This is a variation of an [older question](https://mathoverflow.net/questions/293529/continuous-functions-sharing-a-point) in a more general setting.


Let $V$ denote the set of all functions $f:\omega\to \omega$. We say $f,g\in V$ *share* a point if there is $x\in X$ such that $f(x) = g(x)$ and set $$E = \big\{\{f,g\}: f\neq g\in V\text{ and } f,g \text{ share a point}\big\}.$$ So $G:=(V,E)$ is a simple, undirected graph on $2^\omega$ vertices.


**Question.** If $G' = (2^\omega, E')$ is an uncountable simple undirected graph, is it isomorphic to some induced subgraph of $G$?



Graph of functions sharing a point
Let H be an uncountable independent set. Let G be the graph formed by joining every vertex of some small enough graph F to every vertex of H. (So F could be a singleton, and G could be bipartite if F were independent.) Suppose we could embed G into your shared point graph. Then no two vertices in H share a point, and so the set of points that a vertex f in F shares with an h in H is disjoint from that set shared by f and a distinct point h' in H. This would imply an uncountable nontrivial partition on $\omega$. So you can't have a vertex connected to all members of an uncountable independent set in such an embedded graph.


Gerhard "Reminds Me Of Math 274" Paseman, 2018.02.22


