Subspace

 

Def 1

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

We define subspace topology $\mathscr{T}'$ for $A$ as follow.

$\mathscr{T}' = \left\{U \cap A \; | \; U \in \mathscr{T} \right\}$

 

$(A, \mathscr{T}')$ : subspace of $X$

 

 

 

Thm 1

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

Let $\mathscr{T}'$ be a subspace topology for $A$. Then

$(A, \mathscr{T}')$ is topological space

 

  Show 1 : $\varnothing, A \in \mathscr{T}'$

  $\varnothing = \varnothing \cap A \in \mathscr{T}'$

  $A = X \cap A \in \mathscr{T}'$

 

  end

 

 

  Show 2 : $V_{\alpha} \in \mathscr{T}'$  for $\alpha \in I$  $\Rightarrow$  $\bigcup_{\alpha \in I}  V_{\alpha} \in \mathscr{T}'$

  So there exists $U_{\alpha} \in \mathscr{T}$  for $\alpha \in I$  s.t.

  $V_{\alpha} = U_{\alpha} \cap A$

 

  Therefore

  $\displaystyle \bigcup_{\alpha \in I} V_{\alpha} = \bigcup_{\alpha \in I}  (U_{\alpha} \cap A) = (\bigcup_{\alpha \in I}  U_{\alpha}) \cap A \in \mathscr{T}$

 

  end

 

 

  Show 3 : $V_{1}, \cdots, V_{n} \in \mathscr{T}'$  $\Rightarrow$  $\displaystyle \bigcap_{i=1}^{n} V_{i} \in \mathscr{T}'$

  So there exists $U_{i} \in \mathscr{T}$  for $i=1, \cdots, n$  s.t.

  $V_{i} = U_{i} \cap A$

  Therefore

  $\displaystyle \bigcap_{i=1}^{n}  V_{i} = \bigcap_{i=1}^{n}  (U_{i} \cap A) = (\bigcap_{i=1}^{n}  U_{i}) \cap A \in \mathscr{T}$

 

  end

 

 

 

Thm 2

Let $(X, \mathscr{T})$ be a topological space  and  $(Y, \mathscr{T}')$ be subspace of $X$. Then

1.  $\mathscr{B}$ : base on $X$  $\Rightarrow$  $\mathscr{B}' = \left\{B \cap Y \; | \; B \in \mathscr{B} \right\}$ : base on $Y$

2.  $V$ : open in $Y$  and  $Y$ : open in $X$  $\Rightarrow$  $V$ : open in $X$

3.  $A \subseteq Y$, $A$ : closed in $Y$  $\Leftrightarrow$  $\exists C$ : closed in $X$  s.t.  $A = C \cap Y$

 

pf)

  Show 1 : $\mathscr{B}' \subseteq \mathscr{T}'$

  Since $\mathscr{B} \subseteq \mathscr{T}$,

  $\mathscr{B}' = \left\{B \cap Y \; | \; B \in \mathscr{B} \right\} \subseteq \left\{U \cap Y \; | \; U \in \mathscr{T} \right\} = \mathscr{T}'$

 

  end

 

 

  Show 2 : Every open set of $Y$ is a union of sets in $\mathscr{B}'$

  $\forall V \in \mathscr{T}'$

 

  So there exists $U \in \mathscr{T}$  s.t.

  $V = U \cap Y$

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

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

 

  Let $B'_{\alpha} = B_{\alpha} \cap Y \in \mathscr{B}'$.

  Therefore

  $\displaystyle \bigcup_{\alpha \in I} B'_{\alpha} = \bigcup_{\alpha \in I} (B_{\alpha} \cap Y) = (\bigcup_{\alpha \in I} B_{\alpha}) \cap Y = U \cap Y = V$

 

  end

 

  Since $V \in \mathscr{T}'$, there exists $U \in \mathscr{T}$  s.t.

  $V = U \cap Y$

  Since $U \in \mathscr{T}$ and $Y \in \mathscr{T}$,

  $V \in \mathscr{T}$

 

$\Rightarrow)$

  Since $Y - A \in \mathscr{T}'$, there exists $U \in \mathscr{T}$  s.t.

