I am revising my lecture notes about connectivity, but I am stuck regarding proof of $κ(Q_d) = d$ Then I took a look of the proof by induction in D. West's book. For $d\leq1$, $Q_d$ is a clique with $k+1$ vertices and connectivity $k$.Can anyone visualize that statement about $k+1$ vertices? Also, it would be great if you have another proof or explanation to articulate my understanding of it. Thanks
closed as offtopic by YCor, bof, Chris Godsil, Mark Wildon, Ben Barber Nov 27 '18 at 10:30
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$\begingroup$ Did you mean "$Q_d$ is a clique with $d+1$ vertices and connectivity $d$? $\endgroup$ – bof Nov 24 '18 at 12:52

$\begingroup$ What is hard to visualize about the graphs $Q_0=K_1$ and $Q_1=K_2$? $\endgroup$ – bof Nov 24 '18 at 12:53
You can connect any two vertices by $d$ vertexdisjoint paths directly. If they differ in $k$ coordinates, $k$ paths will be in the $k$cube in which they are opposite (use coordinates in the cyclic order). The rest are formed by moving along extra coordinate axis to an adjacent $k$cube, going to the opposite vertex in it, and moving along the extra axis backwards.
We can define the Hypercube $Q_{k}$ recursively, as follows. We let $Q_{0}=K_{1}$ and $Q_{k}=Q_{k1}\square K_{2}$, for all $k\geq1$. Let us observe that $Q_{k}$, $k\geq1$, is obtained from two disjoint copies of $Q_{k1}$ by joining their corresponding vertices.
Now, we prove that $\kappa(Q_{k})=k$. We proceed by induction on $k$. The equality is obvious for $k=0$ and $k=1$. In such cases the Hypercube is isomorphic to $K_{1}$ and $K_{2}$, respectively. Now let we have the equality for all Hypercubes of order less that $k$ and prove the equality for a Hypercube of order $k$. We now consider the Hypercube $Q_{k}$. It is wellknown that $\kappa(G)\leq \delta(G)$, for all graphs $G$. Therefore, \begin{equation}\label{Q1} \kappa(Q_{k})\leq \delta(Q_{k})=k. \end{equation} Suppose to the contrary that $\kappa(Q_{k})\leq k1$. Let $G_{1}$ and $G_{2}$ be those two copies of $Q_{k1}$ as subgraphs of $Q_{k}$. Let $S$ be a $\kappa(Q_{k})$set. We consider two cases.
Case 1. $S\subseteq V(G_{1})$ or $S\subseteq V(G_{2})$. By symmetry we may assume that $S\subseteq V(G_{2})$. Since $S\leq k1\leq2^{k1}$, the graph $GS$ is obtained from a copy of $Q_{k1}$ by adding $2^{k1}S$ new vertices such that every new vertex has exactly one neighbor in $V(Q_{k1})$ (note that there are some possible edges among the new vertices). So, $GS$ is connected which is a contradiction..
Case 2. $S\cap V(G_{1})\neq\emptyset$ and $S\cap V(G_{2})\neq\emptyset$. Then, $S\cap V(G_{1})<k1$ and $S\cap V(G_{2})<k1$. Therefore, both $G_{1}'=G_{1}(S\cap V(G_{1}))$ and $G_{2}'=G_{2}(S\cap V(G_{2}))$ are connected by the induction hypothesis. On the other hand, there are some edges with one end point in $V(G_{1}')$ and the other in $V(G_{2}')$. This shows that $GS$ is connected, a contradiction.
Therefore, $\kappa(Q_{k})=k$.