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My question is inspired by the fact that there is no universal set (at least in ZF). There are many abstract objects such as group, ring, field, vector space, topology, etc. such that we can say about embedding. In a sense, we can interpret $A$ embeds $B$ as $A$ 'contains' $B$. So it seems that there is no such universal objects (universal group/ring/field...) that embeds every possible objects (group/ring/field ..). How can I prove/disprove it?

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What exactly do you mean: do you want a foo (group/ring/field...) x such that every other foo is a sub-foo of x or that every other foo is an element of x? – Miha Habič Aug 25 '11 at 11:20
@Ilmari Karonen Thanks for grammar correction! I'm still a novice in English so it helps me a lot. – Jineon Baek Aug 25 '11 at 11:21
There is in fact a ‘universal’ ring, but it's not quite the thing you suggest. Rather, it is a ring object in the classifying topos of the theory of rings, and every ring in every topos is the image of the universal ring under a unique geometric morphism. – Zhen Lin Aug 25 '11 at 11:40
If you are working in ZFC, remember that every infinite set can be made into a ring. In particular, if $A$ was a universal ring, how would you embed $P(A)$ into it? – Asaf Karagila Aug 25 '11 at 11:53
every finite group is a subgroup of $\cup S_n$ – yoyo Aug 25 '11 at 22:05
up vote 5 down vote accepted

There are cardinality limits in these cases, too. That is, there exist groups rings fields topologies (etc.) of arbitrarily large cardinal. So no " universal" ones exist.

But for example there are nice things like: a universal separable metric space: that is, a separable metric space such that any other separable metric space is isometric to a subset of it. Or a universal countable group. And so on.

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See e.g. – t.b. Aug 25 '11 at 12:21

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