函數 (Function)

令(X,dx)(X, d_x), (Y,dy)(Y, d_y) 為兩個 metric spaces

函數的定義

  • 函數為兩個集合間的關係

  • ff is a function if ∀x∈X\forall x \in X ∃!y∈Y\exists \; !y \in Y ∋\ni f(x)=yf(x)=y.

  • XX 為 ff的定義域(domain),記為X=domfX=\text{dom} f.

  • f(X)f(X)為ff的值域(range),f(X)={y∈Y∣y=f(x),∀x∈X}f(X) = \{ y\in Y \; \vert \; y=f(x), \forall x \in X\}.

  • 依上述定義,ff可為一對一或是多對一函數,但不可為一對多函數。

    • (one-to-one, bijection) ∀x1,x2∈X\forall x_1, x_2 \in X ∋\ni f(x1)=y1f(x_1)=y_1 and f(x2)=y2f(x_2)=y_2 then y2=y1y_2=y_1 →\rightarrow x1=x2x_1 = x_2.
    • (onto, surjection) ∀y∈Y\forall y \in Y ∃x∈X\exists x \in X ∋\ni f(x)=yf(x)=y.
      • ∵f(X)⊆Y\because f(X) \subseteq Y, 而onto為 f(X)=Yf(X)=Y,此時函數可能為one-to-one或many-to-one.
      • ∃x∈X\exists x\in X ∋\ni f(x)=y1f(x)=y_1 and f(x)=y2f(x)=y_2 but y1≠y2y_1 \neq y_2不符合函數定義。
  • 一對多的情形在代數中稱為映射(mapping)或是關係(relation)。

  • 函數必須是one-to-one時,inverse relation f−1f^{-1} 才是函數。因one-to-one時才可避免invese時one-to-many的情形發生。

  • 若集合XX與集合YY間存在1-1的函數時,稱兩集合有相同的基數(cardinal number),即兩集合的元素個數相同;或稱兩集合等價(equivalent),此時該1-1函數為等價關係(equivalence relation),即滿足reflexive, symmetric, transitive。

  • E.g. 自然數N\mathbb{N}與整數Z\mathbb{Z}為等價集合。

  • E.g. ∀a,b∈R\forall a, b \in \mathbb{R},(a,b)(a,b)與(0,1(0,1為等價集合。
  • E.g. (0,1)(0,1)與實數R\mathbb{R}為等價集合。

Image and pre-image

  • If E⊂XE \subset X, f(E)={f(x)∣ x∈E}f(E) = \lbrace f(x) \vert \ x \in E \rbrace, the f(E)f(E) is the image of EE.

  • If F⊂YF \subset Y, f−1(F)={x∈X∣f(x)∈F}f^{-1}(F) = \lbrace x \in X \vert f(x) \in F \rbrace, f−1f^{-1} is the pre-image (inverse image) of FF

  • E.g. f:{1,2,3}→{a,b,c,d}f:\lbrace 1,2,3 \rbrace \rightarrow \lbrace a,b,c,d \rbrace.

    • f(x)≡{ax=1ax=2cx=3f(x) \equiv \lbrace \begin{array}{ll} a & x=1 \\ a & x=2 \\ c & x=3 \end{array}
    • The image f({2,3})={a,c}f( \lbrace 2,3 \rbrace ) = \lbrace a,c \rbrace.
  • E.g. f:R→Rf: \mathbb{R} \rightarrow \mathbb{R}, and f(x)=x2f(x) = x^2.

    f−1([4,9])={x∈R∣f(x)∈[4,9]}={x∈R∣4≤x2≤9}={x∈R∣2≤x≤3 or −3≤x≤−2}=[−3,−2]∪[2,3]. \begin{array}{rcl} f^{-1}([4,9]) & = & \lbrace x \in \mathbb{R} \vert f(x) \in [4,9] \rbrace \\ & = & \lbrace x \in \mathbb{R} \vert 4 \leq x^2 \leq 9 \rbrace \\ & = & \lbrace x \in \mathbb{R} \vert 2 \leq x \leq 3 \text{ or } -3 \leq x \leq -2 \rbrace \\ & = & [-3, -2] \cup [2,3]. \end{array}

