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For $\alpha=\omega$ the answer is no. If $H=(\omega,E)$ is a linear hypergraph, then $E$ is countable; if $H=(\omega,E)$ is any hypergraph with $E\subseteq[\omega]^\omega$ and $E$ countable, then $\chi(H)\le2$.

For $3\le\alpha\lt\omega$ the answer is yes. It will be convenient to define the hypergraph on a countable vertex set $V\ne\omega$. Let $H=(V,E)$ where $V=[\omega]^{\alpha-1}$ and $E=\{[A]^{\alpha-1}:A\in[\omega]^\alpha\}\subseteq[V]^\alpha$. Then $H$ is a linear hypergraph, and $\chi(G)=\aleph_0$ because $\omega\to(\alpha)^{\alpha-1}_n\ (n\lt\omega)$ by the finite Ramsey theorem.

P.S. $\omega\to (m)^r_n$ is Rado's "arrow notation" for "partition relations"; it means that, for any $n$-coloring of the $r$-element subsets of $\omega$, there is an $m$-element subset of $\omega$ whose $r$-element subsets all have the same color.

For $\alpha=\omega$ the answer is no. If $H=(\omega,E)$ is a linear hypergraph, then $E$ is countable; if $H=(\omega,E)$ is any hypergraph with $E\subseteq[\omega]^\omega$ and $E$ countable, then $\chi(H)\le2$.

For $3\le\alpha\lt\omega$ the answer is yes. It will be convenient to define the hypergraph on a countable vertex set $V\ne\omega$. Let $H=(V,E)$ where $V=[\omega]^{\alpha-1}$ and $E=\{[A]^{\alpha-1}:A\in[\omega]^\alpha\}\subseteq[V]^\alpha$. Then $H$ is a linear hypergraph, and $\chi(G)=\aleph_0$ because $\omega\to(\alpha)^{\alpha-1}_n\ (n\lt\omega)$ by the finite Ramsey theorem.

P.S. $\omega\to (m)^r_n$ is Rado's "arrow notation" for "partition relations"; it means that, for any $n$-coloring of the $r$-element subsets of $\omega$, there is an $m$-element subset of $\omega$ whose $r$-element subsets all have the same color.

For $\alpha=\omega$ the answer is no. If $H=(\omega,E)$ is a linear hypergraph, then $E$ is countable; if $H=(\omega,E)$ is any hypergraph with $E\subseteq[\omega]^\omega$ and $E$ countable, then $\chi(H)\le2$.

For $3\le\alpha\lt\omega$ the answer is yes. It will be convenient to define the hypergraph on a countable vertex set $V\ne\omega$. Let $H=(V,E)$ where $V=[\omega]^{\alpha-1}$ and $E=\{[A]^{\alpha-1}:A\in[\omega]^\alpha\}\subseteq[V]^\alpha$. Then $H$ is a linear hypergraph, and $\chi(G)=\aleph_0$ because $\omega\to(\alpha)^{\alpha-1}_n\ (n\lt\omega)$ by the finite Ramsey theorem.

For $\alpha=\omega$ the answer is no. If $H=(\omega,E)$ is a linear hypergraph, then $E$ is countable; if $H=(\omega,E)$ is any hypergraph with $E\subseteq[\omega]^\omega$ and $E$ countable, then $\chi(H)\le2$.

For $3\le\alpha\lt\omega$ the answer is yes. It will be convenient to define the hypergraph on a countable vertex set $V\ne\omega$. Let $H=(V,E)$ where $V=[\omega]^{\alpha-1}$ and $E=\{[A]^{\alpha-1}:A\in[\omega]^\alpha\}\subseteq[V]^\alpha$. Then $H$ is a linear hypergraph, and $\chi(G)=\aleph_0$ because $\omega\to(\alpha)^{\alpha-1}_n\ (n\lt\omega)$ by the finite Ramsey theorem.

P.S. $\omega\to (m)^r_n$ is Rado's "arrow notation" for "partition relations"; it means that, for any $n$-coloring of the $r$-element subsets of $\omega$, there is an $m$-element subset of $\omega$ whose $r$-element subsets all have the same color.

For $\alpha=\omega$ the answer is no. If $H=(\omega,E)$ is a linear hypergraph, then $E$ is countable; if $H=(\omega,E)$ is any hypergraph with $E\subseteq[\omega]^\omega$ and $E$ countable, then $\chi(H)\le2$.

For $3\le\alpha\lt\omega$ the answer is yes. Let $H=(V,E)$ where $V=[\omega]^{\alpha-1}$ and $E=\{[A]^{\alpha-1}:A\in[\omega]^\alpha\}$. Then $H$ is a linear hypergraph, and $\chi(G)=\aleph_0$ because $\omega\to(\alpha)^{\alpha-1}_n\ (n\lt\omega)$, which is part of the finite Ramsey theorem.

For $\alpha=\omega$ the answer is no. If $H=(\omega,E)$ is a linear hypergraph, then $E$ is countable; if $H=(\omega,E)$ is any hypergraph with $E\subseteq[\omega]^\omega$ and $E$ countable, then $\chi(H)\le2$.

For $3\le\alpha\lt\omega$ the answer is yes. It will be convenient to define the hypergraph on a countable vertex set $V\ne\omega$. Let $H=(V,E)$ where $V=[\omega]^{\alpha-1}$ and $E=\{[A]^{\alpha-1}:A\in[\omega]^\alpha\}\subseteq[V]^\alpha$. Then $H$ is a linear hypergraph, and $\chi(G)=\aleph_0$ because $\omega\to(\alpha)^{\alpha-1}_n\ (n\lt\omega)$ by the finite Ramsey theorem.