I have a function $\delta:\mathbb{R}\rightarrow [0,1].$
We obtain this funtion pointwise as follows: For each point $y\in\mathbb{R}$, $\delta(y)$ is a real number in $[0,1]$. More explicitely, $\delta(y)$ can be any number in $[0,1]$ given a specific $y$.
Once $\delta$ is specified for all $y\in\mathbb{R}$, then $\delta$ will have as many elements as ${\mathbb{R}}$. As we are free to chose any $\delta(y)\in[0,1]$ given a specific $y$, $\delta$ can be obtained in uncountably many ways.
The set $\Delta$ is composed of any possible construction of $\delta$.
Below is a discrete example when $y$ and $\delta$ are defined as sets. In the original question $y\in\mathbb{R}$ and $\delta$ was a continuos function on real numbers.
Discrete example: Let $y:=\{0,1,2\}$ and $\delta(y)\in\{0.1,0.5\}$, Then
$\Delta_1=\{0.1,0.1,0.1\}$
$\Delta_2=\{0.1,0.1,0.5\}$
$\Delta_3=\{0.1,0.5,0.1\}$
.. ..
$\Delta_8=\{0.5,0.5,0.5\}$
and we have $\Delta=\{\Delta_1,\Delta_2...,\Delta_8\}$
I must prove that the set $\Delta$ of all possible $\delta(y)\in \Delta$ is convex and compact.
Convexity: For any given $\alpha\in[0,1]$
$\delta^{'}(y)=\alpha\delta(y)+(1-\alpha)\delta(y)$ is also a valid function in $\Delta$, therefore $\Delta$ is a convex set.
Compactness: If I can show that for each open cover of $\Delta$ there exists a subcover, then I am done but I dont know how to show it or if there is something simpler.
I will be very grateful for any help,
Thanks in advance.