From 5581a9ca04b76d420e92187c8d462a3ebe5620d7 Mon Sep 17 00:00:00 2001 From: gapry Date: Sun, 14 Jan 2024 23:12:47 +0800 Subject: [PATCH 1/9] BabyRudin: Solution: Ex05 in Ch01 --- Books/BabyRudin/Chapter01/ex05.tex | 14 ++++++++++++++ 1 file changed, 14 insertions(+) create mode 100644 Books/BabyRudin/Chapter01/ex05.tex diff --git a/Books/BabyRudin/Chapter01/ex05.tex b/Books/BabyRudin/Chapter01/ex05.tex new file mode 100644 index 0000000..eb48688 --- /dev/null +++ b/Books/BabyRudin/Chapter01/ex05.tex @@ -0,0 +1,14 @@ +\subsection*{Exercise 05} +\begin{flushleft} +$\text{(1) } \exists\ \alpha \in R \ni x \geq \alpha \ \forall x \in A$ \\ +$\text{(2) } \exists\ \alpha \in R \ni -x \leq -\alpha \ \forall -x \in -A$ \\ +\end{flushleft} + +\begin{flushleft} +(1) $\implies \alpha = \inf( A)$ \\ +(2) $\implies -\alpha = \sup(-A)$ +\end{flushleft} + +\begin{flushleft} +Obviously, $a = -(-a) \implies \inf(A) = -\sup(-A)$ +\end{flushleft} From 9ebad839d17bb967d19e127108154e2101057b6e Mon Sep 17 00:00:00 2001 From: gapry Date: Sun, 14 Jan 2024 23:12:47 +0800 Subject: [PATCH 2/9] BabyRudin: Solution: Ex05 in Ch01 --- Books/BabyRudin/Chapter01/ex05.gapry.tex | 14 ++++++++++++++ 1 file changed, 14 insertions(+) create mode 100644 Books/BabyRudin/Chapter01/ex05.gapry.tex diff --git a/Books/BabyRudin/Chapter01/ex05.gapry.tex b/Books/BabyRudin/Chapter01/ex05.gapry.tex new file mode 100644 index 0000000..583f06f --- /dev/null +++ b/Books/BabyRudin/Chapter01/ex05.gapry.tex @@ -0,0 +1,14 @@ +\subsection*{Exercise 05 (Gapry)} +\begin{flushleft} +$\text{(1) } \exists\ \alpha \in R \ni x \geq \alpha \ \forall x \in A$ \\ +$\text{(2) } \exists\ \alpha \in R \ni -x \leq -\alpha \ \forall -x \in -A$ \\ +\end{flushleft} + +\begin{flushleft} +(1) $\implies \alpha = \inf( A)$ \\ +(2) $\implies -\alpha = \sup(-A)$ +\end{flushleft} + +\begin{flushleft} +Obviously, $a = -(-a) \implies \inf(A) = -\sup(-A)$ +\end{flushleft} From 5ac08d416256d13d912576299414afa3b24646dc Mon Sep 17 00:00:00 2001 From: gapry Date: Mon, 29 Jan 2024 18:17:02 +0800 Subject: [PATCH 3/9] BabyRudin: remove the redundant file --- Books/BabyRudin/Chapter01/ex05.tex | 14 -------------- 1 file changed, 14 deletions(-) delete mode 100644 Books/BabyRudin/Chapter01/ex05.tex diff --git a/Books/BabyRudin/Chapter01/ex05.tex b/Books/BabyRudin/Chapter01/ex05.tex deleted file mode 100644 index eb48688..0000000 --- a/Books/BabyRudin/Chapter01/ex05.tex +++ /dev/null @@ -1,14 +0,0 @@ -\subsection*{Exercise 05} -\begin{flushleft} -$\text{(1) } \exists\ \alpha \in R \ni x \geq \alpha \ \forall x \in A$ \\ -$\text{(2) } \exists\ \alpha \in R \ni -x \leq -\alpha \ \forall -x \in -A$ \\ -\end{flushleft} - -\begin{flushleft} -(1) $\implies \alpha = \inf( A)$ \\ -(2) $\implies -\alpha = \sup(-A)$ -\end{flushleft} - -\begin{flushleft} -Obviously, $a = -(-a) \implies \inf(A) = -\sup(-A)$ -\end{flushleft} From 47989efb55a8e4fb85b6e684dc9cf211b4d88091 Mon Sep 17 00:00:00 2001 From: gapry Date: Mon, 29 Jan 2024 18:52:05 +0800 Subject: [PATCH 4/9] BabyRudin: Solution: Ex05 in Ch01: Update --- Books/BabyRudin/Chapter01/ex05.gapry.tex | 17 +++++++---------- 1 file changed, 7 insertions(+), 10 deletions(-) diff --git a/Books/BabyRudin/Chapter01/ex05.gapry.tex b/Books/BabyRudin/Chapter01/ex05.gapry.tex index 583f06f..6cb77be 100644 --- a/Books/BabyRudin/Chapter01/ex05.gapry.tex +++ b/Books/BabyRudin/Chapter01/ex05.gapry.tex @@ -1,14 +1,11 @@ \subsection*{Exercise 05 (Gapry)} -\begin{flushleft} -$\text{(1) } \exists\ \alpha \in R \ni x \geq \alpha \ \forall x \in A$ \\ -$\text{(2) } \exists\ \alpha \in R \ni -x \leq -\alpha \ \forall -x \in -A$ \\ -\end{flushleft} \begin{flushleft} -(1) $\implies \alpha = \inf( A)$ \\ -(2) $\implies -\alpha = \sup(-A)$ -\end{flushleft} +Since $A \subset R$, and $A$ is bounded below, we know that $\inf A$ exists. +Let's claim $\alpha = \inf A$, so that $A = \{x \in R\ |\ x \ge \alpha \}$. +It follows that $-A = \{-x \in R\ | -x \le -\alpha,\ x \in A\}$. +Since we know that $\alpha$ is the greatest lower bound, +the negation of inequality $x \ge \alpha$, which is $-x \le -\alpha$, shows $-\alpha$ the least upper bound, that is , $-\alpha = \sup(-A)$. \\ -\begin{flushleft} -Obviously, $a = -(-a) \implies \inf(A) = -\sup(-A)$ -\end{flushleft} +According to 1.14 Proposition (d), $\alpha = -(-\alpha)$, we can conclude that $\inf A = -\sup(-A)$. +\end{flushleft} \ No newline at end of file From 46321db5f6b73df4aa3bd68393faf5683f6a8807 Mon Sep 17 00:00:00 2001 From: gapry Date: Sun, 14 Jan 2024 23:12:47 +0800 Subject: [PATCH 5/9] BabyRudin: Solution: Ex05 in Ch01 --- Books/BabyRudin/Chapter01/ex05.gapry.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/Books/BabyRudin/Chapter01/ex05.gapry.tex b/Books/BabyRudin/Chapter01/ex05.gapry.tex index 6cb77be..68a9249 100644 --- a/Books/BabyRudin/Chapter01/ex05.gapry.tex +++ b/Books/BabyRudin/Chapter01/ex05.gapry.tex @@ -8,4 +8,4 @@ \subsection*{Exercise 05 (Gapry)} the negation of inequality $x \ge \alpha$, which is $-x \le -\alpha$, shows $-\alpha$ the least upper bound, that is , $-\alpha = \sup(-A)$. \\ According to 1.14 Proposition (d), $\alpha = -(-\alpha)$, we can conclude that $\inf A = -\sup(-A)$. -\end{flushleft} \ No newline at end of file +\end{flushleft} From a5548468ccf3ca861e1e9ecca08f8cc0c7bc0da0 Mon Sep 17 00:00:00 2001 From: gapry Date: Sun, 14 Jan 2024 23:12:47 +0800 Subject: [PATCH 6/9] BabyRudin: Solution: Ex05 in Ch01 --- Books/BabyRudin/Chapter01/ex05.tex | 14 ++++++++++++++ 1 file changed, 14 insertions(+) create mode 100644 Books/BabyRudin/Chapter01/ex05.tex diff --git a/Books/BabyRudin/Chapter01/ex05.tex b/Books/BabyRudin/Chapter01/ex05.tex new file mode 100644 index 0000000..eb48688 --- /dev/null +++ b/Books/BabyRudin/Chapter01/ex05.tex @@ -0,0 +1,14 @@ +\subsection*{Exercise 05} +\begin{flushleft} +$\text{(1) } \exists\ \alpha \in R \ni x \geq \alpha \ \forall x \in A$ \\ +$\text{(2) } \exists\ \alpha \in R \ni -x \leq -\alpha \ \forall -x \in -A$ \\ +\end{flushleft} + +\begin{flushleft} +(1) $\implies \alpha = \inf( A)$ \\ +(2) $\implies -\alpha = \sup(-A)$ +\end{flushleft} + +\begin{flushleft} +Obviously, $a = -(-a) \implies \inf(A) = -\sup(-A)$ +\end{flushleft} From 32153088aea246b2efccd640e26be220926374af Mon Sep 17 00:00:00 2001 From: gapry Date: Sun, 14 Jan 2024 23:12:47 +0800 Subject: [PATCH 7/9] BabyRudin: Solution: Ex05 in Ch01 --- Books/BabyRudin/Chapter01/ex05.gapry.tex | 13 ++++++------- 1 file changed, 6 insertions(+), 7 deletions(-) diff --git a/Books/BabyRudin/Chapter01/ex05.gapry.tex b/Books/BabyRudin/Chapter01/ex05.gapry.tex index 68a9249..6bcabbf 100644 --- a/Books/BabyRudin/Chapter01/ex05.gapry.tex +++ b/Books/BabyRudin/Chapter01/ex05.gapry.tex @@ -1,11 +1,10 @@ \subsection*{Exercise 05 (Gapry)} - \begin{flushleft} -Since $A \subset R$, and $A$ is bounded below, we know that $\inf A$ exists. -Let's claim $\alpha = \inf A$, so that $A = \{x \in R\ |\ x \ge \alpha \}$. -It follows that $-A = \{-x \in R\ | -x \le -\alpha,\ x \in A\}$. -Since we know that $\alpha$ is the greatest lower bound, -the negation of inequality $x \ge \alpha$, which is $-x \le -\alpha$, shows $-\alpha$ the least upper bound, that is , $-\alpha = \sup(-A)$. \\ - +Since $A \subset R$, and $A$ is bounded below, we know that $\inf A$ exists.\\ +\vspace{10px} +Let's claim $\alpha = \inf A$, so for all $x \in A$, we have $x \ge \alpha$. It follows that for all $-x \in -A$, we have $-x \le -\alpha$.\\ +\vspace{10px} +Since we know that $\alpha$ is the greatest lower bound of $A$, the negation of inequality $x \ge \alpha$, which is $-x \le -\alpha$, it shows $-\alpha$ is the least upper bound of $-A$, that is, $-\alpha = \sup(-A)$. \\ +\vspace{10px} According to 1.14 Proposition (d), $\alpha = -(-\alpha)$, we can conclude that $\inf A = -\sup(-A)$. \end{flushleft} From b66a01f2a4b1d7bdec968273ead0fcab1f37fd3c Mon Sep 17 00:00:00 2001 From: gapry Date: Mon, 29 Jan 2024 18:17:02 +0800 Subject: [PATCH 8/9] BabyRudin: remove the redundant file --- Books/BabyRudin/Chapter01/ex05.tex | 14 -------------- 1 file changed, 14 deletions(-) delete mode 100644 Books/BabyRudin/Chapter01/ex05.tex diff --git a/Books/BabyRudin/Chapter01/ex05.tex