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陶哲轩实分析3.4及习题-Analysis I 3.4

2019年4月3日2019年12月13日christangdt20条评论

逆像inverse image是一个之前学得很模糊的概念，这一节阐述的相对清楚了，并且也说了逆像可以对任意一个函数存在。此外这一节还涉及幂集，即子集的集合的概念。集合论很多概念确实需要很严格的区分，非常容易混淆。

Supplement from text: Define $f(S)$ using the axiom of specification (Axiom 3.5).
Solution: For a function $f:X\to Y,f(S):={f(x):x\in S}$
Use Axiom 3.6, let $P(x,y)$ be the property which states that $y=f(x)$ for any $x\in S$ , then for each $x\in S$ there’s at most one $y$ for which $P(x,y)$ is true. So there exists a set ${y:P(x,y) \text{ is true for }x\in A}$ , which is ${y:y=f(x),x\in S}$ .
Use Axiom 3.5, let $P(x)$ be the property: $\exists z\in X$ ,s.t. $f(z)=x$ , then for each $x\in Y$ , we have the set ${x\in A:P(x) \text{ is true}}$ , which is the desired set.

Exercises 3.4.1. Let $f:X\to Y$ be a bijective function, and let $f^{-1}:Y\to X$ be its inverse. Let $V$ be any subset of $Y$ . Prove that the forward image of $V$ under $f^{-1}$ is the same set as the inverse image of $V$ under $f$ ; thus the fact that both sets are denoted by $f^{-1}(V)$ will not lead to any inconsistency.

Solution: Let the forward image of $V$ under $f^{-1}$ to be $A={f^{-1}(v):v\in V}$ , and the inverse image of $V$ under $f$ to be $B={x\in X:f(x)\in V}$ . Then we have

$x\in B\iff f(x)=v\in V\iff x=f^{-1}(v)\iff x\in A$

$\blacksquare$

Exercises 3.4.2. Let $f:X\to Y$ be a function from one set $X$ to another set $Y$ , let $S$ be a subset of $X$ , and let $U$ be a subset of $Y$ . What, in general, can one say about $f^{-1}(f(S))$ and $S$ ? What about $f(f^{-1}(U))$ and $U$ ?

Solution: In general we can say that $f^{-1}(f(S))\supseteq S$ and $f(f^{-1}(U))\subseteq U$ , $f$ can at most be equal but usually shrinks, while $f^{-1}$ can at least be equal and usually enlarges.
To prove, notice that if $x\in S$ , then $f(x)\in f(S)$ , which means $x\in f^{-1}(f(S))$ , on the other hand, if $y\in f(f^{-1}(U))$ , then $\exists x\in f^{-1}(U),f(x)=y$ , but $x\in f^{-1}(U)\iff f(x)=y\in U$ , thus the conclusion holds.

$\blacksquare$

Exercises 3.4.3. Let $A,B$ be two subsets of a set $X$ , and let $f:X\to Y$ be a function. Show that $f(A\cap B)\subseteq f(A)\cap f(B)$ , that $f(A)\backslash f(B)\subseteq f(A\backslash B),f(A\cup B)=f(A)\cup f(B)$ . For the first two statements, is it true that the $\subseteq$ relation can be improved to $=$ ?

Solution:
Statement I: we have

$\begin{aligned}y\in f(A\cap B)&\implies (\exists x\in A\cap B,y=f(x))\\&\implies(x\in A,y=f(x))\wedge (x\in B,y=f(x))\\&\implies (y\in f(A))\wedge(y\in f(B))\\&\implies y\in f(A)\cap f(B)\end{aligned}$

Statement II: we have

$\begin{aligned}y\in f(A)\backslash f(B)&\implies (\exists x\in A,f(x)=y)\wedge (y\notin f(B))\\&\implies(\exists x\in A,f(x)=y)\wedge (x\notin B)\\&\implies (\exists x\in A,x\notin B,f(x)=y)\\&\implies y\in f(A\backslash B)\end{aligned}$

