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 $G$. Is the following statement true?
$\exists$ $\mathfrak B\subseteq\mathfrak A$, $\mathfrak B\neq\emptyset$ such that $\mathfrak B$ is a minimal vertex cover of $H$.
My guess is that it is true. Argument:
Since $H$ can be obtained from $G$ by deleting some edges, $\exists$ some $\mathfrak B'\subseteq\mathfrak A$ such that $\mathfrak B'$ is a vertex cover of $H$ (since $V(H)\subseteq V(G)$). Since $\mathfrak B'$ is a vertex cover of $H$, $\exists$ $\mathfrak B\subseteq\mathfrak B'$ such that $\mathfrak B$ is a minimal vertex cover of $H$.
Is the above argument correct?
Thank you for your reply.