We prove that the infinite composition of continuous surjective mappings defined in compact set is a surjective mapping if the infinite composition is a well-defined mapping.

1. Infinite composition of continuous surjective mappings
LOHANS DE OLIVEIRA MIRANDA
Federal University of Piauí, Brazil
lohansmiranda@gmail.com
Abstract. We prove that the infinite composition of continuous surjective mappings defined in compact
set is a surjective mapping if the infinite composition is a well-defined mapping.
This is joint work with Lossian Barbosa Bacelar Miranda (IFPI, Brazil).
Proposition. Let 𝐶 be a compact set in any topological space and 𝑓𝑖: 𝐶 → 𝐶, 𝑖 ∈ ℕ∗
, a sequence of
continuous surjective mappings. Suppose there is F(x) = lim
𝑖→+∝
( 𝑓1 ∘ 𝑓2 ∘ … ∘ 𝑓𝑖)(𝑥) for any 𝑥 ∈ 𝐶.
Then 𝐹: 𝐶 → 𝐶 is a surjective mapping.
Proof. Let 𝑦𝜖𝐶. ∀𝑖𝜖ℕ∗
, ∃𝑥𝑖 ∈ 𝐶 | (𝑓1 ∘ 𝑓2 ∘ … ∘ 𝑓𝑖)(𝑥𝑖) = 𝑦. Since {𝑥𝑖}𝑖∈ℕ∗ is a sequence in the
compact set 𝐶, then there is a subsequence {𝑥𝑖 𝑘
} 𝑘∈ℕ∗ converging to some element 𝑥 ∈ 𝐶 ( lim
𝑘→+∝
𝑥𝑖 𝑘
=
𝑥 ∈ 𝐶 ) and such that (𝑓1 ∘ 𝑓2 ∘ … ∘ 𝑓𝑖 𝑘
)(𝑥𝑖 𝑘
) = 𝑦, ∀𝑘 ∈ ℕ∗
. Since {(𝑓1 ∘ 𝑓2 ∘ … ∘ 𝑓𝑖)(𝑥𝑖)}𝑖𝜖ℕ∗ is a
sequence which converges to F(x) = lim
𝑖→+∝
( 𝑓1 ∘ 𝑓2 ∘ … ∘ 𝑓𝑖)(𝑥), its subsequence { (𝑓1 ∘ 𝑓2 ∘ … ∘
𝑓𝑖 𝑘
)(𝑥𝑖 𝑘
)} 𝑘∈ℕ∗ also converges for 𝐹(𝑥). However, this sequence is constant and equal to {𝑦}𝑖𝜖ℕ∗.
Therefore, 𝐹(𝑥) = 𝑦.
Observation. From this simple proposition, which we do not know if it has already been demonstrated,
many interesting results can be proved.