Statement III: we have

$\begin{aligned}y\in f(A\cup B)&\iff (\exists x\in A\cup B,y=f(x))\\&\iff (x\in A,y=f(x))\vee (x\in B,y=f(x))\\&\iff (y\in f(A))\vee (y\in f(B))\\&\iff y\in f(A)\cup f(B)\end{aligned}$

The first two statements can’t be improved to $=$ , we give counterexamples:
Statement I: $A=\{1,2\},B=\{1,3\},f:\{1,2,3\}\to \{4,5\}:f(1)=4,f(2)=f(3)=5$ , then $f(A\cap B)=f(\{1\})=\{4\},f(A)\cap f(B)=\{4,5\}$ .
Statement II: $A=\{1,2\},B=\{1\},f:\{1,2\}\to 3:f(1)=f(2)=3$ , then $f(A)\backslash f(B)=\emptyset$ , and $f(A\backslash B)=f(\{2\})=\{3\}$ .

$\blacksquare$

Exercises 3.4.4. Let $f:X\to Y$ be a function from one set $X$ to another set $Y$ , and let $U,V$ be subsetes of $Y$ . Show that $f^{-1}(U\cup V)=f^{-1}(U)\cup f^{-1}(V)$ , that $f^{-1}(U\cap V)=f^{-1}(U)\cap f^{-1}(V)$ , and that $f^{-1}(U\backslash V)=f^{-1}(U)\backslash f^{-1}(V)$ .

Solution:
Statement I:

$\begin{aligned}x\in f^{-1}(U\cup V)&\iff f(x)\in U\cup V\\&\iff (f(x)\in U)\vee (f(x)\in V)\\&\iff (x\in f^{-1}(U))\vee (x\in f^{-1}(V))\\&\iff x\in f^{-1}(U)\cup f^{-1}(V)\end{aligned}$

Statement II:

$\begin{aligned}x\in f^{-1}(U\cap V)&\iff f(x)\in U\cap V\\&\iff (f(x)\in U)\wedge (f(x)\in V)\\&\iff (x\in f^{-1}(U))\wedge (x\in f^{-1}(V))\\&\iff x\in f^{-1}(U)\cap f^{-1}(V)\end{aligned}$

Statement III:

$\begin{aligned}x\in f^{-1}(U\backslash V)&\iff (f(x)\in U)\wedge (f(x)\notin V)\\&\iff (x\in f^{-1}(U))\wedge (x\notin f^{-1}(V))\\&\iff x\in f^{-1}(U)\backslash f^{-1}(V)\end{aligned}$

$\blacksquare$

Exercises 3.4.5. Let $f:X\to Y$ be a function from one set $X$ to another set $Y$ . Show that $f(f^{-1}(S))=S$ for every $S\subseteq Y$ if and only if $f$ is surjective. Show that $f^{-1}(f(S))=S$ for every $S\subseteq X$ if and only if $f$ is injective.

Solution: In Exercise 3.4.2, I’ve already showed that

$f(f^{-1}(S))\subseteq S,\quad \forall S\subseteq Y \\ f^{-1}(f(S))\supseteq S,\quad \forall S\subseteq X$

So $f(f^{-1}(S))=S,\forall S\subseteq Y$ is equivalent to $S\subseteq f(f^{-1}(S))$ , or $\forall y\in S,y\in f(f^{-1}(S))$ , this statement means $\forall y\in S,\exists x\in f^{-1}(S)\subseteq X$ ,s.t. $f(x)=y$ , thus $f$ is surjective. And if $f$ is surjective, then given $S\subseteq Y$ , we have $\forall y\in S,\exists x\in X,f(x)=y$ , so that $y\in f(f^{-1}(S))$ .
The second statement:

$\begin{aligned}f^{-1}(f(S))=S,\forall S\subseteq X&\iff f^{-1}(f(S))\subseteq S,\forall S\subseteq X\\&\iff \left(x\in f^{-1}(f(S))\implies x\in S,\forall S\subseteq X\right)\end{aligned}$

