Product Space

 

Thm 1

Let $(X, \mathscr{T}_{1})$, $(Y, \mathscr{T}_{2})$ be topological space. Then

$\mathscr{B} = \left\{ U \times V \; | \; U \in \mathscr{T}_{1}, \; V \in \mathscr{T}_{2} \right\}$ is a base for a topology on $X \times Y$

( $X \times Y = \left\{(x, y) \; | \; x \in X, \; y \in Y \right\}$ )

 

  Show 1 : The union of the members of $\mathscr{B}$ is $X \times Y$

  Since $X \in \mathscr{T}_{1}$  and  $Y \in \mathscr{T}_{2}$,

  $X \times Y \subseteq \displaystyle \bigcup_{U \in \mathscr{T}_{1}, \; V \in \mathscr{T}_{2}} U \times V \subseteq X \times Y$

 

  end

 

 

  Show 2 : 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.  $x \in B_{x} \subseteq B_{1} \cap B_{2}$

  $\forall B_{1}, B_{2} \in \mathscr{B}$

  So there exists $U_{1}, U_{2} \in \mathscr{T}_{1}$  and  $V_{1}, V_{2} \in \mathscr{T}_{2}$  s.t.

  $B_{1} = U_{1} \times V_{1}$,     $B_{2} = U_{2} \times V_{2}$

 

  $\forall x \in B_{1} \cap B_{2}$

 

  Let $B_{x} = B_{1} \cap B_{2}$

  Therefore

  $B_{1} \cap B_{2} = (U_{1} \times V_{1}) \cap (U_{2} \cap V_{2}) = (U_{1} \cap U_{2}) \times (V_{1} \times V_{2}) \in \mathscr{B}$

 

  end

 

 

 

Def 1

Let $(X, \mathscr{T}_{1})$, $(Y, \mathscr{T}_{2})$ be topological space.

Let $\mathscr{T}$ be the topology generated by $\mathscr{B} = \left\{ U \times V \; | \; U \in \mathscr{T}_{1}, \; V \in \mathscr{T}_{2} \right\}$

 

$\mathscr{T}$ : product topology

$(X \times Y, \mathscr{T})$ : product space

 

 

 

Def 2

Let $(X, \mathscr{T}_{1})$, $(Y, \mathscr{T}_{2})$ be topological space  and  $f: X \to Y$.

$f$ is called open map if

$U \in \mathscr{T}_{1}$  $\Rightarrow$  $f(U) \in \mathscr{T}_{2}$

$f$ is called closed map if

$X \setminus U \in \mathscr{T}_{1}$  $\Rightarrow$  $Y \setminus f(X \setminus U) \in \mathscr{T}_{2}$

 

 

 

Prop 1

Let $X$, $Y$ be topological space  and  $X \times Y$ be product space.

Let projective map $P_{1} : X \times Y \to X$  and  $P_{2} : X \times Y \to Y$ by

$P_{1}(x, y) = x$,     $P_{2}(x, y) = y$

Then

$P_{1}$  and  $P_{2}$ are continuous and open map

 

  Show 1 : $P_{1}, P_{2}$ is continuous

  Let $U$ be an open set in $X$

 

  By definition of $P_{1}$,

  $P_{1}^{-1}(U) = U \times Y$

  $\therefore$  $P_{1}^{-1}(U)$ is open set in $X \times Y$

  $\therefore$  $P_{1}$ is continuous

 

  Similarly we can show $P_{2}$ is continuous

 

  end

 

 

  Show 2 : $P_{1}, P_{2}$ is open map

  Let $W$ be an open set in $X \times Y$

  So there exists $U_{\alpha}, V_{\alpha}$  for $\alpha \in I$  s.t.

  $U_{\alpha}$, $V_{\alpha}$ are open set in $X$, $Y$, respectively

  &     $W = \displaystyle \bigcup_{\alpha \in I} U_{\alpha} \times V_{\alpha}$

 

  By definition of $P_{1}$,

  $P_{1}(W) = P_{1}(\displaystyle \bigcup_{\alpha \in I} U_{\alpha} \times V_{\alpha})$

                 $= \displaystyle \bigcup_{\alpha \in I} P(U_{\alpha} \times V_{\alpha})$

                 $= \displaystyle \bigcup_{\alpha \in I} U_{\alpha}$

 

  $\therefore$  $P_{1}(W)$ is open set in $X$

  $\therefore$  $P_{1}$ is open map

 

  Similarly we can show $P_{2}$ is open map.

