Let $[a,b]$ be an interval in real line . Given any function $f:[a,b]\to \mathbb R$ and set $A \subseteq [a,b]$ of size $n+1$, there exists a unique polynomial $p_{f,A,n}(x)$ of degree $n$ such that $f(a)=p_{f,A,n}(a),\forall a\in A$. Such a polynomial is called the interpolating polynomial of $f$ with nodes $A$. Due to this, given $A\subseteq [a,b]$ with $n+1=|A|$, we can consider a linear map $X_{A,n} : C[a,b]\to C[a,b]$ such that $X_{A,n}(f)=p_{A,f,n}$ . Then $X_{A,n}$ is a continuous map and also a projection i.e. $X^2_{A,n}=X_{A,n}$. It can be shown that there is a constant $c$ such that $||X_{A,n}||_{\infty} \ge c \ln (n)$ for any set $A\subseteq [a,b]$ of size $n+1$, where $||.||_{\infty}$ is the $L^{\infty}$ operator norm induced from the $L^{\infty}$ norm on $C[a,b]$. Since $(C[a,b],||.||_{\infty})$ is a Banach space, it follows from Uniform boundedness principle, that given any sequence of subsets $\{A_n\}_{n=1}^\infty$ of $[a,b]$ with each $A_n$ having size $n$, $\exists f\in C[a,b]$ such that $||X_{A_n,n}(f) - f||_{\infty} $ does not converge to $0$.

This can be rephrased as: Given any sequence $\{A_n\}_{n=1}^\infty$ of subsets of $[a,b]$, with each $A_n$ of size $n+1$, there exists a continuous function $f$ on $[a,b]$ such that the sequence of interpolating polynomials $p_{f,A_n,n}(x)$ doesn't converge to $f$ uniformly on $C[a,b]$.

My question is the following: Given any sequence $\{A_n\}_{n=1}^\infty$ of subsets of $[a,b]$, with each $A_n$ of size $n+1$, does there exist a continuous function $f$ on $[a,b]$ such that the sequence of interpolating polynomials $p_{f,A_n,n}(x)$ doesn't converge to $f$ point-wise at atleast one point ?