Now if $f$ is injective, then if $x\in f^{-1}(f(S))$ , we have $f(x)\in f(S)$ , so suppose $x\notin S$ , we shall have another $x'\in S,f(x')=f(x)$ , a contradiction to injectivity. On the other hand, if we let $S$ be the singleton set $\{a\},a\in X$ , then $f(S)=f(\{a\})=\{f(a)\}$ , thus $x\in f^{-1} \big(f(\{a\})\big)\implies x\in \{a\}$ or $f(x)=f(a)\implies x=a$ . This means $f$ is injective.

$\blacksquare$

Exercises 3.4.6. Prove Lemma 3.4.9.

Solution: We start with the set $\{0,1\}^X$ , which is the set of all functions $f:X\to \{0,1\}$ , for each $f$ , the object $f^{-1}(\{1\})$ forms a subset of $X$ , moreover there’s at most one such set, by the replacement axiom, there exists a set

$A=\big\{Y:Y=f^{-1}(\{1\}),f\in \{0,1\}^X \big\}$

This set is exactly the set we want to prove, since every $Y\in A$ is a subset of $X$ , and for any subset $Y\subseteq X$ , we can define a function $f:X\to {0,1}$ as follows:

$f(x)=\begin{cases}1,\quad x\in Y\\0,\quad x\notin Y\end{cases}$

and $f^{-1}(\{1\})=Y$ .

$\blacksquare$

Exercises 3.4.7. Let $X,Y$ be sets. Define a partial function from $X$ to $Y$ to be any function $f:X'\to Y'$ whose domain $X'$ is a subset of $X$ , and whose range $Y'$ is a subset of $Y$ . Show that the collection of all partial functions from $X$ to $Y$ is itself a set.

Solution: By lemma 3.4.9 we have two sets:

$A=\{X': X'\subseteq X\},\quad B=\{Y':Y'\subseteq Y\}$

each containing all subsets of $X$ and $Y$ .
By the power set axioms, we have, for $X_0\subseteq X,Y_0\subseteq Y$ :

$Y_0^{X_0}=\{\text{functions with domain }X_0 \text{ and range }Y_0\}$

By the replacement axiom, we can have the set

$A_{Y_0}=\{Y_0^{X'}:X'\subseteq X\}$

which is the collection of all sets which contains functions with range $Y_0\subseteq Y$ , and domain to be an arbitrary subset of $X$ .
By the union axiom, we can have the set

$\bigcup_{X'\in A}A_{Y_0}=\{f:f\in Y_0^{X'},X'\in A\}$

By the union axiom again, we can have the set

$\bigcup_{Y'\in B}\bigcup_{X'\in A}A_{Y'}=\{f:f\in Y'^{X'},X'\in A,Y'\in B\}$

$\blacksquare$

Exercises 3.4.8. Show that Axiom 3.4 can be deduced from Axiom 3.1, Axiom 3.3 and Axiom 3.11.

Solution: Given any two sets $A,B$ , By axiom 3.3, we can have a set $C={A,B}$ . $C$ is a set whose elements are sets. By axiom 3.11, we can have a set $\cup C$ , and for any object $x$ :

$x\in \bigcup C \iff (x\in S \text{ for some }S\in C)$

But $S$ can only be $A$ or $B$ , thus

$x\in \bigcup C \iff (x\in A)\vee (x\in B)$

This is the definition of $A\cup B$ .

$\blacksquare$

Exercises 3.4.9. Show that if $\beta$ and $\beta'$ are two elements of a set $I$ , and to each $\alpha \in I$ we assign a set $A_{\alpha}$ , then

$\{x\in A_{\beta}:x\in A_{\alpha} \text{ for all }\alpha \in I \}=\{x\in A_{\beta'}:x\in A_{\alpha} \text{ for all }\alpha \in I \}$

and so the definition of $\bigcap_{\alpha \in I}A_{\alpha}$ defined in (3.3) does not depend on $\beta$ . Also explain why (3.4) is true.

