I'm reading through Dixon and Mortimer. Exercise 2.4.3 asks you to prove that if $|\Omega|<\infty$, then every subgroup of $\hbox{Sym}(\Omega)$ is closed. The definition of a closed permutation group $G$ is that it is the full automorphism group of a set of relations on $\Omega$.
My proof goes as follows. Assume w.l.o.g. $|\Omega|=n $ and let $H\leq \hbox{Sym}(\Omega)$ be arbitrary. I take the orbit under $H$ of the $n$-tuple $(1,2,\dots,n)$. Then this orbit $R$ is an $n$-ary relation, and if $R^{\sigma}=R$, we have in particular that for some $\tau\in H$ , $i^{\sigma} = i^{\tau}$ for all $i\in\Omega$. But then $\sigma=\tau\in H$, which means that any subgroup leaving $R$ invariant must be a subgroup of $H$.
My question is. Why wouldn't this work analogously in the infinite case? Is it disproved by some cardinality argument or did I make a mistake in my proof?