# Topological Disconnected Graphs

How can I define a topology on a complete graph such that the connectedness of the subgraphs of it and the connectedness of the sets on the topological space become equivalent?

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What's wrong with the usual way way to think of a graph as a topological space? You have a point for each vertex, and given vertices $v_1$ and $v_2$, if $[v_1,v_2]$ is an edge of the graph, then you attach an interval $[0,1]$ to $\{v_1,v_2\}$ by identifying $v_1$ with $0$ and $v_2$ with $1$. –  Arturo Magidin Jan 23 '12 at 19:53
There is nothing wring with it, but it doesn't actually give the solution to the problem that I want to solve. The question is to show either a graph or it's complement is connected. I know that there is a classical graph theoretic proof for this theorem, but I want to show it by defining a topology on the complete graph of order $n$ and use connectedness of $K_n$. @ArturoMagidin –  Keivan Feb 15 '12 at 0:10