    • 若令f:R+→Rf: \mathbb{R}^{+} \rightarrow \mathbb{R}, 則f−1([4,9])=[2,3]f^{-1}([4,9]) = [2,3].
  • Theorem: Every infinite subuset of a countable set SS is countable.
  • 由定理可知,一個可數的集合之子集合必為可數集合,而不可為不可數集合。

連續函數 (Continous function)

  • f:X→Yf: X \rightarrow Y is a function, ff is continuous at a point x∈Xx \in X if

    • ∀ϵ>0\forall \epsilon > 0 ∃ δ>0\exists \ \delta > 0 ∋\ni dY(f(x1),f(x2))<δd_Y(f(x_1), f(x_2)) < \delta ∀dX(x1,x2)<ϵ\; \forall d_X(x_1, x_2) < \epsilon .
  • 如果ff在集合SS中每一點都連續時,稱ff is continous on SS。

  • 下圖為X=Y=RX=Y=\mathbb{R}的範例。

    • Epsilon delta表示法

函數的最小上界與最大下界性質

    • Theorem:雙變數連續函數的最小上界的交換性
    • 函數 f:X×Y→Rf: X \times Y \rightarrow \mathbb{R} 有上界。
    • 則supx,yf(x,y)=supxsupyf(x,y)=supysupxf(x,y) \sup_{x,y} f(x,y) = \sup_x \sup_y f(x,y) = \sup_y \sup_x f(x,y) .
    • Theorem:雙變數函數最小上界與最大下界的不等式
    • 函數f:X×Y→R f: X \times Y \rightarrow \mathbb{R} 之值域為有界集合。
    • 則supyinfxf(x,y)≤infxsupyf(x,y) \sup_y \inf_x f(x,y) \leq \inf_x \sup_y f(x,y) .

函數的極限 (Limit of functions)

  • Let E⊆X E \subseteq X, and a function f:E→Yf: E \rightarrow Y.
  • pp is a limit point of EE, and it is written limx→pf(x)=q\lim_{x \rightarrow p} f(x)=q if

    • ∀ϵ>0\forall \epsilon > 0 ∃δ>0\exists \delta > 0 ∋\ni dY(f(x),q)<ϵd_Y(f(x), q) < \epsilon ∀x∈E \forall x \in E dX(x,p)<δd_X(x,p) < \delta.
  • 注意,p∈Xp \in X,但pp不一定在集合EE中。(如2\sqrt{2}不在有理數集合中)。

  • For every sequence pn∈E{p_n} \in E such that pn≠pp_n \neq p and limn→∞pn=p\lim_{n \rightarrow \infty} p_n = p, then limx→pf(x)=q \lim_{x \rightarrow p} f(x) = q ⇔\Leftrightarrow limn→∞f(pn)=q\lim_{n \rightarrow \infty} f(p_n) = q.

  • If ff has two limit points at p1p_1 and p2p_2, then p1=p2p_1=p_2. (極限點唯一存在)

  • Theorem: f:X→Yf: X \rightarrow Y is continuous function if f−1(U)f^{-1}(U) is open set for every open set U⊂YU \subset Y.
  • Proof:

    • "⇒\Rightarrow" Suppose UU is open in YY and ff is continuous.

    • ∴∀y∈U\therefore \forall y \in U ∃ϵy>0\exists \epsilon_y > 0 ∋Nϵy(y)⊂U\ni N_{\epsilon_y}(y) \subset U and

    • ∀x∈f−1({y})\forall x \in f^{-1}(\lbrace y \rbrace) ∃δx>0\exists \delta_x > 0 ∋Nδx(x)⊂f−1(Nϵy(y))⊂f−1(U)\ni N_{\delta_x} (x) \subset f^{-1}(N_{\epsilon_y}(y)) \subset f^{-1}(U).

    • ∴f−1(U)=∪x∈f−1(U)Nδx(x)\therefore f^{-1}(U) = \cup_{x \in f^{-1}(U)}N_{\delta_x}(x) is open.

    • "⇐\Leftarrow" ∀x∈X\forall x \in X andϵ>0\epsilon > 0, f−1(Nϵf(x))f^{-1}(N_{\epsilon}f(x)) is open and contains xx, so it conains some ball about xx, this means ff is continuous at xx.

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