\magnification 1200 \parindent 0pt \parskip 12pt \hsize 7.2truein \vsize 9.7 truein \hoffset -0.45truein \def\w{\vskip -2pt} \def\u{\vskip -7pt} \def\v{\vskip -6pt} \def\h{\hskip 10pt} Thm 46.8: If $Y$ is a metric space, compact open topology = topology of compact convergence on $C(X, Y)$. Show: ${\cal T}_{CO} \subset {\cal T}_C$ Take $ S(C, U) = \{ g ~|~ g \in C(X, Y), g(C) \subset U\}$ Take $g \in S(C, U)$. Thus $g: X \rightarrow Y$, $g(C) \subset U$. We will find an $r > 0$ such that $B_C(g, r) \subset S(C, U)$. $g$ continuous, $C$ compact implies $g(C)$ compact. For each $y \in g(C)$ there exists $B(y, 2r_y) \subset U$. $g(C) \subset \cup_{y \in g(C)} B(y,r_y)$. $g(C)$ compact implies there exists $y_1, ..., y_n$ such that $g(C) \subset \cup_{i = 1}^n B(y_i,r_{y_i})$. Let $r = min\{r_{y_1}, ..., r_{y_n} \}$ Let $f \in B_C(g, r) = \{ h ~|~ sup\{d(h(x), g(x)) ~|~ x \in C\} < r \}$ Show $f \in S(C, U) = \{ g ~|~ g \in C(X, Y), g(C) \subset U\}$ Note $f \in C(X, Y)$ by hypothesis. We need to show $f(C) \subset U$. Take $x \in C$. $f \in B_C(g, r)$ implies $d(f(x), g(x)) < r.$ $g(x)\in B(y_i,r_{y_i})$ for some $i$. Thus $d(g(x), y_i) < r_{y_i} \leq r}.$ Hence $d(f(x), y_i) < 2r \leq 2 r_{y_i}$. Thus $f(x) \in $B(y_i, 2r_{y_i}) \subset U$. Hence $f(C) \subset U$ and $f \in S(C, U)$. Thus ${\cal T}_{CO} \subset {\cal T}_C$ Show: ${\cal T}_{C} \subset {\cal T}_{CO}$ Show $B_C(h, \epsilon) \in {\cal T}_{CO}$ Take $f \in B_C(h, \epsilon)$ and find $U \in {\cal T}_{CO}$ such that $f \in U \subset B_C(h, \epsilon)$ \centerline{---------------------------------------------------------------------------- -------------} Claim 1: there exists an $r > 0$ such that $B_C(f, r) \subset B_C(h, \epsilon)$. Proof 1: $B_C(h, \epsilon)$ is not a ball of radius $\epsilon$ centered at $h$ in $C(X, Y)$, but $B_C(h, \epsilon)$ is a ball of radius $\epsilon$ centered at $h$ in $C(C, Y)$. Thus by p. 120 lemma, there exists an $r > 0$ such that $B_C(f, r) \subset B_C(h, \epsilon)$. Proof 2 (emulate proof of p. 120 lemma): $f \in B_C(h, \epsilon)$ implies $ sup\{d(h(x), f(x)) ~|~ x \in C\} = R < \epsilon$. Let $r = {\epsilon - R \over 2}$. Then if $g \in B_C(f, r)$, then $d(g(x), h(x)) \leq d(g(x), f(x)) + d(f(x), h(x)) < {\epsilon - R \over 2} + R = {\epsilon + R \over 2}$. Thus $sup\{d(h(x), g(x)) ~|~ x \in C\} \leq {\epsilon + R \over 2}< \epsilon$. Thus $g \in B_C(h, \epsilon)$ and $B_C(f, r) \subset B_C(h, \epsilon)$. \centerline{---------------------------------------------------------------------------- -------------} We will find compact sets $C_i$ and open sets $U_i$ such that $f \in \cap_{i = 1}^n S(C_i, U_i) \subset B_C(f, r) \subset B_C(h, \epsilon)$ $f(C) \subset \cup_{y \in f(C)} B(y,{r \over 4})$. $f(C)$ compact implies there exists $y_1, ..., y_n$ such that $f(C) \subset \cup_{i = 1}^n B(y_i,{r \over 4})$. Let $C_i = f^{-1}({\overline{B(y_i,{r \over 4}) }) \cap C$. $f$ continuous implies $C_i$ is closed in $C$. Since $C_i$ is a closed subset of $C$ compact, $C_i$ is compact. \centerline{---------------------------------------------------------------------------- -------------} Claim 2: $f \in \cap_{i = 1}^n S(C_i, B(y_i,{r})) \subset B_C(f, r)$. If $x \in C_i = f^{-1}({\overline{B(y_i,{r \over 4}) }) \cap C$, $f(x) \subset \overline{B(y_i,{r \over 4})} \subset B(y_i,{r \over 2})$ by Claim 3. Thus $f \in S(C_i, B(y_i,{r \over 2}))$ for all $i = 1, ..., n$ \centerline{---------------------------------------------------------------} Claim 3: If $r < s$, then $\overline{B(a, r)} \subset B(a, s)$. First not $\{x ~|~ d(x, a) > r\}$ is open: Suppose $d(z, a) > r$. Then $B(a, d(z, a) - r) \subset \{x ~|~ d(x, a) > r\}$ for is $y \in B(a, d(z, a) - r), $d(y, a) \geq d(z, a) - d(z, y) > d(z, a) - (d(z, a) - r) = r$. Thus $\{x ~|~ d(x, a) > r\}^C = $\{x ~|~ d(x, a) \leq r\}$ is closed. Thus $\overline{B(a, r)} \subset {x ~|~ d(x, a) \leq r\} \subset B(a, s)$ \centerline{---------------------------------------------------------------} We will now show $\cap_{i = 1}^n S(C_i, B(y_i,{r \over 2})) \subset B_C(f, r)$ Take $g \in \cap_{i = 1}^n S(C_i, B(y_i,{r \over 2}))$. We will show sup\{d(f(x), g(x)) ~|~ x \in C\} < r$ Take $x \in C \subset \cup_{i=1}^n C_i$. then $x \in C_j$ for some $j$. Hence $g(x) \in B(y_j,{r \over 2})$. Thus $d(g(x), y_j) < {r \over 2}$ Since $f \in \cap_{i = 1}^n S(C_i, B(y_i,{r}))$, $f(x) \in B(y_j,{r \over 4})$ Thus $d(f(x), g(x)) \leq d(f(x), y_j) + d(y_j, g(x)) < {r \over 4} + {r \over 2} = {3r \over 4}$. Thus $sup\{d(f(x), g(x)) ~|~ x \in C\} \leq {3r \over 4} < r$. Hence $g \in B_C(f, r)$ and $\cap_{i = 1}^n S(C_i, B(y_i,{r \over 2})) \subset B_C(f, r)$ \centerline{---------------------------------------------------------------------------- -------------} \end