Independence of points from same underlying function I have a function $y=sin(x)$
I sample the points $ x_i=\{0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1\}$ and
$$ y_i=sin(x_i) + \epsilon \  \ \epsilon \sim iid\ Normal$$
Now given only the set $\{x_i,y_i\}$ I want to uncover the underlying functional form $f(x)$ given as
$$y=f(x)+\epsilon \ \ \ \ \epsilon \sim iid\ Normal  $$
My question is : ARE $y_i$'s independent ?
My intuition is that $y_1$ and $y_2$ are not independent. This is because given a new ${x}$ say ${x_\star}$ we can use $y_i$'s to learn some information about ${y_\star}$. 
Does the fact that the input points come from the same underlying function make them somehow correlated or posses commonalities ? and if not why do most methods (ex: Non-parametric splines, Gaussian process) use the $y_i$'s to get information and learn ${y_\star}$ ? Isn't the fact that if samples are independent then they can't learn new information from each other ?
 A: If I interpret the comments correctly, you sample the $x_i$ independently of each other. Denote the random sequence that you sample by $X_1, X_2, \dots$ with realizations in the set $S = \{0, 0.1, \dots 1 \}$. The fact that you use an iid-sample from $S$ amounts to saying that for all $x_i, x_j \in S$, $\mathbb{P}(X_i = x_i | X_j = x_j) = \mathbb{P}(X_i = x_i) = 1/11$. (This is the definition of statistical independence that holds for the sequence $X_1, X_2, \dots$ by assumption of an iid sample from $S$.)
Now consider any arbitrary function $f:S \to \mathbb{R}$. As long as the function $f$ has a unique inverse $f^{-1}$ on $S$, it holds that each point in $S$ has a unique point in $\mathbb{R}$ it is associated with. Thus, it holds that the exact same probablity mass is on the event $\{f(X_i) = f(x_i) \}$ as on the event $\{X_i = x_i \}$. In other words, it holds that $\mathbb{P}(f(X_i) = f(x_i)) = \mathbb{P}(X_i = x_i)$! 
Another way to look upon this is to say that if $f$ has a unique inverse on $S$, sampling at random with replacement from $S$ and then plugging it into $f$ is identical to sampling at random with replacement from $f(S) = \{f(0), f(0.1), \dots f(1) \}$, as there is a one-to-one correspondence between the two procedures.
Clearly, the function $f = \sin(x)$ satisfies the condition outlined in the paragraph on $S$. We thus have that the above paragraph can be applied to $f$. Hence, $y_i$ is a sum of two independent random variables, namely $f(X_i)$ and $\epsilon_i$. It follows that $y_i$ and $y_j$ are independent.
