2

Homework 2

Week 2

October 25, 2024

98

1. (25 points) Compute the supremum and the infimum of the following sets. Justify.

   (a) \(A=\left\{\frac{(-1)^{n}}{n}:n\in\mathbb{N}\right\}\)

   \(infA=-1,supA=0.5\)

   Since \(|\frac{(-1)^{n}}{n}|=\frac{1}{n}\), then \(-1\leq\frac{1}{n}\leq1\Rightarrow-1\leq\frac{(-1)^{n}}{n}\leq1\). Thus \(-1\) is a lower bound for \(A\).

   Since \(-1\in A\) (when \(n=1\)), then by proposition \(-1\) is an infimum

   When \(n\) is odd, \(\frac{(-1)^{n}}{n}=\frac{-1}{n}<0\). When \(n\) is even, \(\frac{(-1)^{n}}{n}=\frac{1}{n}>0\).

   Thus the supremum must be positive and \(n\) must be even when it is supremum.

   Then \(B=\left\{\frac{1}{n}:n=2k,k\in\mathbb{N}\right\}\) and \(supA=supB\)

   Since \(n\geq 2\), then \(\frac1n\leq \frac12\). Thus \(\frac12\) is an upper bound.

   Since \(\frac12\in B\) when (\(n=2\)), then by proposition \(\frac12\) is an supremum for \(B\)​.

   Therefore, \(\frac12\) is an supremum for \(A\)​

   (b) \(A=\left\{ 3 - \frac{1}{n} : n \in \mathbb{N} \right\}\)

   \(supA=3,infA=2\)​

   Since \(n\geq1\Rightarrow\frac{1}{n}\leq1\Rightarrow-\frac{1}{n}\geq-1\Rightarrow3-\frac{1}{n}\geq2\), thus \(2\) is a lower bound.

   Since \(2\in A\) (when \(n=1\)), thus by proposition \(2\) is an infimum

   Since \(-\frac1n<0\Rightarrow\) \(3-\frac1n<3\), thus \(3\) is an upper bound.

   Let \(b\) be another upper bound. We need to prove \(b\geq 3\)

   Suppose \(b< 3\)​

   We know \(b\) is an upper bound, then \(b\geq3-\frac{1}{n}\Rightarrow3-b\leq\frac{1}{n},\forall n\)​

   However since \(3-b>0\), by archimedean property, there must exist \(3-b>\frac1n\)​

   Contradiction!

2. (25 points) Prove that \(\mathbb{R} - \mathbb{Z}\) is dense in \(\mathbb{R}\). wrong

   The question is to prove \(\forall x,y\in\mathbb{R},x<y,\exists q\in\mathbb{R}-\Z\) such that \(x<q<y\)

   Then if we can prove \(\forall x,y\in\mathbb{R}-\mathbb{Z},x<y,\exists q\in\mathbb{R}-\mathbb{Z}\) such that \(x<q<y\)

   However, even if we get \(\forall x,y\in\mathbb{R}-\mathbb{Z}\subseteq\mathbb{R}\), but \(x,y\) are still in the \(\R-\Z\)

   Problem! we can fix it by discuss when x,y is integer then....

   Then since \(\forall x,y\in\mathbb{R}-\mathbb{Z}\subset\mathbb{R}\), we can prove the original problem (\(\mathbb{R} - \mathbb{Z}\) is dense in \(\mathbb{R}\))

   Thus our goal is to prove \(\forall x,y\in\mathbb{R}-\mathbb{Z},x<y,\exists q\in\mathbb{R}-\mathbb{Z}\) such that \(x<q<y\)

   Then we consider two cases

   If there is no integer in the interval of \(x\) and \(y\), then we can choose \(q=\frac{x+y}{2}\in\mathbb{R}-\mathbb{Z}\).

