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 [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,

$${\gamma }_{st}\left(G\right)\le \alpha \left(G\right)+\gamma \left(G\right).$$

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 $\gamma \left(G\right)$-set. Let I be an $\alpha \left(G\right)$-set such that $|D\cap I|$ is at its maximum among all $\alpha \left(G\right)$-sets. Notice that for any $x\in D\cap I$,

$$epn(x,D\cup I)\cup ipn(x,D\cup I)\subseteq epn(x,I).$$

(1)

We next define a set $S\subseteq V\left(G\right)$ of minimum cardinality among the sets satisfying the following properties.

(a)

$D\cup I\subseteq S$.

(b)

For every vertex $x\in D\cap I$,

(b1)

if $epn(x,D\cup I)\ne ⌀$, then $S\cap epn(x,D\cup I)\ne ⌀$;

(b2)

if $epn(x,D\cup I)=⌀$, $ipn(x,D\cup I)\ne ⌀$ and $epn(x,I)\backslash ipn(x,D\cup I)\ne ⌀$, then either $epn(x,I)\backslash D=⌀$ or $S\cap epn(x,I)\backslash D\ne ⌀$;

(b3)

if $epn(x,D\cup I)=⌀$ and $epn(x,I)=ipn(x,D\cup I)\ne ⌀$, then $S\cap N\left(epn\right(x,I\left)\right)\backslash \left\{x\right\}\ne ⌀$;

(b4)

if $epn(x,D\cup I)=ipn(x,D\cup I)=⌀$, then $N\left(x\right)\backslash (D\cup I)=⌀$ or $S\cap N\left(x\right)\backslash (D\cup I)\ne ⌀$.

Since D and I are dominating sets, from (a) and (b) we conclude that S is a TDS. From now on, let $v\in V\left(G\right)\backslash S$. Observe that there exists a vertex $u\in N\left(v\right)\cap I\subseteq N\left(v\right)\cap S$, as $I\subseteq S$ is an $\alpha \left(G\right)$-set. To conclude that S is a STDS, we only need to prove that $S^{\prime }=(S\backslash \left\{u\right\})\cup \left\{v\right\}$ is a TDS of G.

First, notice that every vertex in $V\left(G\right)\backslash N\left(u\right)$ is dominated by some vertex in $S^{\prime }$, because S is a TDS of G. Let $w\in N\left(u\right)$. Now, we differentiate two cases with respect to vertex u.

• Case 1. $u\in I\backslash D$. If $w\notin D$, then there exists some vertex in $D\subseteq S^{\prime }$ which dominates w, as D is a dominating set. Suppose that $w\in D$. If $w\in ipn(u,D\cup I)$, then $I^{\prime }=(I\cup \left\{w\right\})\backslash \left\{u\right\}$ is an $\alpha \left(G\right)$-set such that $|D\cap I^{\prime }|>|D\cap I|$, which is a contradiction. Hence, $w\notin ipn(u,D\cup I)$, which implies that there exists some vertex in $(D\cup I)\backslash \left\{u\right\}\subseteq S^{\prime }$ which dominates w.
• Case 2. $u\in I\cap D$. We first suppose that $w\notin D$. If $w\notin epn(u,D\cup I)$, then w is dominated by some vertex in $(D\cup I)\backslash \left\{u\right\}\subseteq S^{\prime }$. If $w\in epn(u,D\cup I)$, then by $\left(\mathrm{b}1\right)$ and the fact that in this case all vertices in $epn(u,D\cup I)$ form a clique, w is dominated by some vertex in $S\backslash \left\{u\right\}\subseteq S^{\prime }$. From now on, suppose that $w\in D$. If $w\notin ipn(u,D\cup I)$, then there exists some vertex in $(D\cup I)\backslash \left\{u\right\}\subseteq S^{\prime }$ which dominates w. Finally, we consider the case in that $w\in ipn(u,D\cup I)$.

We claim that $ipn(u,D\cup I)=\left\{w\right\}$. In order to prove this claim, suppose that there exists $w^{\prime }\in ipn(u,D\cup I)\backslash \left\{w\right\}$. Notice that $w^{\prime }\in D$. By (1) and the fact that all vertices in $epn(u,I)$ form a clique, we prove that $w{w}^{\prime }\in E\left(G\right)$, and so $w\notin ipn(u,D\cup I)$, which is a contradiction. Therefore, $ipn(u,D\cup I)=\left\{w\right\}$ and, as a result,

$$epn(u,D\cup I)\cup \left\{w\right\}\subseteq epn(u,I).$$

(2)

In order to conclude the proof, we consider the following subcases.

Subcase 2.1. $epn(u,D\cup I)\ne ⌀$. By (2), $\left(\mathrm{b}1\right)$, and the fact that all vertices in $epn(u,I)$ form a clique, we conclude that w is adjacent to some vertex in $S\backslash \left\{u\right\}\subseteq S^{\prime }$, as desired.

Subcase 2.2. $epn(u,D\cup I)=⌀$ and $epn(u,I)\backslash \left\{w\right\}\ne ⌀$. By (2), $\left(\mathrm{b}2\right)$, and the fact that all vertices in $epn(u,I)$ form a clique, we show that w is dominated by some vertex in $S\backslash \left\{u\right\}\subseteq S^{\prime }$, as desired.

Subcase 2.3. $epn(u,D\cup I)=⌀$ and $epn(u,I)=\left\{w\right\}$. In this case, by $\left(\mathrm{b}3\right)$ we deduce that w is dominated by some vertex in $S\backslash \left\{u\right\}\subseteq S^{\prime }$, as desired.

According to the two cases above, we can conclude that $S^{\prime }$ is a TDS of G, and so S is a STDS of G. Now, by the the minimality of $\left|S\right|$, we show that $\left|S\right|\le |D\cup I|+|D\cap I|=\left|D\right|+\left|I\right|$. Therefore, ${\gamma }_{st}\left(G\right)\le \left|S\right|\le \left|I\right|+\left|D\right|=\alpha \left(G\right)+\gamma \left(G\right)$, which completes the proof. □

The authors would like to apologize for any inconvenience caused to the readers by these changes. The changes do not affect the scientific results.