Correction: Cabrera Martínez et al. On the Secure Total Domination Number of Graphs. Symmetry 2019, 11 , 1165

The authors wish to make the following corrections on paper [...]

The authors wish to make the following corrections on paper [1]: (1) Eliminate Lemma 1 because we have found that this lemma is not correct.
(2) Theorem 3 states that for any graph G with no isolated vertex, The result is correct, but the proof uses Lemma 1. For this reason, we propose the following alternative proof for Theorem 3.
Proof. Let D be a γ(G)-set. Let I be an α(G)-set such that |D ∩ I| is at its maximum among all α(G)-sets. Notice that for any x ∈ D ∩ I, We next define a set S ⊆ V(G) of minimum cardinality among the sets satisfying the following properties.
Since D and I are dominating sets, from (a) and (b) we conclude that S is a TDS. From First, notice that every vertex in V(G) \ N(u) is dominated by some vertex in S , because S is a TDS of G. Let w ∈ N(u). Now, we differentiate two cases with respect to vertex u. Case 2. u ∈ I ∩ D. We first suppose that w / ∈ D. If w / ∈ epn(u, D ∪ I), then w is dominated by some vertex in (D ∪ I) \ {u} ⊆ S . If w ∈ epn(u, D ∪ I), then by (b1) and the fact that in this case all vertices in epn(u, D ∪ I) form a clique, w is dominated by some vertex in S \ {u} ⊆ S . From now on, suppose that w ∈ D. If w / ∈ ipn(u, D ∪ I), then there exists some vertex in (D ∪ I) \ {u} ⊆ S which dominates w. Finally, we consider the case in that w ∈ ipn(u, D ∪ I).
We claim that ipn(u, D ∪ I) = {w}. In order to prove this claim, suppose that there exists w ∈ ipn(u, D ∪ I) \ {w}. Notice that w ∈ D. By (1) and the fact that all vertices in epn(u, I) form a clique, we prove that ww ∈ E(G), and so w / ∈ ipn(u, D ∪ I), which is a contradiction. Therefore, ipn(u, D ∪ I) = {w} and, as a result, epn(u, D ∪ I) ∪ {w} ⊆ epn(u, I). (2) In order to conclude the proof, we consider the following subcases.
Subcase 2.1. epn(u, D ∪ I) = ∅. By (2), (b1), and the fact that all vertices in epn(u, I) form a clique, we conclude that w is adjacent to some vertex in S \ {u} ⊆ S , as desired.
Subcase 2.3. epn(u, D ∪ I) = ∅ and epn(u, I) = {w}. In this case, by (b3) we deduce that w is dominated by some vertex in S \ {u} ⊆ S , as desired.
According to the two cases above, we can conclude that S is a TDS of G, and so S is a STDS of G. Now, by the the minimality of |S|, we show that |S| ≤ |D ∪ I| + |D ∩ I| = |D| + |I|. Therefore, γ st (G) ≤ |S| ≤ |I| + |D| = α(G) + γ(G), which completes the proof.