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.

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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: $A/I\bigotimes M\cong M/IM \cong (A/I)M$ .
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:

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

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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.

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