交、并、补集的像、原像

hbghlyj

Wilson A. Sutherland - Introduction to Metric and Topological Spaces-Oxford University Press (2009)
Proposition 3.7 映射$f:X\to Y$, 对每个$i\in I$指定$X$的一个子集$A_i$和$Y$的一个子集$C_i$. 那么,
\begin{align*}\mathmakebox[16em][c]{f\left(\bigcup_{i \in I} A_{i}\right)=\bigcup_{i \in I} f\left(A_{i}\right)}&\mathmakebox[16em][c]{f\left(\bigcap_{i \in I} A_{i}\right) \subseteq \bigcap_{i \in I} f\left(A_{i}\right)} \\ \mathmakebox[16em][c]{f^{-1}\left(\bigcup_{i \in I} C_{i}\right)=\bigcup_{i \in I} f^{-1}\left(C_{i}\right)}&\mathmakebox[16em][c]{f^{-1}\left(\bigcap_{i \in I} C_{i}\right)=\bigcap_{i \in I} f^{-1}\left(C_{i}\right)}\end{align*}
证明$\def\ds{\displaystyle}$ (S.3)

1) $\ds y \in f\left(\bigcup_{i \in I} A_i\right)$ $⇔y=f(x)$对某个$\ds x \in \bigcup_{i \in I} A_i$ $⇔y=f(x) \in f\left(A_{i_0}\right)$对某个$i_0 \in I$ $\ds⇔y \in \bigcup_{i \in I} f\left(A_i\right)$	2) $\ds y \in f\left(\bigcap_{i \in I} A_i\right)$ $⇒y=f(x)$对某个$\ds x \in \bigcap_{i \in I} A_i$ $⇒x \in A_i\;\forall i \in I$ $⇒y \in f\left(A_i\right)\;\forall i \in I$
3) $\ds x \in f^{-1}\left(\bigcup_{i \in I} C_i\right)$ $\ds⇔f(x) \in \bigcup_{i \in I} C_i $ $⇔f(x) \in C_{i_0}$对某个$i_0 \in I$ $⇔x \in f^{-1}\left(C_{i_0}\right)$对某个$i_o \in I$ $⇔\ds x \in \bigcup_{i \in I} f^{-1}\left(C_i\right)$	4) $\ds x \in f^{-1}\left(\bigcap_{i \in I} C_{i}\right)$ $⇔\ds f(x) \in \bigcap_{i \in I} C_{i}$ $⇔f(x) \in C_{i}\;∀i∈I$ $⇔x \in f^{-1}\left(C_{i}\right)\;∀i∈I$ $⇔\ds x \in \bigcap_{i \in I} f^{-1}\left(C_{i}\right)$ $⇔\ds f^{-1}\left(\bigcap_{i \in I} C_{i}\right)\subseteq\bigcap_{i \in I} f^{-1}\left(C_{i}\right)$

关于2的真包含的例子:
令$f:x\mapsto x^2,A_1=\{-1,0\},A_2=\{1,0\}$.
$f(A_1\cap A_2)=f(\{0\})=\{0\}$
$f(A_1)\cap f(A_2)=\{0,1\}\cap\{0,1\}=\{0,1\}$

hbghlyj

Proposition 3.8 映射$f:X\to Y$, 设$B\subset X$, $D\subset Y$, 则
$$f(X \backslash B) \supseteq f(X) \backslash f(B), \quad f^{-1}(Y \backslash D)=X \backslash f^{-1}(D)$$This follows by taking $A =X,C=Y$ in the next proposition (for the second part of Proposition 3.8 we use also $f^{-1}(Y)=X$).
Proposition 3.9 映射$f:X\to Y$, 设$A,B\subset X$, $C,D\subset Y$, 则
$$f(A \backslash B) \supseteq f(A) \backslash f(B),\quad f^{-1}(C \backslash D)=f^{-1}(C) \backslash f^{-1}(D)$$
1)
Let $y \in f(A) \backslash f(B)$. Then $y=f(a)$ for some $a \in A$, but $y \neq f(b)$ for any $b \in B$. Hence we must have $a \in A \backslash B$, so $y \in f(A \backslash B)$ as required.
2)
Suppose that $x \in f^{-1}(C \backslash D)$. Then $f(x) \in C \backslash D$. So $f(x) \in C$ but $f(x) \notin D$. Hence $x \in f^{-1}(C)$ but $x \notin f^{-1}(D)$, so $x \in f^{-1}(C) \backslash f^{-1}(D)$. This proves $f^{-1}(C \backslash D) \subseteq f^{-1}(C) \backslash f^{-1}(D)$.
Suppose $x \in f^{-1}(C) \backslash f^{-1}(D)$. Then $x \in f^{-1}(C)$ but $x \notin f^{-1}(D)$. So $f(x) \in C$ but $f(x) \notin D$. Hence $f(x) \in C \backslash D$, so $x \in f^{-1}(C \backslash D)$. This proves $f^{-1}(C) \backslash f^{-1}(D) \subseteq f^{-1}(C \backslash D)$.
These two together give $f^{-1}(C) \backslash f^{-1}(D)=f^{-1}(C \backslash D)$ as required.

