several whimsy fragments of propositions (1)

Bernstein’s theorem denotes that the existence of subsets such that X\sim Y',\ X'\sim Y,\ X'\subset X,\ Y'\subset Y is equivalent to X\sim Y. Here we have a good example of a real open interval (-1,1) having bijection to a real closed interval [-1,1] since f:(-1,1)\subset [-1,1];\ f(x)=x,\ g:[-1,1]\subset (-1,1);\ g(x)=x/2 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 F:(-1,1)\sim [-1,1] as:

    \[F(a)=\begin{cases} a & (a\in \displaystyle{\bigcup_{k=0}^\infty \{x\in R : 2^{-(k+1)}<|x|<2^{-k} \} }) \\ 2a & (a\in \displaystyle{\bigcup_{k=1}^\infty \{-2^{-k},2^{-k} \} }) \end{cases}\]

, 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 \{A_k\}_{1\leq k},\ \{B_k\}_{0\leq k} as following corresponding to two injection maps f and g.

    \[\begin{array}{lcl} B_0 = B-f((-1,1)) &=& \{-1,1\} \\ A_1 = g(B_0) &=& \{-1/2,1/2 \} \\ & \ddots & \\ A_n=B_n &=& \{-2^{-n},2^{-n}\} \end{array}\]

Let we define some symbols as A_*=\bigcup_{n=1}^\infty A_n,\ B_*=B_0\cup A_*,\ A^*=A_*^c,\ B^*=B_*^c({}^c indicates the complementary set), because f(A^*)=B^*,\ g(B_*)=A_* are generally established, we obtain bijections A^* \overset{F^*}{\sim} B^*,\ B_* \overset{G_*}{\sim} A_* by limiting the codomains. By definition, we know A=A_*\sqcup A^*,\ B=B_*\sqcup B^*(direct sum). Thus, the bijection has successfully constituted with maps F^*,\ G_*^{-1} depending on the belongings of an element of A.


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/I\bigotimes M\cong M/IM \cong (A/I)M.
A primary ideal I=(x)\subset A can be considered as an A-Module, let’s say M, the equation above concludes A/(x)\bigotimes (x)\cong (x)/(x^2) \cong (A/(x))(x). Two right modules are not trivial, yet the most left module is since x\in A leads to A/(x)\bigotimes (x) \ni \bar{a}\otimes x = x\bar{a}\otimes 1 = 0\ (\because xa\in (x)) 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.

A/(x)\bigotimes (x) 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 A\overset{u}{\cong} Ax (x is not a zero divisor), 1_A corresponds to x in (x). Thus we should cautiously note that A/(x)\bigotimes (x) is generated by a\otimes x\ (a\notin (x)) over A.

This denotes that if and only if a\in (x)\backslash \{0\}, nontrivial element of each space becomes zero as following:

    \[0=(a\mod x)\otimes x=ax \mod x^2 = (a\mod x)x\]

prop 2.15. In a direct system (M_i,\mu_{ij}) over direct set I, for any elements of the direct limit \displaystyle{\lim_\rightarrow M_i = M}, there exists i\in I and x_i\in M_i such that the element is represented as a form of \mu_i(x_i) (\mu_i is the restriction to M_i of a projection map). In addition to it, if \mu_i(x_i)=0 there exists j\geq i and M_j such that \mu_{ij}(xi)=0.

pf 2.15. Let we define some symbols as: C=\bigoplus_{i\in I}M_i,\ D=(x_i-\mu_{ij}(x_i))_{x_i\in M_i},\ \mu:C\rightarrow C/D=M. There is a representation such as \forall x\in M,\ \exists \{x_{\alpha_j}\}_{j\in \{1,\ldots, n\}},\ x = \mu(\sum_j^n x_{\alpha_j})=\mu(x'). Taken i\geq \alpha_j\ (\forall j\in \{1,\ldots,n\}), let y be y = \sum_j^n \mu_{\alpha_j,i}(x_{\alpha_j})-x_{\alpha_j} \in D, we obtain:

    \[x = \mu(x')=\mu(x'+y) = \mu(\sum_j^n \mu_{\alpha_j,i}(x_{\alpha_j})) = \mu_i(x_i)\ (x_i\in M_i)\]

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

    \[\exists k\geq i,j,\ \mu_{ik}(x_i)=\mu_{jk}(x_j) \Leftrightarrow \mu(x_i-x_j)=0\]

So we now prove this first. Taken k\geq i,j such that x_i\in M_i,\ x_j\in M_j,\ \mu_{ik}(x_i)=\mu_{jk}(x_j). We instantly achieve

    \[x_i-x_j = (x_i-\mu_{ik}(x_i)) - (x_j-\mu_{jk}(x_j)) \in D\ \cdots \ (*)\]

, thus \mu(x_i-x_j) = 0. Conversely \mu(x_i-x_j) = 0 denotes that there exist finite direct set Λ, an element of M_\alpha, and a family of homomorphism (x_\alpha, \mu_{\alpha,\beta{\alpha}})_{\alpha\in \Lambda} that meet the conditions of:

    \[x_i-x_j = \sum_{\alpha\in \Lambda} x_\alpha-\mu_{\alpha,\beta(\alpha)}(x_\alpha)\in D\]

(Here we don’t use x_i-x_j\in M_i+M_j but we do the left-hand side belongs to D).

By the way, when we take k\geq i,j,\alpha,\beta(\alpha), the homomorphism of A-Module \psi: \bigoplus_{\alpha\in \Lambda} M_\alpha \rightarrow M_k is compatibly defined as \psi(\sum_{\alpha\in \Lambda} x_\alpha) = \sum_{\alpha\in \Lambda} \mu_{\alpha k}(x_\alpha).

Then applying to the equation (*),

    \[\begin{array}{lcl} \mu_{ik}(x_i)-\mu_{jk}(x_j) &=& \sum_{\alpha\in\Lambda}\mu_{\alpha k}(x_\alpha)-\mu_{\beta(\alpha)k}\circ \mu_{\alpha\beta(\alpha)}(x_\alpha) \\ &=& \sum_{\alpha\in\Lambda}\mu_{\alpha k}(x_\alpha) - \mu_{\alpha k}(x_\alpha) \\ &=& 0 \end{array}\]

The lemma is completed above. Now when we assume \mu_i(x_i)=0, this denotes that \mu(x_i-0_k)=0 is achieved against for any zeros 0_k\in M_k (note: \mu_i is the restriction of \mu). The lemma denotes that there exists j in condition of j\geq k,i, we get \mu_{ij}(x_i)=\mu_{kj}(0_k)=0. The proof has been shown.