Let $G$ be a graph. A subset $C \subseteq V(G)$ is a vertex cover of $G$ if for each $e \in E(G)$, $e\cap C \neq \phi$. If $C$ is minimal with respect to inclusion, then $C$ is called minimal vertex cover of $G$.
Let $G$ be a finite simple undirected graph and $H$ be a subgraph of $G$ (not necessarily induced). Assume that $H$ contains at least one edge. Let $\mathfrak A$ be a minimal vertex cover of $H$. Is the following statement true?
$\exists$ $\mathfrak B\subseteq V(G)$ with $\mathfrak A\subseteq\mathfrak B$ such that $\mathfrak B$ is a minimal vertex cover of $G$.
My argument in support of the above statement is:
$G$ is obtained from $H$ by adding some edges. We first note that we don't need to consider those edges of $G$ whose at least one end point is in $\mathfrak A$. Consider $e\in E(G)\setminus E(H)$ such that both end points of $e$ are not in $\mathfrak A$. Take one end point of $e$ (say $x$) and consider $\mathfrak A'=\mathfrak A\cup\{x\}$. If $H'$ is the subgraph of $G$, where $V(H')=V(G)$ and $E(H')=E(H)\cup e\cup X$, where $X$ is the set of all edges of $G$ whose atleast one end point is in $\mathfrak A$, then $\mathfrak A'$ is a minimal vertex cover of $H'$.
Replacing $H$ by $H'$ and $\mathfrak A$ by $\mathfrak A'$ and continuing the above process we eventually get a minimal vertex cover of $G$ which contains $\mathfrak A$.
Is the above argument correct?
Thank you in advance.
The result is false.
Let $G$ be the line graph with vertices $a,b,c,d$ in order along the line.
Let $H$ be formed by deleting edge $b-c$. Then $\{a,d\}$ is a minimal cover set for $H$. However, the minimal cover sets for $G$ are $\{b,d\}$,$\{b,c\}$ and $\{a,c\}$.
$\mathfrak A\cup(V(G)\setminus V(H))$ is a vertex cover for $G$. Any subset which is also a cover set for $G$ must cover every edge of $H$. None of these edges have an endpoint in $V(G)\setminus V(H)$ and so must be covered by $\mathfrak A$. Therefore $\mathfrak A$ is in a minimal cover set for $G$.