hbghlyj

根据1#证明的“并集的原像等于原像的并集”,
$\cal B$是拓扑$Y$的基, 为了检查$f:X\to Y$的连续性, 只需检查每个$B∈\cal B$的原像在$X$中是开集.

“紧集的连续像是紧集”也是根据1#证明的.

hbghlyj

See this page

hbghlyj

2018 toplectnotes17 (1).pdf 末页
Let $X$ and $Y$ be sets, $\left\{U_i\right\}_{i \in I}$ a set of subsets of $X$ and $\left\{V_j\right\}_{j \in J}$ a set of subsets of $Y$.
(1) De Morgan laws
\begin{aligned}
& X ∖ \bigcap_{i \in I} U_i=\bigcup_{i \in I}\left(X ∖ U_i\right) \\
& X ∖ \bigcup_{i \in I} U_i=\bigcap_{i \in I}\left(X ∖ U_i\right)
\end{aligned}(2) Distributivity of $\bigcap$ over $\bigcup$
$$
A \cap\left(\bigcup_{i \in I} U_i\right)=\bigcup_{i \in I}\left(A \cap U_i\right)
$$
(3) Images and inverse images
Let $f: X \rightarrow Y$ be a map. Recall that for any subset $A$ in $Y$ its inverse image or pre-image is
$$
f^{-1}(A)=\{x \in X ; f(x) \in A\} .
$$
Then
\begin{align*}
f(U) \subseteq V &\Leftrightarrow U \subseteq f^{-1}(V) \\
f^{-1}(Y ∖ V)&=X ∖ f^{-1}(V)\\
f\left(\bigcup_{i \in I} U_{i}\right) & =\bigcup_{i \in I} f\left(U_{i}\right) \\ f\left(\bigcap_{i \in I} U_{i}\right) & \subseteq \bigcap_{i \in I} f\left(U_{i}\right) \\ f^{-1}\left(\bigcup_{j \in J} V_{j}\right) & =\bigcup_{j \in J} f^{-1}\left(V_{j}\right) \\ f^{-1}\left(\bigcap_{j \in J} V_{j}\right) & =\bigcap_{j \in J} f^{-1}\left(V_{j}\right)
\end{align*}If $A \cap B=\emptyset$, $A, B$ subsets of $Y$, then $f^{-1}(A) \cap f^{-1}(B)=\emptyset$. Indeed assume that there exists $x \in f^{-1}(A) \cap f^{-1}(B)$. Then $f(x) \in A$ and $f(x) \in B$, which contradicts $A \cap B=\emptyset$.

hbghlyj

hbghlyj 发表于 2022-10-17 16:44
Proposition 3.7

从4)推出2): 令$C_i=f(A_i),$
\[f^{-1}\left(\bigcap_{i \in I}f(A_i)\right)=\bigcap_{i \in I} f^{-1}\left(f(A_i)\right)\]对两边用$f$
\[\bigcap_{i \in I}f(A_i)=f\left(\bigcap_{i \in I} f^{-1}\left(f(A_i)\right)\right)\]
使用$f^{-1}(f(S))\supset S$
\[\bigcap_{i \in I}f(A_i)\supset f\left(\bigcap_{i \in I} A_i\right)\]

hbghlyj

若$f$为单射，则2的等号成立。
因为对任何$f$，$f^{-1}$一定为单射，所以2能推出4成立。