Solution: Let $M=\{x\in A_{\beta}:x\in A_{\alpha} \text{ for all }\alpha\in I\},N=\{x\in A_{\beta'}:x\in A_{\alpha} \text{ for all }\alpha\in I\}$ .
Suppose $x\in M$ , then $x\in A_{\beta}$ and $x\in A_{\alpha}$ for all $\alpha\in I$ , since $\beta'\in I$ , we know $x\in A_{\beta'}$ , thus $x\in N$ . By the same logic we have $x\in N\implies x\in M$ .
To see why (3.4) is true, by definition:

$y\in \bigcap_{\alpha\in I}A_{\alpha} \iff (y\in A_{\beta})\wedge (y\in A_{\alpha} \text{ for all } \alpha\in I)$

Since $(\beta\in I)\wedge (y\in A_{\alpha}, \forall \alpha\in I)\implies (y\in A_{\beta})$ , we have

$(y\in A_{\beta})\wedge (y\in A_{\alpha} \text{ for all }\alpha\in I)\iff y\in A_{\alpha}, \forall \alpha\in I$

$\blacksquare$

Exercises 3.4.10. Suppose that $I$ and $J$ are two sets, and for all $\alpha \in I\cup J$ let $A_{\alpha}$ be a set. Show that $\left(\bigcup_{\alpha \in I}A_{\alpha} \right)\cup \left( \bigcup_{\alpha \in J}A_{\alpha} \right)=\bigcup_{\alpha \in I\cup J}A_{\alpha}$ . If $I$ and $J$ are non-empty, show that $\left(\bigcap_{\alpha \in I}A_{\alpha} \right)\cap \left( \bigcap_{\alpha \in J}A_{\alpha} \right)=\bigcap_{\alpha \in I\cup J}A_{\alpha}$ .

Solution: We have

$\begin{aligned} x\in (\cup_{\alpha\in I}A_{\alpha})\cup(\cup_{\alpha\in J}A_{\alpha})&\iff (x\in \cup_{\alpha\in I}A_{\alpha})\vee (x\in \cup_{\alpha\in J}A_{\alpha})\\&\iff (x\in A_m,m\in I)\vee (x\in A_n,n\in J)\\&\iff x\in A_m,m\in I\cup J \\&\iff x\in \bigcup_{\alpha\in I\cup J}A_{\alpha}\end{aligned}$

And if $I$ and $J$ are nonempty:

$\begin{aligned}x\in (\cap_{\alpha\in I}A_{\alpha})\cap (\cap_{\alpha\in J}A_{\alpha})&\iff (x\in \cap_{\alpha\in I}A_{\alpha})\wedge (x\in \cap_{\alpha\in J}A_{\alpha})\\&\iff (x\in A_m,\forall m\in I)\wedge (x\in A_n,\forall n\in J)\\&\iff x\in A_m,\forall m\in I\cup J \\&\iff x\in \bigcap_{\alpha\in I\cup J}A_{\alpha}\end{aligned}$

$\blacksquare$

Exercises 3.4.11. Let $X$ be a set, let $I$ be a non-empty set, and for all $\alpha \in I$ let $A_{\alpha}$ be a subset of $X$ . Show that

$X\backslash \bigcup_{\alpha \in I}A_{\alpha}=\bigcap_{\alpha \in I}(X\backslash A_{\alpha})$

and

$X\backslash \bigcap_{\alpha \in I}A_{\alpha}=\bigcup_{\alpha \in I}(X\backslash A_{\alpha})$

This should be compared with de Morgan’s laws in Proposition 3.1.28 (although one cannot derive the above identities directly from de Morgan’s laws, as $I$ could be infinite).