 

  end

 

 

 

Thm 2

Let $X$, $Y$ be topological space  and  $(X \times Y, \mathscr{T})$ be product space. Then

Let projective map $P_{1} : (X \times Y, \mathscr{T}_{2}) \to X$  and  $P_{2} : (X \times Y, \mathscr{T}_{2}) \to Y$ by

$P_{1}(x, y) = x$,     $P_{2}(x, y) = y$

Then

$P_{1}$  and  $P_{2}$ are continuous  $\Rightarrow$  $\mathscr{T} \subseteq \mathscr{T}_{2}$

 

# The product topology is the smallest topology that projective maps continuous

 

  Let $\mathscr{B}$ be base for $\mathscr{T}$

  Then it is sufficient to show

  $\mathscr{B} \subseteq \mathscr{T}_{2}$

 

 

  $\forall B \in \mathscr{B}$

  So there exists $U$, $V$  s.t.

  $U$, $V$ are open set in $X$, $Y$, respectively

  &     $B = U \times V$

 

  By definition of $P_{1}$  and  $P_{2}$,

  $P_{1}^{-1}(U) = U \times Y$,     $P_{2}^{-1}(V) = X \times V$

 

  Therefore

  $P_{1}^{-1}(U) \cap P_{2}^{-1}(V) = (U \times Y) \cap (X \times V) = (U \cap X) \times (Y \cap V) = U \times V$

  Since $P_{1}$  and  $P_{2}$ are continuous,

  $P_{1}^{-1}(U), \; P_{2}^{-1}(V) \in \mathscr{T}_{2}$

 

  $\therefore$  $B = U \times V \in \mathscr{T}_{2}$

  $\therefore$  $\mathscr{B} \subseteq \mathscr{T}_{2}$

 

 

 

Thm 3

Let $X$, $Y$, $Z$ be topological space  and  $X \times Y$ be product space.

Let projective map $P_{1} : X \times Y \to X$  and  $P_{2} : X \times Y \to Y$ by

$P_{1}(x, y) = x$,     $P_{2}(x, y) = y$

T.F.A.E.

$(1)$  $f : Z \to X \times Y$ is continuous

$(2)$  $P_{1} \circ f : Z \to X$  and  $P_{2} \circ f : Z \to X$ are continuous

 

pf)

$(1) \Rightarrow (2)$

  It is clear by Prop 1 and Thm 1.

 

$(2) \Leftarrow (1)$

  Let $\mathscr{B}$ be base for product topology on $X \times Y$

  $\forall B \in \mathscr{B}$

  So there exists $U$, $V$  s.t.

  $U$, $V$ are open set in $X$, $Y$, respectively

  &     $B = U \times V$

 

  By definition of $P_{1}$  and  $P_{2}$,

  $P_{1}^{-1}(U) = U \times Y$,     $P_{2}^{-1}(V) = X \times V$

 

  Therefore

  $P_{1}^{-1}(U) \cap P_{2}^{-1}(V) = (U \times Y) \cap (X \times V) = (U \cap X) \times (Y \cap V) = U \times V$

 

 

  Since $U \times V = P_{1}^{-1}(U) \cap P_{2}^{-1}(V)$,

  $f^{-1}(B) = f^{-1}(U \times V)$

                   $= f^{-1}(P_{1}^{-1}(U) \cap P_{2}^{-1}(V))$

                   $= f^{-1}(P_{1}^{-1}(U)) \cap f^{-1}(P_{2}^{-1}(V))$

                   $= (P_{1} \circ f)^{-1}(U) \cap (P_{2} \circ f)^{-1}(V)$

  By assumption, 

  $(P_{1} \circ f)^{-1}(U)$  and  $(P_{2} \circ f)^{-1}(V)$ are open set in $Z$

 

  $\therefore$  $f^{-1}(B)$ is open set in $Z$

 

  By Thm 2 - 2,

  $f$ is continuous

 

 

 

Def 3

Let $X$ be a set. We define

$\Delta (X) = \left\{(x, x) \; | \; x \in X \right\}$

 

 

 

Lemma 1

Let $X$ be a set  and  $x, y \in X$. Then

$x \neq y$  $\Leftrightarrow$  $(x, y) \in (X \times X) \setminus \Delta (X)$

 

pf)

$\Rightarrow)$

  Since $x, y \in X$,

  $(x, y) \in X \times X$

 

  Since $x \neq y$,

  $(x, y) \notin \Delta (X)$

 

  $\therefore$  $(x, y) \in (X \times X) \setminus \Delta (X)$

 

$\Leftarrow)$

  Since $(x, y) \in (X \times X) \setminus \Delta (X)$,

  $(x, y) \in X \times X$  and  $(x, y) \notin \Delta (X)$

 

  $\therefore$  $x \neq y$

 

 

 

Lemma 2

Let $X$, $U$, $V$ be sets  and  $U, V \subseteq X$. Then

$U \cap V = \varnothing$  $\Leftrightarrow$  $U \times V \subseteq (X \times X) \setminus \Delta (X)$

 

pf)

$\Rightarrow)$

  $\forall (x, y) \in U \times V$

 

  Since $U, V \subseteq X$,

  $x, y \in X$

 

  Since $U \cap V = \varnothing$

  $x \neq y$

 

  By Lemma 1,

  $(x, y) \in X \times X \setminus \Delta (X)$

 

$\Leftarrow)$

  Suppose : $U \cap V \neq \varnothing$

  So there exists $a$  s.t.

