Continuous

 

Def 1

Let $(X, \mathscr{T})$, $(Y, \mathscr{T}')$ be topological space. We define

$f : (X, \mathscr{T}) \to (Y, \mathscr{T}')$ is continuous

$\Leftrightarrow$  For any open set $U$ in $Y$,  $f^{-1}(U)$ is open set in $X$ 

 

 

 

Thm 1

Let $X$, $Y$, $Z$ be topological space. Then

$f : X \to Y$  and  $g : Y \to Z$ are continuous  $\Rightarrow$  $g \circ f$ is continuous

 

  Let $U$ be open in $Z$

  $(g \circ f)^{-1}(U) = f^{-1} \circ g^{-1} (U) = f^{-1}(g^{-1}(U))$

 

  Since $g$ is continuous,

  $g^{-1}(U)$ is open in $Y$

  Since $f$ is continuous,

  $f^{-1}(g^{-1}(U))$ is open in $X$.

  $\therefore$  $g \circ f$ is continuous

 

 

 

Thm 2

Let $X$, $Y$ be topological space  and  $f : X \to Y$

T.F.A.E

$(1)$  $f$ is continuous

$(2)$  Let $\mathscr{B}$ be a base of $Y$. For any $B \in \mathscr{B}$, $f^{-1}(B)$ is open set in $X$

$(3)$  For any $A \subseteq X$,  $f(\overline{A}) \subseteq \overline{f(A)}$

$(4)$  For any $C \subseteq Y$,  $\overline{f^{-1}(C)} \subseteq f^{-1}(\overline{C})$

$(5)$  For any closed set $C$ of $Y$, $f^{-1}(C)$ is closed set in $X$

 

pf)

$(1) \Rightarrow (2)$

  It is clear.

 

$(2) \Rightarrow (3)$

  $\forall A \subseteq X$

 

  $\forall y \in f(\overline{A})$

  So there exists $x \in \overline{A}$  s.t.

  $f(x) = y$

  Let $U$ be an open set containing $y$.

 

 

  Claim : $U \cap f(A) \neq \varnothing$

  By Thm 1, there exists $B \in \mathscr{B}$  s.t.

  $y \in B \subseteq U$

  $\therefore$  $x \in f^{-1}(B)$

 

  By assumption,

  $f^{-1}(B)$ is open in $X$.

  Since $x \in \overline{A}$,

  $f^{-1}(B) \cap A \neq \varnothing$

  $\therefore$  $B \cap f(A) \neq \varnothing$

 

  Since $B \subseteq U$,

  $U \cap f(A) \neq \varnothing$

 

  end

 

  $\therefore$  $y \in \overline{f(A)}$

  $\therefore$  $f(\overline{A}) \subseteq \overline{f(A)}$

 

$(3) \Rightarrow (4)$

  $\forall C \subseteq Y$

 

  $\forall x \in \overline{f^{-1}(C)}$

  So $f(x) \in f(\overline{f^{-1}(C)})$

 

  By assumption,

  $f(\overline{f^{-1}(C)}) \subseteq \overline{f(f^{-1}(C))} \subseteq \overline{C}$

 

  Since $f(f^{-1}(C)) \subseteq C$, by Cor 4

  $\overline{f(f^{-1} (C))} \subseteq \overline{C}$

  $\therefore$  $f(x) \in \overline{C}$

  $\therefore$  $x \in f^{-1}(\overline{C})$

 

  $\therefore$  $\overline{f^{-1}(C)} \subseteq f^{-1}(\overline{C})$

 

$(4) \Rightarrow (5)$

  Let $C$ be a closed set in $Y$.

  So $\overline{C} = C$

 

  By assumption,

  $\overline{f^{-1}(C)} \subseteq f^{-1}(\overline{C}) = f^{-1}(C)$

 

  Since $f^{-1}(C) \subseteq \overline{f^{-1}(C)}$,

  $\overline{f^{-1}(C)} = f^{-1}(C)$

 

  $\therefore$  $f^{-1}(C)$ is closed set in $X$

 

$(5) \Rightarrow (1)$

  Let $U$ be an open set in $Y$.

  So $Y \setminus U$ is closed set in $Y$

 

  By assumption,

  $f^{-1}(Y \setminus U)$ is closed set in $X$

  Since $f^{-1}(Y \setminus U) = f^{-1}(Y) \setminus f^{-1}(U) = X \setminus f^{-1}(U)$,

  $f^{-1}(U)$ is open set in $X$

 

 

 

Def 2

Let $(X, \mathscr{T})$, $(Y, \mathscr{T}')$ be topological space  and  $x_{0} \in X$. We define

$f : X \to Y$ is continuous at $x_{0}$

$\Leftrightarrow$  $\forall V \in \mathscr{T}'$ containing $f(x_{0})$,  $\exists U \in \mathscr{T}$ containing $x_{0}$  s.t.  $f(U) \subseteq V$

 

 

 

Thm 3

Let $(X, \mathscr{T})$, $(Y, \mathscr{T}')$ be topological space  and  $f : X \to Y$. Then

$f$ is continuous

$\Leftrightarrow$  $f$ is continuous for every point $x \in X$

 

pf)

$\Rightarrow)$

  $\forall x \in X$

 

  $\forall V \in \mathscr{T}'$ containing $f(x)$

  So $x \in f^{-1}(V)$

 

  Since $f$ is continuous,

  $f^{-1}(V)$ is open set in $X$

 

  By Prop 2, there exists $U \in \mathscr{T}$  s.t.

  $x \in U \subseteq f^{-1}(V)$

 

  $\therefore$  $f(U) \subseteq V$

  $\therefore$  $f$ is continuous at $x$

 

$\Leftarrow)$

  $\forall V \in \mathscr{T}'$

 

 

  Case 1 : $f^{-1}(V) = \varnothing$

  $\therefore$  $f^{-1}(V) \in \mathscr{T}$

 

  end

 

 

  Case 2 : $f^{-1}(V) \neq \varnothing$

  For each $x \in f^{-1}(V)$,

  since $f(x) \in V$  and  $f$ is continuous at $x$, there exists $U_{x} \in \mathscr{T}$ containing $x$  s.t.

  $f(U_{x}) \subseteq V$

 

  Since $x \in U_{x} \subseteq f^{-1}(V)$,

 

  $f^{-1}(V) = \displaystyle \bigcup_{x \in f^{-1}(V)} \left\{x \right\} \subseteq \displaystyle \bigcup_{x \in f^{-1}(V)} U_{x} \subseteq f^{-1}(V)$

 

  $\therefore$  $f^{-1}(V) = \displaystyle \bigcup_{x \in f^{-1}(V)} U_{x} \in \mathscr{T}$

 

  end

 

  $\therefore$  $f$ is continuous