Solution: $I$ is not empty, thus we have $r\in I$ , and $A_r\subseteq X$ , thus

$\begin{aligned}x\in X\backslash \bigcup_{\alpha\in I}A_{\alpha}&\iff (x\in X)\wedge (x\notin \cup_{\alpha\in I}A_{\alpha}) \\&\iff (x\in X)\wedge (x\notin A_{\alpha},\forall \alpha\in I) \\&\iff x\in X\backslash A_{\alpha},\forall \alpha\in I \\&\iff x\in \{x\in X\backslash A_r:x\in X\backslash A_{\alpha},\forall \alpha\in I\}\\&\iff x\in \bigcap_{\alpha\in I}(X\backslash A_{\alpha}) \end{aligned}$

And:

$\begin{aligned}x\in X\backslash \bigcap_{\alpha\in I}A_{\alpha}&\iff (x\in X)\wedge (x\notin \cap_{\alpha\in I}A_{\alpha})\\&\iff (x\in X)\wedge (x\notin \{x\in A_r:x\in A_{\alpha},\forall \alpha\in I\})\\&\iff (x\in X)\wedge \exists \beta\in I,x\notin A_{\beta}\\&\iff (\exists \beta\in I,x\in X\backslash A_{\beta})\\&\iff x\in \bigcup_{\alpha\in I}(X\backslash A_{\alpha})\end{aligned}$

$\blacksquare$

陶哲轩实分析3.3及习题-Analysis I 3.3

2019年4月1日2019年12月13日christangdt8条评论

函数是很经典的概念。这一节用有条件限制集合的方式定义函数非常新颖，读了这一章对函数的相等、复合函数，比较经典的injective\surjective\bijective概念都能有很清晰的认识。

Exercise 3.3.1. Show that the definition of equality in Definition 3.3.7 is reflexive, symmetric and transitive. Also verify the substitution property: if $f,\widetilde{f}:X\to Y$ and $g,\widetilde{g}:Y\to Z$ are functions such that $f=\widetilde{f}$ and $g=\widetilde{g}$ , then $g\circ f=\widetilde{g}\circ \widetilde{f}$ .

Solution:
Reflexive: $\forall x\in X,f(x)=f(x)$
Symmetric:
$\begin{aligned}(f=g)&\implies \big(\forall x\in ,f(x)=g(x)\big)\\&\implies \big(\forall x\in X,g(x)=f(x)\big)\\&\implies g=f \end{aligned}$
Transitive:
$\begin{aligned}(f=g)\wedge (g=h)&\implies \forall x\in X,\big(f(x)=g(x))\wedge (g(x)=h(x)\big)\\&\implies \forall x\in X,f(x)=h(x)\\&\implies f=h\end{aligned}$
Substitution property: $\forall x\in X$ , we have
$f(x)=\widetilde f (x)\in Y \implies g(f(x))=\widetilde g\big (\widetilde f (x)\big) \implies (g\circ f)(x)=\left(\widetilde g\circ \widetilde f\right)(x)$

$\blacksquare$

Exercise 3.3.2. Let $f:X\to Y$ and $g:Y\to Z$ be functions. Show that if $f$ and $g$ are both injective, then so is $g\circ f$ ; similarly, show that if $f$ and $g$ are both surjective, then so is $g\circ f$ .

Solution:
Case I: suppose $x,x'\in X, (g\circ f)(x)=(g\circ f)(x')$ , then $g\left(f(x)\right)=g\left(f(x')\right)$ , which means $f(x)=f(x')$ , which means $x=x'$ , so $g\circ f$ is injective.
Case II: let $z\in Z$ , then as $g$ is surjective, we have $y\in Y,g(y)=z$ . For this $y$ , as $f$ is surjective, we have $x\in X,f(x)=y$ . Together we have $g\left(f(x)\right)=(g\circ f)(x)=z$ .

$\blacksquare$

Exercise 3.3.3. When is the empty function injective? surjective? bijective?

Solution:Empty function $f:\emptyset \to X$ is always injective, it’s surjective and bijective if $X=\emptyset$ .

