陶哲轩实分析 习题解答

习题 3.3

3.3.1

(1) 证明自反性

?x∈X,f(x)=f(x)<script id="MathJax-Element-1" type="math/tex"> \forall x \in X, f(x) = f(x) </script>
所以 f=f<script id="MathJax-Element-2" type="math/tex">f = f</script>

(2) 证明对称性
假设 f=g<script id="MathJax-Element-3" type="math/tex">f = g</script>
那么 ?x∈X,f(x)=g(x)<script id="MathJax-Element-4" type="math/tex">\forall x \in X, f(x) = g(x)</script>
所以 ?x∈X,g(x)=f(x)<script id="MathJax-Element-5" type="math/tex">\forall x \in X, g(x) = f(x)</script>
所以 g=f<script id="MathJax-Element-6" type="math/tex">g = f</script>

(3) 传递性
假设 f=g,g=h<script id="MathJax-Element-7" type="math/tex">f =g, g = h</script>
那么有
?x∈X,f(x)=g(x) ?x∈X,g(x)=h(x)<script id="MathJax-Element-8" type="math/tex"> \forall x \in X, f(x) = g(x) \ \forall x \in X, g(x) = h(x) </script>
所以?x∈X,f(x)=h(x)<script id="MathJax-Element-9" type="math/tex">\forall x \in X, f(x) = h(x)</script>
所以 f=h<script id="MathJax-Element-10" type="math/tex">f = h</script>

3.3.2

(1) 证明。当 f<script id="MathJax-Element-11" type="math/tex">f</script> 和 g<script id="MathJax-Element-12" type="math/tex">g</script> 都是单射时。g°f<script id="MathJax-Element-13" type="math/tex">g \circ f</script> 也是单射。
反证法： 设存在不同样的 x1<script id="MathJax-Element-14" type="math/tex">x_1</script> 和 x2<script id="MathJax-Element-15" type="math/tex">x_2</script>，满足 (g°f)x1=(g°f)x2<script id="MathJax-Element-16" type="math/tex">(g \circ f) x_1 = (g \circ f) x_2</script>。
已知 f<script id="MathJax-Element-17" type="math/tex">f</script> 是单射。所以 f(x1)≠f(x2)<script id="MathJax-Element-18" type="math/tex">f (x_1) \neq f (x_2)</script>
设 y1=f(x1)<script id="MathJax-Element-19" type="math/tex">y_1 = f(x_1)</script>。 y2=f(x2)<script id="MathJax-Element-20" type="math/tex">y2 = f(x_2)</script>。 y1≠y2<script id="MathJax-Element-21" type="math/tex">y_1 \neq y_2</script>。
那么 g(y1)=g(y2)<script id="MathJax-Element-22" type="math/tex">g(y_1) = g(y_2)</script> 这与g<script id="MathJax-Element-23" type="math/tex">g</script> 是单射矛盾。
所以g°f<script id="MathJax-Element-24" type="math/tex">g \circ f</script> 也是单射。

(2) f<script id="MathJax-Element-25" type="math/tex">f</script>、g<script id="MathJax-Element-26" type="math/tex">g</script> 是满射时，g°f<script id="MathJax-Element-27" type="math/tex">g \circ f</script> 也是满射。

由于 g<script id="MathJax-Element-28" type="math/tex">g</script> 是满射，所以对随意的 z∈Z<script id="MathJax-Element-29" type="math/tex">z \in Z</script>， 存在 y∈Y<script id="MathJax-Element-30" type="math/tex">y \in Y</script> 使得 g(y)=z<script id="MathJax-Element-31" type="math/tex">g(y) = z</script>
由于 f<script id="MathJax-Element-32" type="math/tex">f</script> 是满射，所以存在 x∈X<script id="MathJax-Element-33" type="math/tex">x \in X</script> 满足 f(x)=y<script id="MathJax-Element-34" type="math/tex">f(x)= y</script>
所以对于随意的 z∈Z<script id="MathJax-Element-35" type="math/tex">z \in Z</script> 都存在 x∈X<script id="MathJax-Element-36" type="math/tex">x \in X</script> 满足 (g°f)(x)=z<script id="MathJax-Element-37" type="math/tex">(g \circ f) (x) = z</script>
所以g°f<script id="MathJax-Element-38" type="math/tex">g \circ f</script> 是满射

