Order 1 Simple Graph is Unique up to Isomorphism
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Theorem
Let $G_1 = \struct {\map V {G_1}, \map E {G_1} }$ and $G_2 = \struct {\map V {G_2}, \map E {G_2} }$ be simple graphs of order $1$.
Then $G_1$ and $G_2$ are isomorphic.
Proof
There is only one bijection from $\map V {G_1}$ to $\map V {G_2}$.
There are no vertices adjacent to the sole vertex in $\map V {G_1}$
There are no vertices adjacent to the sole vertex in $\map V {G_2}$.
Hence the result, vacuously.
$\blacksquare$
Sources
- 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.2$: Isomorphic Graphs