$\blacksquare$

Exercise 3.3.4. In this section we give some cancellation laws for composition. Let $f:X\to Y,\widetilde{f}:X\to Y,g:Y\to Z, \widetilde{g}:Y\to Z$ be functions. Show that if $g\circ f=g\circ \widetilde{f}$ and $g$ is injective, then $f=\widetilde{f}$ . Is the same statement true if $g$ is not injective? Show that if $g\circ f=\widetilde{g}\circ f$ and $f$ is surjective, then $g=\widetilde{g}$ . Is the same statement true if $f$ is not surjective?

Solution: First, for $x\in X$ , we have $g(f(x))=g\left(\widetilde f(x)\right)$ , and since $g$ is injective, $f(x)=\widetilde f(x)$ .
If $g$ is not injective, the statement does not need to be true.
Next, $\forall y\in Y$ , as $f$ is surjective, $\exists x\in X,f(x)=y$ , for this $x$ we have $g(f(x))=\widetilde g(f(x))$ , thus $g(y)=\widetilde g(y)$ .
If $f$ is not surjective, this statement does not need to be true.

$\blacksquare$

Exercise 3.3.5. Let $f:X\to Y$ and $g:Y\to Z$ be functions. Show that if $g\circ f$ is injective, then $f$ must be injective. Is it true that $g$ must also be injective? Show that if $g\circ f$ is surjective, then $g$ must be surjective. Is it true that $f$ must also be surjective?

Solution: $\forall x,x'\in X$ , suppose $f(x)=f(x')$ , then $g(f(x))=g(f(x'))$ , as $g\circ f$ is injective, we have $x=x'$ , this shows that $f$ is injective. $g$ doesn’t have to be injective.
Next, if $g\circ f$ is surjective, let $\forall z\in Z$ , then $\exists x\in X, g(f(x))=z$ , we know that $y=f(x)\in Y$ is the element that makes $g$ surjective. $f$ doesn’t have to be surjective.

$\blacksquare$

Exercise 3.3.6. Let $f:X\to Y$ be a bijective function, and let $f^{-1}:Y\to X$ be its inverse. Verify the cancellation laws $f^{-1}(f(x))=x$ for all $x\in X$ and $f(f^{-1}(y))=y$ for all $y\in Y$ . Conclude that $f^-1$ is also invertible, and has $f$ as its inverse (thus $(f^{-1})^{-1}=f$ ).

Solution: First show that $f^{-1} (f(x))=x,\forall x\in X$ . Since $f$ is bijective, there’s $y\in Y,f(x)=y$ . Also by definition of $f^{-1}$ , we have $f^{-1} (y)=x$ , this is what we want to prove.
Next show that $f(f^{-1} (y))=y,\forall y\in Y$ . Since $f$ is bijective, there’s exactly one $x\in X,f(x)=y$ , this value of $x$ is denoted $f^{-1} (y)$ , so $f(x)=f\left(f^{-1} (y)\right)=y$ .

$\blacksquare$

Exercise 3.3.7. Let $f:X\to Y$ and $g:Y\to Z$ be functions. Show that if $f$ and $g$ are bijective, then so is $g\circ f$ , and we have $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$ .

Solution: Use Exercise 3.3.2, we can easily show that $g\circ f$ is bijective. Now let $\forall z\in Z$ .
As $g\circ f$ is bijective, there’s a $x\in X,(g\circ f)(x)=z$ , so $(g\circ f)^{-1} (z)=x$ , and we also have $(g\circ f)(x)=g(f(x))=z$ . Because $g$ is bijective, there’s exactly one $y\in Y,g(y)=z$ , thus we can get $g^{-1} (z)=y=f(x)$ . Then as $f$ is bijective, we have
$(f^{-1}\circ g^{-1})(z)=f^{-1} \left(g^{-1} (z)\right)=f^{-1} (y)=f^{-1} \big(f(x)\big)=x=(g\circ f)^{-1} (z)$
Since $z$ is arbitrary, we know $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$ .

