phaiakia

FRA.1.3.14

noreply@blogger.com (Lindsey K. Gamard) — Sat, 25 Jan 2014 01:41:00 +0000

FRA.1.3.8

noreply@blogger.com (Lindsey K. Gamard) — Sat, 25 Jan 2014 01:38:00 +0000

FRA.1.2.3

noreply@blogger.com (Lindsey K. Gamard) — Sat, 25 Jan 2014 01:35:00 +0000

DFAA.10.2.12

noreply@blogger.com (Lindsey K. Gamard) — Sat, 25 Jan 2014 01:30:00 +0000

DFAA.10.2.10

noreply@blogger.com (Lindsey K. Gamard) — Sat, 25 Jan 2014 01:27:00 +0000

DFAA.10.2.9

noreply@blogger.com (Lindsey K. Gamard) — Sat, 25 Jan 2014 01:26:00 +0000

DFAA.10.2.4

noreply@blogger.com (Lindsey K. Gamard) — Sat, 25 Jan 2014 01:24:00 +0000

DFAA.10.1.19

noreply@blogger.com (Lindsey K. Gamard) — Sat, 25 Jan 2014 01:21:00 +0000

DFAA.10.1.8

noreply@blogger.com (Lindsey K. Gamard) — Sat, 25 Jan 2014 01:18:00 +0000

DFAA.10.1.9

noreply@blogger.com (Lindsey K. Gamard) — Wed, 15 Jan 2014 16:54:00 +0000

Solution. Let $N, \; M, \; R$ be as in the problem statement. To show that $A = \{ r \in R \mid rn = 0 \text{ for all } n \in N \}$, the annihilator of $N$ in $R$, is a 2-sided ideal of $R$, we must show that $A$ is a subring of $R$ (i.e., $A$ is a subgroup of $R$ that is closed under multiplication), and that $A$ satisfies the left and right absorption properties. First, since $0m = 0$ for all $m \in M$ ($\dagger$), and $N \subseteq M$, then $0 \in A$, so $A$ is nonempty. Now let $r,s \in A$. Then for all $n \in N$, since distribution holds in a module, $$(r-s)n = rn - sn = 0-0 = 0,$$ so $A$ satisfies the subgroup criterion, and $A$ is a subgroup of the abelian group $R$ under addition. Furthermore, for all $n \in N$, $$(rs)n = r(sn) = r(0) = r(n - n) = rn - rn = 0 - 0 = 0,$$ and likewise $(sr)n = 0$, so $A$ is closed under the multiplication in $R$. Thus $A$ is a subring of $R$. It remains to show that for all $r \in R$ and $a \in A$, $ra \in A$ and $ar \in A$. We observe that for any arbitrary $n \in N \subseteq M$, $$(ra)n = r(an) = r(0) = r(n - n) = rn - rn = m - m = 0,$$ since $rn = m$ for some $m \in M$. Furthermore, for all $r \in R$, $rn \in N$ by definition. Let $rn = \tilde{n}$. Then $$(ar)n = a(rn) = a \tilde{n} = 0,$$ since $a \in A$. Thus $A$ satisfies the left and right absorption properties, so $A$ is an ideal of $R$. $\square$

$\dagger$ Since $R$ is a ring with 1, and $R$ is an abelian group under addition, then $m = 1m = (1+0)m = 1m + 0m = m + 0m$. Subtracting $m$ from the far right and far left sides of this equation yields $0m = 0$.

DFAA.10.1.7

noreply@blogger.com (Lindsey K. Gamard) — Wed, 15 Jan 2014 16:11:00 +0000

Solution. Let $R, \; M,$ and each $N_i$ be as in the problem statement. By the Submodule Criterion, it suffices to show that $N = \bigcup_{i = 1}^{\infty} N_i$ is nonempty, and that $x+ry \in N$ for all $r \in R$ and for all $x,y \in N$. Since each $N_i$ is a submodule of $M$, then each $N_i$ is nonempty, so for each $i$, there is some $x_i \in N_i$. It follows immediately that since $N$ is the union of all the $N_i$'s, that $x_i \in N$ for all $i$, so $N$ is nonempty. Now let $x,y \in N$ and $r \in R$ be arbitrary. Then $x \in N_j$ and $y \in N_k$ for some $k,j \in \N$. Let $\ell = \max \{ j,k \}$; it follows that $x,y \in N_{\ell}$ since $N_j \subseteq N_{\ell}$ and $N_k \subseteq N_{\ell}$. Since $N_{\ell}$ is a submodule of $M$, then $x+ry \in N_{\ell}$. Furthermore, $N_{\ell} \subseteq N$, so $x+ry \in N$. Therefore, $N$ is a submodule of $M$. $\square$

