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@ -23,7 +23,7 @@ $X^{X}$ is a compact Hausdorff space.
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\item $X^X \ni f \mapsto f \circ f_0$
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\item $X^X \ni f \mapsto f \circ f_0$
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is continuous:
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is continuous:
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Consider $\{f : f f_0 \in U_{\epsilon}(x,y)\}$.
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Consider $\{f : f \circ f_0 \in U_{\epsilon}(x,y)\}$.
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We have $ff_0 \in U_{\epsilon}(x,y)$
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We have $ff_0 \in U_{\epsilon}(x,y)$
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iff $f \in U_\epsilon(x,f_0(y))$.
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iff $f \in U_\epsilon(x,f_0(y))$.
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\item Fix $x_0 \in X$.
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\item Fix $x_0 \in X$.
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@ -108,7 +108,7 @@ This will follow from the following lemma:
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is continuous.
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is continuous.
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Since $T_n$ is compact,
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Since $T_n$ is compact,
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we have that $\{(x,t) \mapsto tx : t \in T_n\}$
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we have that $\{(x,t) \mapsto tx : t \in T_n\}$
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is equicontinuous for all $n$.
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is equicontinuous.
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So there is $\epsilon > 0$ such that
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So there is $\epsilon > 0$ such that
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$d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a -b$
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$d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a -b$
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for all $t \in T_n$.
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for all $t \in T_n$.
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