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enter image description here∗Show that both the graphs above are isomorphic to the following graph: the vertex set is {1,2,...,5} (unordered pairs of numbers), and two vertices {i,j} and {k,l} (i,j,k,l ∈ {1,2,...,5}) form an edge if and only if {i, j} ∩ {k, l} = ∅.

In this question, I need some help with the notation. I think {i,j} means an edge, but it says vertex from the question. Also, {i,j} and {k,l} are disjoint if they were to form an edge, does this mean {1,2} and {3,4}, {1,5},{2,3}?

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    Looks like you aren't including the who definition. The vertex set is the unordered pairs of $\{1,2,3,4,5\}$. So there are $10$ vertices of this graph: $\{1,2\},\{1,3\},\dots,\{4,5\}$. The edges are then a pair of vertices, or a pair of pairs. – 2017-02-18
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    @ Thomas Andrews so, that means pick 2 pairs of unordered vertices, like {1,3} and {2,4}, as long as they are disjoint, then they form 2 edges respectively? – 2017-02-18
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    No, $v_1=\{1,3\}$ and $v_2=\{2,4\}$ are vertices, and the edge between them is $\{v_1,v_2\}=\{\{1,3\},\{2,4\}\}$. So there is only one edge. – 2017-02-18
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    @ Thomas Andrews Sorry to be so slow, I am influenced by my textbook and it points me to believe {1,3} is an edge. Now, in a graph, each vertice is only labeled as {1,2,3.....,n}, how can 4 vertices form 1 edge? please help me understand! – 2017-02-18
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    That's probably you misreading the text, but I can't tell, because (1) I don't know the text, and (2) you haven't told us the exact text. – 2017-02-18
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    I added the details, also the textbook is Invitation to Discrete Mathematics 2e. – 2017-02-18
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    Somehow the 10 vertex points are being mapped to 10 pairs {1,2}{1,3}{1,4}{1,5}{2,3}{2,4}{2,5}{3,4}{3,5}{4,5} and the edges are the pairs of of pairs that are disjoint ({1,2} {3,5}) is an edge ({3,4}{1,5}) is another. Each vertex belongs to three edges. The three edges of {2,5} for example are ({2,5},{1,3})({2,5},{1,4}) ad ({2,5}{3,4}). the graph representation is very counter-intuative and there is nothing in it to show a point representing a pair or how the lines between them represent the pair of pairs being dijoint (and a lack of a line represents having an intersection). – 2017-02-18
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    Okay, take the circle graph. Vertex in the center is {1,2}. It is connected to the three vertices in clockwise direction on the circle: {3,4}, {3,5}, and {4,5}. The nine verices on te circle in clockwise order are: {3,4}{2,5}{1,4}{3,5}{2,4}{1,3}{4,5}{2,3} and {1,5}. and there are 15 edges. ({1,2},{3,4}), ({1,2}{3,5}) and ({1,2},{4,5}) are the three edges from the center vertex. – 2017-02-18

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I think that in the expression $$ \{i,j\} \cap \{k,l\} $$ you are to think of the two pieces in braces as sets, not edges. That intersection is empty just when $$ i \ne k, i \ne l, j \ne k, j \ne l . $$

The graph has $10$ vertices, labeled by the $10$ pairs you can build from the set $\{1,2,3,4,5\}$. When thought of as vertices of the graph, $\{1,2\}$ is connected to $\{3,4\}$ by an edge, but not connected to $\{1,4\}$.

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    @ Ethan Bolker I understand this, however, I just don't know how to form an edge from 4 vertices. i,j,k,l, or how do I label a vertice with 2 numbers i,j? it just doesn't make sense to me. – 2017-02-18
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    i,j,k,l are not vertices. They aren't anything. The vertices are {i,j} and {k,l} and the edge is ({i,j},{k,l}) . A vertex is a pair of numbers. – 2017-02-18
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The Petersen graph is the complement of the line graph of $$K_{5}$$. It is also the Kneser graph $$KG_{5,2}$$; this means that it has one vertex for each 2-element subset of a 5-element set, and two vertices are connected by an edge if and only if the corresponding 2-element subsets are disjoint from each other.

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