  $a \in U \cap V$

  $\therefore$  $(a, a) \in U \times V$  and 

 

  By assumption,

  $(a \times a) \in (X \times X) \setminus \Delta (X)$

  $\therefore$  $(a, a) \notin \Delta(X)$

  $\therefore$  $a \neq a$

 

  Contradiction

 

 

 

Thm 4

Let $X$ be a topological space.

T.F.A.E.

$(1)$  $X$ is Hausdorff

$(2)$  $\Delta(X)$ is closed set in $X \times X$

 

pf)

$(1) \Rightarrow (2)$

  $\forall (x, y) \in (X \times X) \setminus \Delta (X)$

  By Lemma 1,

  $x \neq y$

 

  Since $X$ is Hausdorff, there exists open set $U$, $V$ in $X$  s.t.

  $x \in U$,     $y \in V$,     $U \cap V = \varnothing$

 

  By Lemma 2,

  $U \times V \subseteq (X \times X) \setminus \Delta (X) $

  $\therefore$  $(x, y) \in U \times V \subseteq (X \times X) \setminus \Delta (X)$

 

  Since $U \times V$ is open set in $X \times X$, by Prop 2

  $(X \times X) \setminus \Delta (X)$ is open in $X \times X$

  $\therefore$  $\Delta(X)$ is closed set in $X \times X$

 

$(2) \Leftarrow (1)$

  $\forall x, y \; (x \neq y) \in X$

  By Lemma 1,

  $(x, y) \in (X \times X) \setminus \Delta(X)$

 

  Since $X \times X \setminus \Delta(X)$ is open set, there exists open set $U$, $V$ in $X$  s.t.

  $(x, y) \in U \times V \subseteq (X \times X) \setminus \Delta (X)$

  $\therefore$  $x \in U$,  $y \in V$

 

  By Lemma 2,

  $U \cap V = \varnothing$

 

  $\therefore$  $X$ is Hausdorff

 

 

 

Thm 5

Let $(X, d)$ be a metric space. Then

$d : X \times X \to \mathbb{R}$ is continuous

( Consider $X \times X$ as product space  and  $\mathbb{R}$ as usual space )

 

  Let $\mathscr{T}$, $\mathscr{T}'$ be topology in $X \times X$, $\mathbb{R}$, respectively.

  $\forall (x_{0}, y_{0}) \in X \times X$

 

  $\forall V \in \mathscr{T}'$ containing $d(x_{0}, y_{0})$

 

  By Thm 2, there exists $r >0$  s.t.

  $B_{r}^{E}(d(x_{0}, y_{0})) \subseteq V$

  ( $B^{E}$ is open ball by Euclidean metric )

 

  Let $W = B_{\frac{r}{2}}(x_{0}) \times B_{\frac{r}{2}}(y_{0})$

  ( $B$ is open ball by $d$ )

  $\therefore$  $(x_{0}, y_{0}) \in W$  and  $W \in \mathscr{T}$

 

 

  Claim : $d(W) \subseteq B_{r}^{E}(d(x_{0}, y_{0}))$

  $\forall w \in d(W)$

  So there exists $x, y \in W$  s.t.

  $w = d(x, y)$

  $\therefore$  $d(x_{0}, x) < \dfrac{r}{2}$  and  $d(y_{0}, y) < \dfrac{r}{2}$

 

  Since $d$ is metric,

  $d(x, y) \leq d(x, x_{0}) + d(x_{0}, y_{0}) + d(y_{0}, y) < r + d(x_{0}, y_{0})$

  $d(x_{0}, y_{0}) \leq d(x_{0}, x) + d(x, y) + d(y, y_{0}) < r + d(x, y)$

 

  $\therefore$  $\left|d(x, y) - d(x_{0}, y_{0}) \right| < r$

  $\therefore$  $w = d(x, y) \in B_{r}^{E}(d(x_{0}, y_{0}))$

 

  end

 

  $\therefore$  $(x_{0}, y_{0}) \in d(W) \subseteq V$

  $\therefore$  $d$ is continuous at $(x_{0}, y_{0})$

 

  By Thm 3,

  $d$ is continuous