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TutorialsWorshopsHomework 3.pdf

Homework 1

Week 1

October 18, 2024

1. (25 points) Show that

   \(\left| \frac{1}{a} \right| = \frac{1}{|a|}\), if \(a \neq 0\).

   Proof

   \(a\)​	\(\frac1a\)​	\(\|a\|\)​	\(\frac{1}{\left\|a\right\|}\)​	\(\|\frac{1}{a}\|\)​
   \(>0\)​	\(>0\)​	\(a\)​	\(\frac1a\)​	\(\frac1a\)​
   \(<0\)​	\(<0\)​	\(-a\)​	\(-\frac{1}{a}\)​	\(-\frac1a\)​

   Since we consider all the cases, from the table we can conclude that \(\left|\frac{1}{a}\right|= \frac{1}{|a|}\), if \(a \neq 0\) is true.

2. (25 points) Let \(a, b \in \mathbb{R}\) and \(b \geq 0\). Prove that \(|a| \geq b\) if and only if \(a \leq -b\) or \(a \geq b\).

   Proof

   \(\Rightarrow\)) N.T.P. \(|a|\geq b\Rightarrow a\leq -b\) or \(a\geq b\)

   If \(a\geq0\), \(|a|=a\geq b\)

   If \(a<0\), \(|a|=-a\geq b\Rightarrow-a\cdot\left(-1\right)\leq b\cdot\left(-1\right)\Rightarrow a\leq-b\)

   Thus \(a \leq -b\) or \(a \geq b\)

   \(\Leftarrow\)) N.T.P. \(a\leq-b\) or \(a\geq b\Rightarrow|a|\geq b\)

   Since \(a\leq -b\), then \(-1\cdot a\geq -1\cdot(-b)"sign~rules"\Rightarrow-a\geq b\Rightarrow b\leq -a\)

   Since \(a\geq b\), then \(b\leq a\)

   Thus \(b\leq |a|\) because we've concluded \(b\leq a\) and \(b\leq -a\)​

   Thus \(|a|\geq b\)​

3. (25 points) Let \(A \subseteq \mathbb{R}\) be finite and non-empty. What is \(\sup A\) and \(\inf A\)? Justify.

   \(supA=maxA\) and \(infA=minA\)

   Claim: Since \(A\) is finite and non-empty, then \(A\) must have maximum and minimum.

   If not, then \(A\) has two cases when it does not have max and min.

   1. If \(A\) is convergent to an value, then there are infinite elements converges to the limit. Contradiction!
   2. If \(A\) goes to infinity, then it must have infinite elements. Contradiction!

   Thus \(A\) has a maximum and minimum.

   Since we know if a set has a maximum and minimum, it must have a \(supA=maxA\)​ and \(infA=minA\)

4. (25 points) Let \(A\) be a nonempty set which is bounded below. Let \(B := \{ b \in \mathbb{R} : b \text{ is a lower bound for } A \}\). Show that \(\sup B = \inf A\).

   Analysis: Although we usually like to list definition and observe what we want, but there are too many definition will cause a mess.

   So we need to do some converts

   We have a set \(B\) and \(\sup B\), we need to prove \(\inf A=\sup B\).

   Then we can see \(\sup B\) as an element only, and to prove it is \(\inf A\)​

   Proof

   To prove it, we need to prove \(\sup B\) is the lower bound of \(A\)

   Since \(B\) is the set of lower bounds for \(A\) and \(\sup B\in B\), then \(\sup B\) is the lower bound of \(A\)

   Then we need to prove if there is another lower bound of \(A\), then it is less than \(\sup B\)

   Take \(L\) as another lower bound for \(A\), then \(L\in B\)

   Since \(\sup B\) is the supremum of \(B\), then \(\sup B\geq b,\forall b\in B\)

   Then \(\sup B\geq b\)

   Finally, we finish the proof

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