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There are many examples throughout mathematics of abstracting the formal properties of a "familiar" structure, but then having a theorem stating that all models of the abstract axioms embed into one of the original "familiar" structures. Examples include

  • every group is a subgroup of a symmetric group;
  • every small category is concretizable (essentially by the Yoneda lemma);
  • every manifold smoothly embeds into real space;
  • every abelian category is a full subcategory of a category of modules, with exact inclusion;
  • every topos can be embedded in a Boolean one;
  • ...

Is there such an embedding theorem for [some nice subcategory of] the category of topological spaces? Here I am not sure what the "familiar" structures would be.

More generally, can we formulate a metatheorem on when "every object behaving formally like a concrete structure embeds into one"?

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Are you asking for theorems of the form "every nice topological space embeds into some even nicer topological space" or for theorems of the form "every nice subcategory of $\text{Top}$ embeds into some even nicer category"? – Qiaochu Yuan May 26 '13 at 4:38
One theorem of the first form is "every second-countable Tychonoff space embeds into $[0, 1]^{\mathbb{N}}$." – Qiaochu Yuan May 26 '13 at 4:40
The concept of universa space is possibly what you are looking for. $X$ is universal for a class of topological space if each of them embeds into it as a clossed subspace. This plays an important role, particularly in descriptive topology. As a starting point you could consult "Analytic sets" by C.A. Rogers (ed.) for the following important universal spaces---countable products of the closed unit interval (already mentioned), the integers (i.e., the irrationals), the two point space (the Cantor set) and the real line. – jbc May 26 '13 at 4:54
I think this question should be Community Wiki. – André Henriques May 26 '13 at 10:25
It's not true that every topos can be embedded in a boolean one – that would imply every topos is boolean! Rather, every topos can be covered by a boolean one. – Zhen Lin May 26 '13 at 10:50

3 Answers 3

up vote 1 down vote accepted

I think this question is closely related to reflective subcategories and the adjoint functor theorems. Adjoint functors do not guarantee you to have real “embeddings”, however, usually this part is easy to prove.

Let me restate some results given by Joseph Van Name:

The category of compact Hausdorff spaces is a reflective subcategory of $\mathbf{Top}$. It follows from the special adjoint functor theorem by observing that $[0,1]$ is a cogenerator of compact Hausdorff spaces. The proof of the SAFT gives you an embedding (only in the completely regular case (where $[0,1]$ is still a cogenerator) it is strictly an embedding) into a product of unit intervals.

The category of Banach spaces with linear contractions as morphisms is a reflective subcategory (by only considering the unit balls) of the category of bounded metric spaces. The real numbers are a cogenerator in the category of Banach spaces (by the Hahn-Banach theorem), the SAFT gives you an embedding into a product of copies of $\mathbb{R}$ (which is an $\ell^\infty$-space in the category of Banach spaces).

$\left\{0,1\right\}$ is a cogenerator of totally disconnected spaces. By the SAFT you get that the category of Stone spaces (compact, totally disconnected, Hausdorff) is a reflective subcategory of $\mathbf{Top}$. Well, for non-totally-disconnected spaces your adjoint does not give you an embedding, but the corresponding Stone space still carries the information about the Boolean algebra of clopen sets, thus it remains useful.

The category of compact topological groups is a reflective subcategory of the category of topological groups: Topological groups can be “embedded” (it is not really an embedding) into its Bohr compactification. Since there is no small cogenerating set, you cannot apply the special adjoint functor theorem, thus you do not get the Bohr compactification as a product of very simple groups, but you still get a nice, compact group carrying all the information about finite-dimensional unitary representations of your original group.

To provide a general theorem: If you have a complete, locally small category with a cogenerator $I$, then for every object $X$ the map given by $\prod_{f\in\mathrm{Hom}(X,I)} f$ is a monomorphism. However, this is not enough in the topological case, since we want to have embeddings=extremal monomorphisms, not just injections=monomorphisms. The adjoint functor theorems are probably the most fruitful generalisation of such a criterion.

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Products of topological spaces give us nice embedding characterizations of spaces. Let $X$ be a topological space. Then we say that a topological space $Y$ is $X$-compact if $Y$ is isomorphic to a closed subspace of some product $X^{I}$, so the products $X^{I}$ are in some sense the universal $X$-compact spaces. We say that a space $Y$ is $X$-completely regular if $Y$ is isomorphic to a subspace of some product $X^{I}$. Some nice topological properties may be characterized using $X$-compactness and $X$-complete regularity.

For instance, the $[0,1]$-completely regular spaces are simply the completely regular spaces. The $\{0,1\}$-completely regular spaces are the zero-dimensional spaces. A topological space is compact and Hausdorff if and only if it is $[0,1]$-compact. A topological space is a Boolean space(i.e. compact and zero-dimensional) if and only if it is $\{0,1\}$-compact. A topological space whose cardinality is below the first measurable cardinal is realcompact ($\mathbb{R}$-compact spaces are called realcompact spaces) if and only if it can be given a compatible complete uniformity. Furthermore, a topological space whose cardinality is below the first measurable cardinal is $\mathbb{N}$-compact if and only if it can be given a compatible complete non-Archimedean uniformity (a non-Archimedean uniform space is a uniform space generated by equivalence relations). I must mention that the restriction that the cardinality of your space is below the first measurable cardinal is a very minor restriction. The existence of a measurable cardinal goes beyond the standard axioms of set theory. Furthermore, measurable cardinals are extremely large if they do exist.

On a related note, every metric space can be isometrically embedded into some space $\ell^{\infty}(A)$ for some set $A$ (See Isbells book on Uniform Space for a proof). Therefore the spaces $\ell^{\infty}(A)$ are in a sense the universal metric spaces. In particular, since every metric space is uniformly homeomorphic to a bounded metric space, every metric space is uniformly homeomorphic to a subspace of the unit ball of $\ell^{\infty}(A)$. Since every uniform space is induced by a family of pseudometrics, we conclude that every uniform space is uniformly homeomorphic to a subspace of a product of the unit balls in some $\ell^{\infty}(A)$. In particular, a topological space is can be given a compatible complete uniformity if and only if it is isomorphic to some closed subspace of some product $\ell^{\infty}(A)^{I}$.

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Every $T_0$ space is, in a canonical way, a subspace of an algebraic lattice equipped with its Scott topology. Some details can be found in this nLab article.

I'm not quite sure what to do with the general question. Some possible inspiration might come from the study of Chu constructions -- see for examples these notes by Vaughan Pratt. Many familiar mathematical objects can be realized concretely as Chu spaces in one way or another.

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