3.3.3

空函数是 f:?→X<script id="MathJax-Element-39" type="math/tex">f:\emptyset \rightarrow X</script>。
当X<script id="MathJax-Element-40" type="math/tex">X</script> 是随意集合时，空函数都是单射。由于没有 x1∈?<script id="MathJax-Element-41" type="math/tex">x_1 \in \emptyset</script>, x2∈?<script id="MathJax-Element-42" type="math/tex">x_2 \in \emptyset</script>, x1≠x2<script id="MathJax-Element-43" type="math/tex">x_1 \neq x_2</script> 满足f(x1)=f(x2)<script id="MathJax-Element-44" type="math/tex">f(x_1) = f(x_2)</script>

当X=?<script id="MathJax-Element-45" type="math/tex">X = \emptyset</script> 时，空函数是满射，也是双射。

3.3.4

(1) g<script id="MathJax-Element-46" type="math/tex">g</script> 是单射，g°f=g°f~<script id="MathJax-Element-47" type="math/tex">g \circ f = g \circ \tilde f</script>, 则 f=f~<script id="MathJax-Element-48" type="math/tex">f = \tilde f</script>
反证法。若 f≠f~<script id="MathJax-Element-49" type="math/tex">f \neq \tilde f</script>, 则存在 x<script id="MathJax-Element-50" type="math/tex">x</script> 使得 f(x)≠f~(x)<script id="MathJax-Element-51" type="math/tex">f(x) \neq \tilde f(x)</script>
设 y=f(x)<script id="MathJax-Element-52" type="math/tex">y = f(x)</script>, y~=f~(x)<script id="MathJax-Element-53" type="math/tex">\tilde y = \tilde f(x)</script>
由于 g<script id="MathJax-Element-54" type="math/tex">g</script> 是单射,所以 g(y)≠g(y~)<script id="MathJax-Element-55" type="math/tex">g(y) \neq g(\tilde y)</script>
所以 (g°f)(x)=(g°f~)(x)<script id="MathJax-Element-56" type="math/tex">(g \circ f) (x) = (g \circ \tilde f) (x)</script> 矛盾.
所以 f=f~<script id="MathJax-Element-57" type="math/tex">f = \tilde f</script>

(2) f<script id="MathJax-Element-58" type="math/tex">f</script> 是满射, g°f=g~°f<script id="MathJax-Element-59" type="math/tex">g \circ f = \tilde g \circ f</script>. 则 g=g~<script id="MathJax-Element-60" type="math/tex">g = \tilde g</script>
反证法: 若g≠g~<script id="MathJax-Element-61" type="math/tex">g \neq \tilde g</script> 则存在 y<script id="MathJax-Element-62" type="math/tex">y</script> 满足 g(y)≠g~(y)<script id="MathJax-Element-63" type="math/tex">g(y) \neq \tilde g(y)</script>
由于 f<script id="MathJax-Element-64" type="math/tex">f</script> 是满射,所以存在 x∈X<script id="MathJax-Element-65" type="math/tex">x \in X</script> 满足 f(x)=y<script id="MathJax-Element-66" type="math/tex">f(x) = y</script>
那么 (g°f)(x)≠(g~°f)(x)<script id="MathJax-Element-67" type="math/tex">(g \circ f) (x) \neq (\tilde g \circ f)(x)</script> 矛盾.
所以 g=g~<script id="MathJax-Element-68" type="math/tex">g = \tilde g</script>

3.3.5

(1) g°f<script id="MathJax-Element-69" type="math/tex">g \circ f</script> 是单射,则 f<script id="MathJax-Element-70" type="math/tex">f</script> 是单射.
反证法: 若 f<script id="MathJax-Element-71" type="math/tex">f</script> 不是单射,则存在不同样的 x1<script id="MathJax-Element-72" type="math/tex">x_1</script> 和 x2<script id="MathJax-Element-73" type="math/tex">x_2</script>,满足 f(x1)=f(x2)=y<script id="MathJax-Element-74" type="math/tex">f(x_1) = f(x_2) = y</script>
设 z=g(y)<script id="MathJax-Element-75" type="math/tex">z= g(y)</script> 则 (g°f)(x1)=(g°f)(x2)<script id="MathJax-Element-76" type="math/tex">(g \circ f) (x_1) = (g \circ f)(x_2)</script>, 与 g°f<script id="MathJax-Element-77" type="math/tex">g \circ f</script> 是单射矛盾.

