How to prove that the space created by pointwise Bernoulli random variables are compact 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.
 A: In the end, it seems that $\Delta$ is simply $\Delta=[0,1]^\mathbb R$ the set of all functions defined on $\mathbb R$ with values in $[0,1]$. Then the convexity of $\Delta$ is trivial since any product of convex sets is convex and $\Delta$ is a product of the convex set $[0,1]$. Is this set $\Delta$ compact? For the product topology, this is Tychonoff theorem.
But now I note that later on in the question, you declare that every $\delta$ in $\Delta$ is a continuous function... Thus, $\Delta$ would be $\Delta=C^0(\mathbb R,[0,1])$. Then the convexity of $\Delta$ is trivial. Is this set $\Delta$ compact? This could depend on the topology you put on $\Delta$ (pointwise convergence? uniform convergence?) but in general the answer shall be no.
Note Two confusions seem to plague your understanding:


*

*First, functions are not numbers. You cannot at the same time pretend that For each point $y\in\mathbb{R}$, $\delta(y)$ is a real number in $[0,1]$ and that $\delta(y)$ is a function defined on $\mathbb R$. If the first assertion holds, then $\delta$ is a function and $\delta(y)$ is a number. Please watch out for this confusion when you write down your definitions.

*Second, one does not check convexity the way you check it: if $\delta$ is in a set $\Delta$, then $\delta'=\alpha\delta+(1-\alpha)\delta$ is also in $\Delta$, always! Simply because $\delta'=\delta$. Convexity asks something else, namely that $\alpha\delta_1+(1-\alpha)\delta_2$ is in $\Delta$, for every $\alpha$ in $[0,1]$ and every $\delta_1$ and $\delta_2$ in $\Delta$.
Finally, note that when I declare that the convexity of any product of convex sets is trivial, I mean it. Please write the definitions down and you should see that this is obvious and that the only hypothesis you need is that $[0,1]$ (the target set) is convex. Ditto for the convexity of $C^0(\mathbb R,[0,1])$.
Edit It appears now that $\Delta$ would be the set of continuous nondecreasing functions $\delta$ defined on $\mathbb R$ such that $0\leqslant\delta\leqslant1$, $\delta(y)=0$ for every $y$ small enough $y$ and $\delta(y)=1$ for every $y$ large enough. Then, convexity is still trivial and compactness still dubious (consider the functions $(\delta_t)_{t\in\mathbb R}$ in $\Delta$ defined by $\delta_0:y\mapsto\max\{0,\min\{y,1\}\}$ and, for every $t$, $\delta_t:y\mapsto\delta_0(y-t)$).
