Bernstein’s theorem denotes that the existence of subsets such that is equivalent to . Here we have a good example of a real open interval (-1,1) having bijection to a real closed interval [-1,1] since are both injective in each direction. We soon know that the specific construction of the bijectional map is commonly not very easy.

As the consequence, we can find such a bijection as:

, where the map is apparently discontinuous and somewhat complex.

With this constitution (as in the proof of Bernstein’s Theorem), we define families of sets as following corresponding to two injection maps f and g.

Let we define some symbols as ( indicates the complementary set), because are generally established, we obtain bijections by limiting the codomains. By definition, we know (direct sum). Thus, the bijection has successfully constituted with maps depending on the belongings of an element of A.

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I had an unsolved problem until right before I wrote this down a minutes ago.

Let A be a commutative ring, M an A-Module, A-Module isomorphisms exist as following: .

A primary ideal can be considered as an A-Module, let’s say M, the equation above concludes . Two right modules are not trivial, yet the most left module is since leads to with the use of bilinearity. In other words, any arbitrary ideal of A is a set of nilpotents, namely, the ring A is absolutely flat? This seemed an odd result.

is a tensor product over A, which allows only an element of A to be transferred to the other space; the `x’ of (x) should not be regarded as an element of A when we see (x) as an A-Module, although x is an element of A for the following reason. As an A-Module (x), we accept (x) is equivalent to Ax. For A-Module isomorphism (x is not a zero divisor), corresponds to x in (x). Thus we should cautiously note that is generated by over A.

This denotes that if and only if , nontrivial element of each space becomes zero as following:

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prop 2.15. In a direct system over direct set I, for any elements of the direct limit , there exists and such that the element is represented as a form of ( is the restriction to of a projection map). In addition to it, if there exists and such that .

pf 2.15. Let we define some symbols as: . There is a representation such as . Taken , let y be , we obtain:

The second half is realized with the help of a lemma:

So we now prove this first. Taken such that . We instantly achieve

, thus . Conversely denotes that there exist finite direct set Λ, an element of , and a family of homomorphism that meet the conditions of:

(Here we don’t use but we do the left-hand side belongs to D).

By the way, when we take , the homomorphism of A-Module is compatibly defined as .

Then applying to the equation (*),

The lemma is completed above. Now when we assume , this denotes that is achieved against for any zeros (note: is the restriction of ). The lemma denotes that there exists j in condition of , we get . The proof has been shown.