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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)$

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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  $B_C(f, r) \subset B_C(h, \epsilon)$.
 
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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.

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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 $i = 1, ..., n$

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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 $\overline{B(a, r)} \subset {x ~|~ d(x, a) \leq r\} \subset B(a, s)$

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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 $\cap_{i = 1}^n S(C_i, B(y_i,{r \over 2})) \subset B_C(f, r)$

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