Let K be the attractor of the following IFSThe main results of this paper are as follows:By virtue of Lemma 2.3, it follows thatLemma 2.12. If c ≥ (1 − λ) 2 , then K + K = [0, 2]. Proof. By Lemmas 2.1 and 2.11. Take I = J = [c − λ, c] ∪ [1 − λ, 1]. Therefore, for f (x, y) = x + y, we have f (I, J) = [2(c − λ), 2c] ∪ [c + 1 − 2λ, 1 + c] ∪ [2(1 − λ), 2]. By Lemma 2.3, we conculde that 2c − (c + 1 − 2λ) = c + 2λ − 1 ≥ 0, 1 + c − (2(1 − λ)) = c + 2λ − 1 ≥ 0.