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First of all, sorry for my English, especially mathematical one. The problem is that I know how to call such things in Ukrainian but unfortunately did not manage to find proper translations to English. So, I will give some brief definitions of terms in order to not confuse you with the names.

So, an algebra (maybe, partial) is a set of elements together with a set of operations defined on these elements $ (A, \Omega) $.

Let $ B \subset A $. Then a closure $[B]_f$ of $A$ by an $n$-ary operation $f \in \Omega$ is a set defined by two rules:

  1. $ B \subseteq [B]_f$
  2. $ \forall (a_1, a_2, \ldots, a_n) \subset [B]_f $ if $f(a_1, a_2, \ldots, a_n)$ is defined then $f(a_1, a_2, \ldots, a_n) \in [B]_f$

A closure $[B]$ of $A$ is $ B^0 \cup B^1 \cup B^2 \ldots $ where $B^0 = B$, $B^{i+1} = \bigcup_{f \in \Omega}[B^i]_f$.

The definition of subalgebra and therefore algebra extension is obvious, I think.

$B$ is a system of generators (I believe, it is close to the basis) for algebra $(A,\Omega)$ if $[B] = A$.

And, finally, subalgebra $(B,\Omega)$ is called maximal subalgebra of $(A,\Omega)$ if there is no subalgebra $(C,\Omega)$ such that $B \subset C \subset A, B \neq A, C \neq A, C \neq B$.

It can be proved that for a algebra with existing finite system of generators each its subalgebra can be extended to some maximal subalgebra. But for algebras with infinite systems of generators, this statement is not always true. And I need a counter-example. Therefore, I need an example of infinite-based algebra and a subalgebra which is impossible to extend to some maximum subalgebra.

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    I thought about infinite-based vector spaces and replacements of all vectors with i-th coordinate equal to some number (maybe, to all numbers <=i) for each i, but I didn't manage to come up with something.2012-02-26
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    Unfortunately, every proper subspace $W$ of a vector space $V$ is contained in a maximal (I assume by this you also mean it doesn't equal the whole thing) subspace; just take a vector $v \in V$ not in $W$, let $B$ be a basis of $W$, extend $B \cup \{ v \}$ to a basis of $V$, and consider the span of this basis minus $v$.2012-02-26
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    Your "creators" are usually called "generators".2012-02-26
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    Anyway, I would recommend you try thinking about groups instead of vector spaces. In fact there is a relatively straightforward example of an abelian group which has no maximal proper subgroups whatsoever.2012-02-26
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    @g.castro: thanks2012-02-26
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    @QiaochuYuan: Thank you, I will think of it.2012-02-26
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    @QiaochuYuan: I believe, you mean an additive group of rational numbers which does not have maximal subgroups at all.2012-02-26

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