(2)g°f<script id="MathJax-Element-78" type="math/tex">g \circ f</script> 是满射,则 g<script id="MathJax-Element-79" type="math/tex">g</script> 是满射.
反证法: 若 g<script id="MathJax-Element-80" type="math/tex">g</script> 不是满射, 则存在 z0<script id="MathJax-Element-81" type="math/tex">z_0</script> 没有不论什么 y∈Y<script id="MathJax-Element-82" type="math/tex">y \in Y</script> 满足 g(y)=z0<script id="MathJax-Element-83" type="math/tex">g(y) = z_0</script>
而我们又知道g°f<script id="MathJax-Element-84" type="math/tex">g \circ f</script> 是满射,则存在 x∈X<script id="MathJax-Element-85" type="math/tex">x \in X</script> , 满足 (g°f)(x)=z0<script id="MathJax-Element-86" type="math/tex">(g \circ f) (x) = z_0</script>
设 y0=f(x)<script id="MathJax-Element-87" type="math/tex">y_0 = f(x)</script> 那么就有 g(y0)=z0<script id="MathJax-Element-88" type="math/tex">g(y_0) = z_0</script> 矛盾.
所以 g<script id="MathJax-Element-89" type="math/tex">g</script> 是满射

3.3.6

(1) 由于 f<script id="MathJax-Element-90" type="math/tex">f</script> 是双射, 对随意的 x∈X<script id="MathJax-Element-91" type="math/tex">x \in X</script>, 都有唯一的y∈Y<script id="MathJax-Element-92" type="math/tex">y \in Y</script> 满足f(x)=y<script id="MathJax-Element-93" type="math/tex">f(x) = y</script>
由 f?1<script id="MathJax-Element-94" type="math/tex">f^{-1}</script> 的定义可知: f?1(y)=x<script id="MathJax-Element-95" type="math/tex">f^{-1} (y) = x</script>
所以: (f?1°f)(x)=f?1(y)=x<script id="MathJax-Element-96" type="math/tex">(f^{-1} \circ f )(x) = f^{-1} (y) = x </script> 对一切 x∈X<script id="MathJax-Element-97" type="math/tex">x \in X</script> 成立.

(2) 由于 f<script id="MathJax-Element-98" type="math/tex">f</script> 是双射, 对随意的 y∈Y<script id="MathJax-Element-99" type="math/tex">y \in Y</script> 都有唯一的 x∈X<script id="MathJax-Element-100" type="math/tex">x \in X</script> 满足 f?1(y)=x<script id="MathJax-Element-101" type="math/tex">f^{-1} (y) = x</script>
又有 f(x)=y<script id="MathJax-Element-102" type="math/tex">f(x) = y</script>. 所以 (f°f?1)(y)=y<script id="MathJax-Element-103" type="math/tex">(f \circ f^{-1}) (y) = y</script> 对随意 y∈Y<script id="MathJax-Element-104" type="math/tex">y \in Y</script>都成立.
所以 f?1<script id="MathJax-Element-105" type="math/tex">f^{-1} </script>是可逆的,且逆为 f<script id="MathJax-Element-106" type="math/tex">f</script>

3.3.7

先证明 g°f<script id="MathJax-Element-107" type="math/tex">g \circ f</script> 是单射. (略)
再证明 g°f<script id="MathJax-Element-108" type="math/tex">g \circ f</script> 是满射. (反证法, 略)

3.3.8

(a) 对一切 x∈X<script id="MathJax-Element-109" type="math/tex">x \in X</script> 有 τX→Y(x)=x<script id="MathJax-Element-110" type="math/tex">\tau_{X \rightarrow Y} (x) = x</script>
对一切 x∈X?Y<script id="MathJax-Element-111" type="math/tex">x \in X \subseteq Y</script> 有 τY→Z(x)=x<script id="MathJax-Element-112" type="math/tex">\tau_{Y \rightarrow Z} (x) = x</script>
所以有 一切 x∈X<script id="MathJax-Element-113" type="math/tex">x \in X</script>, (τY→Z°τX→Y)(x)=x=τX→Z(x)<script id="MathJax-Element-114" type="math/tex">(\tau_{Y \rightarrow Z} \circ \tau_{X \rightarrow Y}) (x) = x = \tau_{X \rightarrow Z}(x)</script>
表明: τY→Z°τX→Y=τX→Z<script id="MathJax-Element-115" type="math/tex">\tau_{Y \rightarrow Z} \circ \tau_{X \rightarrow Y} = \tau_{X \rightarrow Z}</script>

