Every set-theoretic model of an algebraic theory gives rise to notion of (algebraically) "closed subset" in a canonical fashion; namely, the closed subsets are those that cannot be escaped via the operations of the algebra. There is also a corresponding closure operator. Now it is a central tenet of categorial logic that we may consider models of algebraic theories in a wide range of categories. Do such models also give rise to (algebraically) "closed subobjects" and a corresponding closure operator?
If so, I am especially interested in the following question. Are the (algebraically) closed subspaces of an algebraic structure internal to $\mathrm{Top}$ precisely those subspaces that are not only inescapable via the operations, but also, topologically closed?
In short, the answer is no: there exist subgroups of topological groups that are not closed as subspaces. Let $\hat{\mathbb{Z}}$ be the profinite completion of $\mathbb{Z}$, viz $\hat{\mathbb{Z}} = \varprojlim_{n} \mathbb{Z} / n \mathbb{Z}$. Then $\hat{\mathbb{Z}}$ is a compact Hausdorff topological group (by Tychonoff's theorem, say) and the canonical homomorphism $\mathbb{Z} \to \hat{\mathbb{Z}}$ is injective, but its image (which is not discrete) is not a closed subgroup of $\hat{\mathbb{Z}}$. (The closed subgroups of $\hat{\mathbb{Z}}$ are profinite, but there is no way of making $\mathbb{Z}$ into a profinite group.)