FRA.1.2.2

noreply@blogger.com (Lindsey K. Gamard) — Wed, 15 Jan 2014 15:11:00 +0000

$\dagger$ A proof that every open set $F \in \mathcal{F}$ can be written as the countable union of open intervals can be found at: http://math.stackexchange.com/questions/318299/any-open-subset-of-bbb-r-is-a-countable-union-of-disjoint-open-intervals-co

DFAA.10.1.2

noreply@blogger.com (Lindsey K. Gamard) — Tue, 14 Jan 2014 04:54:00 +0000

Solution. Let $R$ be a ring with 1 and let $M$ be a left $R$-module. Recall that $R^{\times}$ is a subset of $R$, and that $R^{\times}$ is a group under multiplication. Define the group action of $R^{\times}$ on $M$ by $r \cdot m = rm$ for $r \in R^{\times}$ and $m \in M$ (i.e. restrict the mapping $R \times M \to M$ from the definition of a module to $R^{\times} \times M \to M$). By the definition of a unital module, for all $r,s \in R$ and $m \in M$, $(rs)m = r(sm)$ and $1m = m$. It follows that $$\begin{align*} r \cdot (s \cdot m) &= r \cdot (sm) \\ &= r(sm) \\ &= (rs)m \\ &= (rs) \cdot m, \end{align*}$$ and that $$\begin{align*} 1 \cdot m &= 1m \\ &= m, \end{align*}$$ as desired. $\square$

DFAA.10.1.4

noreply@blogger.com (Lindsey K. Gamard) — Tue, 14 Jan 2014 04:32:00 +0000

Solution to (a). Let $M$ be the module $R^n$ as in Example 3, and let $I_1,I_2,\dots,I_n$ be left ideals of $R$. We wish to show that the set $N_a = \{ (x_1,x_2,\dots,x_n) \mid x_i \in I_i \}$ is a submodule of $M$. By the Submodule Criterion (Prop. 1), it suffices to show that $N_a \neq \emptyset$ and that for all $r \in R$ and $x,y \in N_a$, $x+ry \in N_a$. First, since each $I_i$ is an ideal, it is nonempty, and so $0 \in I_i$ for all $i$. Thus, $(0,0,\dots,0) \in N_a$, so $N_a \neq \emptyset$. Denote $x = (x_1,x_2,\dots,x_n) \in N_a$ and $y = (y_1,y_2,\dots,y_n) \in N_a$. Since each $I_i$ is a left ideal, then $ry_i \in I_i$ for each $i$, and it follows that $ry = (ry_1,ry_2,\dots,ry_n) \in N_a$. Furthermore, since each $I_i$ is a subgroup of the abelian group $R$ under addition, then each $I_i$ is closed under addition. Hence $x_i + ry_i \in I_i$ for each $i$, and it follows that $(x_1+ry_1,x_2+ry_2,\dots,x_n+ry_n) \in N_a$. Since $x+ry = (x_1+ry_1,x_2+ry_2,\dots,x_n+ry_n)$, this shows that $N_a$ is a submodule of $M$. $\square$

Sketch of (b). Let $N_b = \{ (x_1,x_2,\dots,x_n) \mid x_i \in R \text{ and } x_1+x_2+ \cdots +x_n = 0 \}$. Since $0+0+ \cdots +0 = 0$, then $N_b \neq \emptyset$. Also, for any $r \in R$, $x,y \in N_b$, $$\begin{align*} (x_1+ry_1) + (x_2+ry_2) + \cdots + (x_n+ry_n) &= (x_1+x_2+ \cdots + x_n) + r(y_1+y_2+ \cdots + y_n) \\ &= 0 + r(0) \\ &= 0, \end{align*}$$ so $x+ry \in N_b$. $\square$

DFAA.10.1.1

noreply@blogger.com (Lindsey K. Gamard) — Tue, 14 Jan 2014 03:42:00 +0000

Solution. Let $R$ be a ring with 1 and let $M$ be a left $R$-module. Using the definition of a module, $m = 1m = (0+1)m = 0m + 1m = 0m + m$, and subtracting $m$ from both the far left and far right sides yields $0 = 0m$. Now, $0 = 0m = (1+-1)m = 1m + (-1)m = m + (-1)m$, and it follows that $0-m = m + (-1)m - m$, so $-m = (-1)m$. $\square$