$\blacksquare$

Exercise 3.3.8. If $X$ is a subset of $Y$ , let $\iota_{X\to Y}: X\to Y$ be the inclusion map from $X$ to $Y$ , defined by mapping $x\mapsto x$ for all $x\in X$ , i.e., $\iota_{x\to Y}(x):=x$ for all $x\in X$ . The map $\iota_{X\to X}$ is in particular called the identity map on $X$ .
( a ) Show that if $X\subseteq Y\subseteq Z$ then $\iota_{Y\to Z}\circ \iota_{X\to Y}=\iota_{X\to Z}$ .
( b ) Show that if $f:A\to B$ is any function, then $f=f\circ \iota_{A\to A}=\iota_{B\to B}\circ f$ .
( c ) Show that, if $f:A\to B$ is a bijective function, then $f\circ f^{-1}=\iota_{B\to B}$ and $f^{-1}\circ f=\iota_{A\to A}$ .
( d ) Show that if $X$ and $Y$ are disjoint sets, and $f:X\to Z$ and $g:Y\to Z$ are functions, then there is a unique function $h:X\cup Y\to Z$ such that $h\circ \iota_{X\to X\cup Y}=f$ and $h\circ \iota_{Y\to X\cup Y}=g$ .

Solution:
( a ) $\forall x\in X,\iota_{X\to Z} (x)=x$ , and $(\iota_{Y\to Z}\circ \iota_{X\to Y})(x)=\iota_{Y\to Z}\big(\iota_{X\to Y}(x)\big)=\iota_{Y\to Z)}(x)=x$
( b )If $f:A\to B$ , then for $\forall a\in A,f(a)\in B$ , thus
$(f\circ \iota_{A\to A} )(a)=f(\iota_{A\to A} (a))=f(a),\quad(\iota_{B\to B}\circ f)(a)=\iota_{B\to B} (f(a))=f(a)$
( c )Use exercise 3.3.6, we can get the conclusion.
( d )Define $h:X\cup Y\to Z$ as:

$h(a)=\begin{cases}f(a),&a\in X \\g(a),&a\in Y\end{cases}$

then we have

$(h\circ \iota_{X\to X\cup Y})(x)=h\big(\iota_{X\to X\cup Y}(x)\big)=h(x)=f(x),\quad \forall x\in X \\(h\circ \iota_{Y\to X\cup Y})(y)=h\big(\iota_{Y\to X\cup Y)}(y)\big)=h(y)=g(y),\quad \forall y\in Y$

To prove $h$ is unique, suppose there’s another function $h'$ satisfy the two equal relations, then let $b\in X\cup Y$ , we have either $b\in X$ or $b\in Y$ , only one is valid.
If $b\in X$ , then $(h'\circ \iota_{X\to X\cup Y} )(b)=f(b)$
If $b\in Y$ , then $(h'\circ \iota_{Y\to X\cup Y})(b)=g(b)$
according to the equal relations. On the other hand, we have

$(h'\circ \iota_{X\to X\cup Y})(b)=h'(\iota_{X\to X\cup Y}(b))=h'(b),\quad b\in X$

and

$(h'\circ \iota_{Y\to X\cup Y})(b)=h'(\iota_{Y\to X\cup Y}(b))=h'(b),\quad b\in Y$

Combine the four equations, we get

$h'(b)=\begin{cases}f(b),&b\in X \\ g(b),&b\in Y\end{cases}$

Thus $h'=h$ on $X\cup Y$ .

$\blacksquare$

陶哲轩实分析3.2及习题-Analysis I 3.2

2019年3月25日2020年1月19日christangdt19条评论

这一节讲罗素悖论，比较通俗的一个版本是理发师只给不给自己刮脸的人刮脸，不过直到这一节才学到了集合论比较严谨的表述。这个悖论虽然标了选读，但是对理解集合论公理比较重要，在Zorich的分析中也在非常靠前的地方就介绍了这个悖论。从这一节开始，书里的习题就不是那么好做了。

Exercise 3.2.1. Show that the universal specification axiom, Axiom 3.8, if assumed to be true, would imply Axioms 3.2,3.3,3.4,3.5 and 3.6. (if we assume that all natural numbers are objects, we also obtain Axiom 3.7.)

