
> 
> Is it the case that, for any given knot $K$,
>  there exists some graph $G$ whose every embedding into $\mathbb{R}^3$
>  (or into $\mathbb{S}^3$)
>  contains a cycle that realizes $K$?
> 
> 
> 


I know the famous Conway-Gordon result that the complete graph
on seven vertices $K\_7$ is intrinsically knotted in that every
embedding contains a knotted cycle.
And Joel Foisy proved that $K\_{3,3,1,1}$ is also intrinsically knotted
(*J. Graph Theory*, 2002, [ACM link](http://dl.acm.org/citation.cfm?id=1380507)).


Instead, I am asking a Ramsey-like question: 
Whether it's known that, for every particular knot, there exists some
graph for which that particular knot is unavoidable.



Is every knot unavoidable in the embeddings of some graph?
**Yes.** See this [paper](http://www.ams.org/journals/tran/1991-324-02/S0002-9947-1991-1069741-9/S0002-9947-1991-1069741-9.pdf) of Negami. The main result is that for any fixed knot (or link) of type $k$, there is a constant $R(k)$ such that every straight line embedding of $K\_{R(k)}$ in $\mathbb{R}^3$ contains a knot (or link) equivalent to $k$. 