  $Y - A = U \cap Y$

 

  Let $C = X - U$.

  So $C$ is closed in $X$.

  Also

  $C \cap Y = (X - U) \cap Y = (X \cap Y) - (U \cap Y) = Y - U \cap Y = A$

 

$\Leftarrow)$

  Since $A = C \cap Y$,

  $A \subseteq Y$

 

  Also

  $Y - A = Y - C \cap Y = (X \cap Y) - (C \cap Y)  = (X - C) \cap Y$

 

  Since $X - C$ is open set in $X$,

  $Y - A$ is open set in $Y$

 

  $\therefore$  $A$ is closed set in $Y$

 

 

 

Prop 1

Let $(X, \mathscr{T})$ be a topological space  and  $(A, \mathscr{T}')$ be subspace of $X$.

Suppose $B \subseteq A$. Then

$B$ is subspace of $A$  $\Leftrightarrow$  $B$ is subspace of $X$

 

  It is sufficient to show

  $\left\{U' \cap B \; | \; U' \in \mathscr{T}' \right\} = \left\{U \cap B \; | \; U \in \mathscr{T} \right\}$

 

 

  Show 1 : $\left\{U' \cap B \; | \; U' \in \mathscr{T}' \right\} \subseteq \left\{U \cap B \; | \; U \in \mathscr{T} \right\}$

  $\forall U' \cap B \in \left\{U' \cap B \; | \; U' \in \mathscr{T}' \right\}$

 

  Since $U' \in \mathscr{T}'$, there exists $U \in \mathscr{T}$  s.t.

  $U' = U \cap A$

 

  Since $B \subseteq A$,

  $U' \cap B = U \cap A \cap B = U \cap B$

 

  $\therefore$  $U' \cap B \in \left\{U \cap B \; | \; U \in \mathscr{T}' \right\}$

  $\therefore$  $\left\{U' \cap B \; | \; U' \in \mathscr{T}' \right\} \subseteq \left\{U \cap B \; | \; U \in \mathscr{T} \right\}$

 

  end

 

 

  Show 2 : $\left\{U \cap B \; | \; U \in \mathscr{T} \right\} \subseteq \left\{U' \cap B \; | \; U' \in \mathscr{T}' \right\}$

  $\forall U \cap B \in \mathscr{T}''$

 

  Since $B \subseteq A$,

  $U \cap B = U \cap (A \cap B) = (U \cap A) \cap B$

  Since $U \cap A \in \mathscr{T}'$

  $U \cap B \in \left\{U' \cap B \; | \; U' \in \mathscr{T}' \right\}$

 

  $\therefore$  $\left\{U \cap B \; | \; U \in \mathscr{T} \right\} \subseteq \left\{U' \cap B \; | \; U' \in \mathscr{T}' \right\}$

 

  end

 

 

Def 2

Let $A \subseteq B$

Inclusion map $i : A \to B$ is a map which is defined by

$i(x) = x$

 

 

 

Prop 2

Let $X$ be a topological space  and  $Y$ be subspace of $X$. Then

inclusion map $i: Y \to X$ is continuous

 

  Let $U$ be open set in $X$

 

  Therefore

  $i^{-1}(U) = U \cap Y$ is open in $Y$

 

 

 

Thm 3

Let $(X, \mathscr{T})$ be a topological space  and  $(Y, \mathscr{T}')$ be subspace of $X$. Then

inclusion map $i : (Y, \mathscr{T}_{2}) \to (X, \mathscr{T})$ is continuous  $\Rightarrow$  $\mathscr{T}' \subseteq \mathscr{T}_{2}$

 

# The subspace topology is the smallest topology that makes inclusion map continuous.

 

  $\forall V \in \mathscr{T}'$

  So there exists $U \in \mathscr{T}$  s.t.

  $V = U \cap Y$

 

  Since $i$ is continuous,

  $i^{-1}(U) = U \cap Y = V \in \mathscr{T}_{2}$

 

  $\therefore$  $\mathscr{T}' \subseteq \mathscr{T}_{2}$