(b) 对一切 x∈A<script id="MathJax-Element-116" type="math/tex">x \in A</script> 有 f°τA→A(x)=f(x)<script id="MathJax-Element-117" type="math/tex">f \circ \tau_{A \rightarrow A} (x) = f(x)</script>
所以 f=f°τA→A<script id="MathJax-Element-118" type="math/tex">f = f \circ \tau_{A \rightarrow A}</script>

对一切 x∈A<script id="MathJax-Element-119" type="math/tex">x \in A</script> 有 τB→B°f(x)=τB→B(f(x))=f(x)<script id="MathJax-Element-120" type="math/tex">\tau_{B \rightarrow B} \circ f(x) = \tau_{B \rightarrow B}(f(x)) = f(x)</script>
所以 τB→B°f=f<script id="MathJax-Element-121" type="math/tex">\tau_{B \rightarrow B} \circ f = f</script>

所以f=f°τA→A=τB→B°f<script id="MathJax-Element-122" type="math/tex">f = f \circ \tau_{A \rightarrow A} = \tau_{B \rightarrow B} \circ f</script>

(c) (易证,略)

(d) 反证法: 假设存在两个不同的函数 h1<script id="MathJax-Element-123" type="math/tex">h_1</script> 和 h2<script id="MathJax-Element-124" type="math/tex">h_2</script> 满足 hi°τX→X?Y=f<script id="MathJax-Element-125" type="math/tex">h_i \circ \tau_{X \rightarrow X \bigcup Y} = f</script> 和 hi°τY→X?Y=g<script id="MathJax-Element-126" type="math/tex">h_i \circ \tau_{Y \rightarrow X \bigcup Y} = g</script>

那么必定存在一个 a∈X?Y<script id="MathJax-Element-127" type="math/tex">a \in X \bigcup Y</script> 使得h1(a)≠h2(a)<script id="MathJax-Element-128" type="math/tex">h_1(a) \neq h_2(a)</script>
分两种情况讨论:
a∈X<script id="MathJax-Element-129" type="math/tex">a \in X</script> 时:
h1°τX→X?Y(a)=h1(a)<script id="MathJax-Element-130" type="math/tex">h_1 \circ \tau_{X \rightarrow X \bigcup Y} (a) = h_1(a)</script>
h2°τX→X?Y(a)=h2(a)<script id="MathJax-Element-131" type="math/tex">h_2 \circ \tau_{X \rightarrow X \bigcup Y} (a) = h_2(a)</script>
所以:
h1°τX→X?Y(a)≠h2°τX→X?Y(a)<script id="MathJax-Element-132" type="math/tex">h_1 \circ \tau_{X \rightarrow X \bigcup Y} (a) \neq h_2 \circ \tau_{X \rightarrow X \bigcup Y} (a) </script> 矛盾.

a∈Y<script id="MathJax-Element-133" type="math/tex">a \in Y</script> 时:
h1°τY→X?Y(a)=h1(a)<script id="MathJax-Element-134" type="math/tex">h_1 \circ \tau_{Y \rightarrow X \bigcup Y} (a) = h_1(a)</script>
h2°τY→X?Y(a)=h2(a)<script id="MathJax-Element-135" type="math/tex">h_2 \circ \tau_{Y \rightarrow X \bigcup Y} (a) = h_2(a)</script>
h1°τY→X?Y(a)≠h2°τY→X?Y(a)<script id="MathJax-Element-136" type="math/tex">h_1 \circ \tau_{Y \rightarrow X \bigcup Y} (a) \neq h_2 \circ \tau_{Y \rightarrow X \bigcup Y} (a)</script> 矛盾

所以 仅仅有唯一的函数 h<script id="MathJax-Element-137" type="math/tex">h</script>