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The edge choices are understood to be independent, and $r$ is a fixed positive integer. We prove the hypergraph extension property and then show directly that it determines a unique countable hypergraph isomorphism type.

For a finite $S\subseteq \mathbb{N}$, prescribe a family $\mathcal{A}\subseteq \left(\genfrac{}{}{0ex}{}{S}{r-1}\right)$. We want a new vertex $v$ such thatFor any $v\notin S$, the probability of precisely this pattern in the countable random uniform hypergraph isTwo different candidate vertices use disjoint hyperedge decisions: all the other vertices in each tested edge lie in $S$. The successes are therefore independent random variables with the same positive probability $q$. The probability that the first $N$ candidates all fail is $(1-q{)}^N\to 0$. The same argument after discarding any finite initial set of candidates shows that infinitely many witnesses exist almost surely.

There are only countably many finite subsets of $\mathbb{N}$, and finitely many prescriptions for each. A countable union of null events is null. Consequently the random uniform hypergraph satisfies every such extension prescription simultaneously almost surely.

Now let $H$ and $K$ be two countably infinite $r$-uniform hypergraphs with the hypergraph extension property. Enumerate their vertex sets and construct increasing finite partial hypergraph isomorphisms by the back-and-forth method. At a forward step, take the least unused vertex $u$ of $H$. Its incident hyperedges with each $(r-1)$-subset of the current domain specify a pattern on the matched vertices of $K$. The hypergraph extension property supplies an unused vertex $v$ of $K$ with exactly this pattern. Adding $u\mapsto v$ preserves and reflects every hyperedge in the enlarged domain. At the next step, take the least unused vertex of $K$ and perform the identical construction with $H$ and $K$ interchanged.

The union of the finite maps is defined on every vertex and has every vertex in its range. It is a bijection, and each finite $r$-set appears at some finite stage, so all hyperedges and nonedges are preserved. Thus it is a hypergraph isomorphism. Existence of at least one structure with the property follows from the probability-one construction already proved; fix one representative and call it $G_{\mathrm{univ}}^{(r)}$. ThereforeThe hypergraph isomorphism type does not depend on $p$. In the case $r=2$ it is the Rado graph. For $r=1$, the argument says that the infinitely many selected and unselected singleton vertices determine the same one-uniform structure for every $p\in (0,1)$.

The universal countable uniform hypergraph is the unique, up to hypergraph isomorphism, countably infinite $r$-uniform hypergraph with the hypergraph extension property. To prove uniqueness, enumerate two such structures and construct finite partial hypergraph isomorphisms. Alternately add the least unused vertex of the first enumeration to the domain and the least unused vertex of the second to the range. Prescribe its incidences with every $(r-1)$-subset of the already matched vertices and apply the extension property in the other structure. The resulting union is a bijection preserving every hyperedge and nonedge. Existence follows from the countable random uniform hypergraph construction. A construction with only the forward steps embeds every countable $r$-uniform hypergraph as an induced subhypergraph, explaining the word universal. Its isomorphism type is independent of $p\in (0,1)$.