b/Books/BabyRudin/Chapter01/ex05.tex deleted file mode 100644 index eb48688..0000000 --- a/Books/BabyRudin/Chapter01/ex05.tex +++ /dev/null @@ -1,14 +0,0 @@ -\subsection*{Exercise 05} -\begin{flushleft} -$\text{(1) } \exists\ \alpha \in R \ni x \geq \alpha \ \forall x \in A$ \\ -$\text{(2) } \exists\ \alpha \in R \ni -x \leq -\alpha \ \forall -x \in -A$ \\ -\end{flushleft} - -\begin{flushleft} -(1) $\implies \alpha = \inf( A)$ \\ -(2) $\implies -\alpha = \sup(-A)$ -\end{flushleft} - -\begin{flushleft} -Obviously, $a = -(-a) \implies \inf(A) = -\sup(-A)$ -\end{flushleft} From fbb86f30b3c943f6e13f13850d978d66aab8da5e Mon Sep 17 00:00:00 2001 From: gapry Date: Fri, 23 Feb 2024 05:43:57 +0800 Subject: [PATCH 9/9] BabyRudin: Solution: Update: Ex05 in Ch01 --- Books/BabyRudin/Chapter01/ex05.gapry.tex | 20 ++++++++++++++++---- 1 file changed, 16 insertions(+), 4 deletions(-) diff --git a/Books/BabyRudin/Chapter01/ex05.gapry.tex b/Books/BabyRudin/Chapter01/ex05.gapry.tex index 6bcabbf..ee26367 100644 --- a/Books/BabyRudin/Chapter01/ex05.gapry.tex +++ b/Books/BabyRudin/Chapter01/ex05.gapry.tex @@ -1,10 +1,22 @@ \subsection*{Exercise 05 (Gapry)} \begin{flushleft} -Since $A \subset R$, and $A$ is bounded below, we know that $\inf A$ exists.\\ +For all $x$ in the set $A$, $-x$ is in the set $-A$. +Let's claim $y = -x$, we know that for all $y \in -A$, $y \le sup(-A)$ is true. +We have the following inferences: +\begin{flalign*} + y &\le sup(-A) &\\ + \iff -x &\le sup(-A) &\\ +\implies x &\ge -sup(-A) +\end{flalign*} +It means $-sup(-A)$ is one of the lower bound of the set $A$. \\ \vspace{10px} -Let's claim $\alpha = \inf A$, so for all $x \in A$, we have $x \ge \alpha$. It follows that for all $-x \in -A$, we have $-x \le -\alpha$.\\ +Let's claim $B$ is a set of all lower bound of the set $A$. \\ +Obviously, for all $b$ in the set $B$, $b \le x$ is true for all $x \in A$. \\ \vspace{10px} -Since we know that $\alpha$ is the greatest lower bound of $A$, the negation of inequality $x \ge \alpha$, which is $-x \le -\alpha$, it shows $-\alpha$ is the least upper bound of $-A$, that is, $-\alpha = \sup(-A)$. \\ +Since $b \le x \implies -b \ge -x$ is true, it follows $-b$ will greater than or equal to all element in the set $-A$. It proves $-b \ge sup(-A)$ is true. \\ \vspace{10px} -According to 1.14 Proposition (d), $\alpha = -(-\alpha)$, we can conclude that $\inf A = -\sup(-A)$. +Since $-b \ge sup(-A) \implies b \le -sup(-A)$ is true, it follows $-sup(-A)$ is the upper bound of $b$. \\ +\vspace{10px} +We prove the $x \ge -sup(-A)$ and $b \le -sup(-A)$ are both true, hence $-sup(-A)$ is the greatest lower bound of the set $A$, that is $-sup(-A) = inf(A)$. \end{flushleft} +