Let $ \mathcal{H}$ be the class of all graphs with at most $ 2^{\aleph_0}$ vertices not containing a complete subgraph of size $ \aleph_1$. Show that there is no graph $ H \in \mathcal{H}$ such that every graph in $ \mathcal{H}$ is a subgraph of $ H$.
This is a problem from Miklos Schweitzer competition. I have no idea about starting it. How to proceed? thanks a lot.
Towards a contradiction, suppose $(V, E)$ is universal for $\mathcal{H}$. Define a new graph $(V_1, E_1)$ by letting $V_1 = \{f: (\exists \alpha < \omega_1)(f: \alpha \rightarrow V \text{ is one-one and range of } f \text{ is a complete subgraph of (V, E)})\}$ and $(f, g) \in E_1$ iff $f \cup g \in V_1$. It is easily checked that $(V_1, E_1) \in \mathcal{H}$ (the continuum size restriction is used here). Let $F:V_1 \rightarrow V$ be an injection witnessing the universality of $(V, E)$. Inductively, construct $\langle (f_{\alpha}, v_{\alpha}) : \alpha < \omega_1 \rangle$ such that for each $\alpha$, $F(f_{\alpha}) = v_{\alpha}$ and $f_{\alpha}$'s are increasing. Start by letting $f_0 = 0, v_0 = F(0)$. At stage $\alpha$, define $f_{\alpha}$ by $f_{\alpha}(\beta) = v_{\beta}$ for $\beta < \alpha$ and put $v_{\alpha} = F(f_{\alpha})$. But now $(V, E)$ has a complete subgraph: $\{v_{\alpha}: \alpha < \omega_1\}$.
