# Show that the set $A=\big\{ f_y\,\big|\, y\in[0,1]\big\}$ is compact in ${\mathcal C}[0,1]$.

Let $$f:[0,1]\times[0,1]\to\mathbb R$$ be continuous. For each $$y\in[0,1]$$ define $$f_y:[0,1]\to\mathbb R$$ by $$f_y(x)= f(x,y)$$. Show that the set $$A=\big\{ f_y\,\big|\, y\in[0,1]\big\}$$ is compact in $${\cal C}[0,1]$$.

I tried to use the Arzela-Ascoli Theorem, that is $$A$$ is comapct if and only if $$A$$ is closed, pointwise bounded and equicontinuous.

I managed to show that $$A$$ is pointwise bounded by Extreme Value Theorem. I am not sure how to prove that $$A$$ is closed and equicontinuous.

Hint: Take a sequence of point $$y_n$$, the set of functions $$f_{y_n}$$ is uniformly bounded since $$f$$ is continuous on $$[0,1]\times [0,1]$$ and its image is compact since $$[0,1]\times [0,1]$$ is compact.
It is uniformly equicontinuous continuous since $$f$$ is uniformly continuous as a continuous function defined on a compact.
You can do this directly. Each sequence in $$A$$ has the form $$f_{y(1)},f_{y(2)},\ldots$$ for some sequence $$y(1),y(2),\ldots$$ in $$[0,1]$$. By compactness some subsequence of $$y(n)$$ tends to some $$y \in [0,1]$$. For ease of notation assume $$y(n) \to y$$. Then we claim $$f_{y(n)} \to f_y$$. This can be proved using uniform continuity of $$f$$.
Since the domain of $$f$$ is compact and $$f$$ is continuous, it is uniformly continuous. Therefore, one can show that the function $$y \mapsto f_y : [0, 1] \to C[0, 1]$$ is continuous using a straightforward $$\delta$$-$$\epsilon$$ argument. The image of a compact set under a continuous map is compact; thus, $$\{f_y : y \in [0, 1]\}$$ is compact.