SP.I.1.3

noreply@blogger.com (Lindsey K. Gamard) — Wed, 08 Jan 2014 04:36:00 +0000

Solution. The proof is by induction. For our base case, let $n = 1$. Clearly $\mathsf{P}(A_1) \leq \mathsf{P}(A_1)$. Now we assume that $$\mathsf{P}(A_1 \cup \cdots \cup A_n) \leq \mathsf{P}(A_1) + \cdots + \mathsf{P}(A_n),$$ and we must prove that $$\mathsf{P}(A_1 \cup \cdots \cup A_n \cup A_{n+1}) \leq \mathsf{P}(A_1) + \cdots +\mathsf{P}(A_n) + \mathsf{P}(A_{n+1}).$$

Denote $B = A_1 \cup \cdots \cup A_n$. We observe that $B \cup A_{n+1} = (B \backslash A_{n+1}) + (A_{n+1} \backslash B) + A_{n+1}B$ and that $B = (B \backslash A_{n+1}) + A_{n+1}B$ and $A_{n+1} = (A_{n+1} \backslash B) + A_{n+1}B$. Since each of $B \backslash A_{n+1}, \; A_{n+1} \backslash B, \; \text{and} \; A_{n+1}B$ are finite, we list the elements of each: $$B \backslash A_{n+1} = \{ \omega_1,\dots,\omega_i \},$$ $$A_{n+1}B = \{ \omega_{i+1},\dots,\omega_j \},$$ $$A_{n+1} \backslash B = \{ \omega_{j+1},\dots,\omega_k \},$$ so that $A_1 \cup \cdots \cup A_n \cup A_{n+1} = \{ \omega_1,\dots,\omega_k \}$, $B = \{ \omega_1,\dots,\omega_j \}$, and $A_{n+1} = \{ \omega_{i+1},\dots,\omega_k \}$.

Furthermore, $$\mathsf{P}(B \backslash A_{n+1}) = \sum_{m=1}^i p(\omega_m),$$ $$\mathsf{P}(A_{n+1} B) = \sum_{m=i+1}^j p(\omega_m),$$ $$\mathsf{P}(A_{n+1} \backslash B) = \sum_{m=j+1}^k p(\omega_m).$$ It follows (by the nonnegativity of $p(\omega_m)$) that $$\begin{align} \mathsf{P}(A_1 \cup \cdots \cup A_n \cup A_{n+1}) &= \sum_{m=1}^k p(\omega_m) \\ &= \sum_{m=1}^j p(\omega_m) + \sum_{m=j+1}^k p(\omega_m) \\ &\leq \sum_{m=1}^j p(\omega_m) + \sum_{m=i+1}^k p(\omega_m) \\ &= \mathsf{P}(B) + \mathsf{P}(A_{n+1}) \\ &\leq \mathsf{P}(A_1) + \cdots + \mathsf{P}(A_n) + \mathsf{P}(A_{n+1}), \end{align}$$ as desired. $\square$

SP.I.1.1

noreply@blogger.com (Lindsey K. Gamard) — Tue, 07 Jan 2014 15:59:00 +0000

Partial solution. Let $X$ be a nonempty set and $A,B \subseteq X$ with $A \neq \emptyset$ and $B \neq \emptyset$. Suppose $x \in \overline{A \cup B}$. Then $$x \in X \backslash (A \cup B) \hspace{1cm} \text{by definition}$$ $$\iff x \in X \wedge x \notin (A \cup B) \hspace{1cm} \text{by definition}$$ $$\iff x \in X \wedge \neg (x \in A \vee x \in B) \hspace{1cm} \text{by definition}$$ $$\iff x \in X \wedge (x \notin A \wedge x \notin B) \hspace{1cm} \text{DeMorgan's Laws}$$ $$\iff (x \in X \wedge x \notin A) \wedge (x \in X \wedge x \notin B) \hspace{1cm} \text{idempotency, commutativity, associativity}$$ $$\iff x \in X \backslash A \wedge x \in X \backslash B \hspace{1cm} \text{by definition}$$ $$\iff x \in \overline{A} \wedge x \in \overline{B} \hspace{1cm} \text{by definition}$$ $$\iff x \in \overline{A} \cap \overline{B} \hspace{1cm} \text{by definition}.$$ Hence we may conclude that $\overline{A \cup B} = \overline{A} \cap \overline{B}$. $\square$