Solution:
Axiom 3.2:
let $P(x)$ be the property : $x \text{ is not the same as } x$ . Then $\{x:P(x) \text{ is true}\}$ is empty.
Axiom 3.3:
$a$ is an object then $\{x:x=a\}$ is the singleton set. $a,b$ are objects then $\{x:(x=a)\vee (x=b)\}$ is the pair set.
Axiom 3.4
the set $\{x:(x\in A)\vee (x\in B)\}$
Axiom 3.5:
a direct consequence of Axiom 3.8, as they’re both specification axioms.
Axiom 3.6:
Let $P(y)$ be the property: $\exists x\in A,s.t.P(x,y) \text{ is true}$ . Then the set
$\{y:P(y) \text{ is true}\}=\{y:P(x,y) \text{ is true for some } x\in A\}$
While the left set, according to Axiom 3.8, exists.
Axiom 3.7
we let $P(x)$ be the property of $P_1(x)\wedge P_2(x)$ , where

$P_1(x)=\Big(x=0\Big)\vee \Big((x\neq 0)\wedge (x++\neq 0)\wedge (\exists ! \text{ natural number }y,y++=x)\Big)$

$P_2(x)=\Big(Q(0)\text{ is true}\Big)\wedge \Big((Q(n)\text{ is true}\Rightarrow Q(n++)\text{ is true})\implies Q(x)\text{ is true}\Big),\\ \forall \text{property }Q\text{ pertaining to a natural number }n$

then $N=\{x:P(x) \text{ is true}\}$

$\blacksquare$

Exercise 3.2.2. Use the axiom of regularity (and the singleton set axiom) to show that if $A$ is a set, then $A\notin A$ . Furthermore, show that if $A$ and $B$ are two sets, then either $A\notin B$ or $B\notin A$ (or both).

Solution: We first suppose $A$ is a set. If $A=\emptyset$ , then no element can belong to $A$ , thus $A\notin A$ . Suppose that $A\neq \emptyset,A\in A$ , then due to the singleton set axiom, $\{A\}$ is a set. Thus use the axiom of regularity, there’s an element of $\{A\}$ which is either not a set, or disjoint from $\{A\}$ , however the only element of $\{A\}$ is $A$ , so this means $A$ is disjoint from $\{A\}$ . But as $A\in A,A\in \{A\}$ , we have $A\cap \{A\}=A$ , this is a contradiction.
Further, if $A$ and $B$ are two sets and suppose both $A\in B$ and $B\in A$ are true, then for the set $\{A,B\}$ , due to the axiom of regularity, there’s an element which is disjoint from $\{A,B\}$ . But we have $B\in A\cap \{A,B\}$ and $A\in B\cap \{A,B\}$ , so the axiom of regularity is violated, this is a contradiction.

$\blacksquare$

Exercise 3.2.3. Show (assuming the other axioms of set theory) that the universal specification axiom, Axiom 3.8, is equivalent to an axiom postulating the existence of a “universal set” $\Omega$ consisting of all objects (i.e., for all objects $x$ , we have $x\in \Omega$ ). In other words, if Axiom 3.8 is true, then a universal set exists, and conversely, if a universal set exists, then Axiom 3.8 is true. (This may explain why Axiom 3.8 is called the axiom of universal specification). Note that if a universal set $\Omega$ existed, then we would have $\Omega\in \Omega$ by Axiom 3.1, contradicting Exercise 3.2.2. Thus the axiom of foundation specifically rules out the axiom of universal specification.

Solution: If a universal set $\Omega$ exists, then use Axiom of specification, we can deduce Axiom 3.8 immediately.
On the other hand, if Axiom 3.8 is true, then we can have a set

$\{x:x\notin \emptyset \}$

Then every object belongs to this set.

$\blacksquare$

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