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Let $E$ be a compact space and $X,Y \subset E $ compact subsets of $E$.

I wonder if the family of continous functions $C(X,Y) := \{f:X\rightarrow Y \mid f\ \text{is continuous} \}$ is equicontinuous.

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No, it is not equicontinuous.
Consider the set $$\left\{\sin (k\ \cdot ):[0,1]\to [-1,1]\ \mid k \in \mathbb{N}\right\} \subset C([0,1],[-1,1]).$$ For $\epsilon = 1/2 $ and and every $\delta >0$ will find a $k\in \mathbb{N}$ such that there is some $x \in [0,1]$ with $|x-0|<\delta$ and $|sin(kx)-0|>\epsilon$.

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If this family $C(X,Y)$ is equicontinuous, by Arzelà–Ascoli theorem, the closed unit ball in $C(X,Y)$ is compact. And this compactness result holds if and only if $C(X,Y)$ is a finite dimensional vector space.

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That is not the case in general.

Take $E=X=Y=[0,1]$. Then $C(X,Y)$ contains the functions defined by: $$f_n(x)=\begin{cases} 0 \text{ if } 0 \le x \le \frac{1}{2}\\ n(x-\frac{1}{2}) \text{ for } \frac{1}{2} \le x \le \frac{1}{2} + \frac{1}{n}\\ 1 \text{ if } \frac{1}{2} + \frac{1}{n} \le x \le 1 \end{cases}$$ for $n \ge 1$ integer.

The set $\{f_n \ : \ n \ge 1\}$ is not equicontinuous at $\frac{1}{2}$.

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