\alpha} \in \mathscr{T}$

  It is clear by definition of $\mathscr{T}$.

 

  end

 

 

  Show 3 : $U_{1}, U_{2} \in \mathscr{T}$  $\Rightarrow$  $U_{1} \cap U_{2} \in \mathscr{T}$

  By definition of $\mathscr{T}$, there exists $D_{1}, D_{2} \subseteq J$  s.t.

$U_{1} = \displaystyle \bigcup_{\alpha \in D_{1}} B_{\alpha}$

$U_{2} = \displaystyle \bigcup_{\alpha \in D_{2}} B_{\alpha}$

 

  $\forall x \in U_{1} \cap U_{2}$, there exists $\alpha_{1} \in D_{1}$ and $\alpha_{2} \in D_{2}$  s.t.

  $x \in B_{\alpha_{1}} \cap B_{\alpha_{2}}$

 

  Since $B_{\alpha_{1}}, B_{\alpha_{2}} \in \mathscr{B}$, by assumption $(\mathrm{ii})$ there exists $B_{\alpha_{x}} \in \mathscr{B}$  s.t.

$x \in B_{\alpha_{x}} \subseteq B_{\alpha_{1}} \cap B_{\alpha_{2}}$

 

  Since $B_{\alpha_{1}} \cap B_{\alpha_{2}} \subseteq U_{1} \cap U_{2}$,

  $U_{1} \cap U_{2} = \displaystyle \bigcup_{x \in U_{1} \cap U_{2}} \left\{x \right\} \subseteq \bigcup_{x \in U_{1} \cap U_{2}} B_{\alpha_{x}} \subseteq U_{1} \cap U_{2}$

 

  $\therefore$  $U_{1} \cap U_{2} = \displaystyle \bigcup_{x \in U_{1} \cap U_{2}} B_{\alpha_{x}} \in \mathscr{T}$

 

  end

 

  $\therefore$  $\mathscr{T}$ is a topology on $X$

 

 

 

Ex 1

$X = \mathbb{R}$

$\mathscr{B} = \left\{[a, b) \; | a < b \right\}$

$\Rightarrow$  $\mathscr{B}$ is a base for some topology for $\mathbb{R}$.

 

 

 

Def 2

Topology of Ex 1 is called lower limit topology.

We write $(\mathbb{R}, \text{lower limit topology}) = \mathbb{R}_{l}$.

 

 

 

Def 3

Let $\mathscr{T}_{1}, \mathscr{T}_{2}$ be two topology of $X$.

If $\mathscr{T}_{1} \subseteq \mathscr{T}_{2}$, then $\mathscr{T}_{1}$ is weaker (or coarser) than $\mathscr{T}_{2}$.

( $\mathscr{T}_{2}$ is stronger (or finer) than $\mathscr{T}_{1}$ )

 

 

 

Prop 1

For $\mathbb{R}$,

$\text{usual topology} \subsetneqq \text{lower limit topology}$

 

  Let $\mathscr{T}_{usual}$ be usual topology  and  $\mathscr{T}_{lower}$ be lower limit topology.

  Let $\mathscr{B}$ is base of $\mathscr{T}_{usual}$

 

 

  Show 1 : $\mathscr{T}_{usual} \subseteq \mathscr{T}_{lower}$

  It is sufficient to show $\mathscr{B} \subseteq \mathscr{T}_{lower}$

 

  For $(a, b) \in \mathscr{B}$,

  $(a, b) = \displaystyle \bigcup_{n \in \mathbb{N}} [a + \frac{1}{n}, b) \in \mathscr{T}_{lower}$

 

  end

 

 

  Show 2 : $\mathscr{T}_{usual} \neq \mathscr{T}_{lower}$

  We prove from $[0, 1) \notin \mathscr{T}_{usual}$

 

  Let $U = [0, 1)$.

  For $0 \in U$, there doesn't exist $B \in \mathscr{B}$  s.t.

$0 \in B \subseteq [0, 1)$

  By Thm 1,

  $U \notin \mathscr{T}_{usual} $

 

  end

 

 

 

Def 4

If $(X, \mathscr{T})$ has a countable base, then we call $(X, \mathscr{T})$ is a 2-nd countable space.

 

 

 

Thm 3

$(\mathbb{R}, \text{usual topology})$ is a 2-nd countable space.

 

  Let $\mathscr{B} = \left\{(a, b) \; | \; a < b , \; a, b \in \mathbb{Q} \right\}$

  It is clearly subset of usual topology.

 

  $\forall a, b \in \mathbb{R}$, there exists $a_{n}, b_{n} \in \mathbb{Q}$  s.t.

  $\displaystyle \lim_{n \to \infty} a_{n} = a$,     $\displaystyle \lim_{n \to \infty} b_{n} = b$,     $a < a_{n} < b_{n} < b$

  $\therefore$  $(a, b) = \displaystyle \bigcup_{n \in \mathbb{N}} (a_{n}, b_{n})$

 

  $\therefore$  Every open set of $\mathbb{R}$ is a union of sets in $\mathscr{B}$.

  $\therefore$  $(\mathbb{R}, \text{usual topology})$ is a 2-nd countable space.