   Then \(q=\frac{x+y}{2}<\frac{y+y}{2}=y\) and \(q=\frac{x+y}{2}>\frac{x+x}{2}=x\), thus \(x<q<y\) \(\checkmark\)

   If there are integers in the interval of \(x\) and \(y\), then let \(q=y-\frac{y-\lfloor y\rfloor}{2}\)​

   Prove this \(q\) works.

   \(\lfloor y\rfloor\) is the integer part of \(y\) and \(\lfloor y\rfloor\leq y\), thus \(\frac{y-\lfloor y\rfloor}{2}>0\Rightarrow-\frac{y-\lfloor y\rfloor}{2}<0\Rightarrow y-\frac{y-\lfloor y\rfloor}{2}<y\).

   Then we have \(q<y\)​

   Since \(y>\lfloor y\rfloor\Rightarrow2y-y>\lfloor y\rfloor\Rightarrow2y-y+\lfloor y\rfloor>2\lfloor y\rfloor\Rightarrow y-\frac{y-\lfloor y\rfloor}{2}>\lfloor y\rfloor\), then \(q>\lfloor y\rfloor>x\)​

   Therefore, \(x<q<\lfloor y\rfloor<y\)​

   Therefore, \(\mathbb{R} - \mathbb{Z}\) is dense in \(\mathbb{R}-\Z\).

   Therefore, \(\mathbb{R} - \mathbb{Z}\) is dense in \(\mathbb{R}\).

   Another way

   \(\forall x,y\in\mathbb{R},x<y,\exists q\in\mathbb{R}-\Z\) such that \(x<q<y\)

   \(\begin{cases}\frac{x+y}{2}\text{ if }\frac{x+y}{2}\notin \Z\\\frac{x+y}{2}-\varepsilon \text{ if }\frac{x+y}{2}\in \Z,\varepsilon=\{\frac13,\frac{y-x}{3}\}\end{cases}\)​

3. (25 points) Show that for \(p \geq 1\), \(\lim_{n \to \infty} \frac{1}{n^p} = 0\).

   Since \(n^p\geq n\) and \(p\geq 1\)​\, then \(\frac{1}{n^{p}}<\frac{1}{n}\)

   For \(\forall\varepsilon>0,\exists n_0\in\mathbb{N},n_0>\frac{1}{\varepsilon}\) such that for \(n\geq n_0\) such that \(|\frac{1}{n^{p}}-0|=\frac{1}{n^{p}}<\frac{1}{n}\leq\frac{1}{n_0}<\varepsilon\)

4. (25 points) Prove that if \(\lim_{n \to \infty} a_n = a\), then \(\lim_{n \to \infty} |a_n| = |a|\). Does the converse hold?

   Since \(\lim_{n \to \infty} a_n = a\), then by definition \(\forall\varepsilon>0,\exists n_0\in\mathbb{N}\) such that for \(n\geq n_0\), \(|a_{n}-a|<\varepsilon\)

   Prove this inequality:

   \(|a|=|a-b+b|\leq|a-b|+|b|\Rightarrow|a|-|b|\leq|a-b|\\|b|=|b-a+a|\leq|b-a|+|a|\Rightarrow|b|-|a|\leq|a-b|\)

   Then we can get \(\vert\vert a\vert-\vert b\vert\vert\leq\vert a-b\vert\)

   Thus we can use it here.

   \(\vert\vert a_{n}\vert-\vert a\vert\vert\leq|a_{n}-a|<\varepsilon\Rightarrow\vert\vert a_{n}\vert-\vert a\vert\vert<\varepsilon\)​

   Thus \(\forall\varepsilon>0,\exists n_0\in\mathbb{N}\) such that for \(n\geq n_0\), \(|\vert a_{n}\vert-\vert a\vert\vert<\varepsilon\)

   Therefore, \(\lim_{n \to \infty} |a_n| = |a|\)

   Converse is not true. Because by inequality \(\vert\vert a_{n}\vert-\vert a\vert\vert\leq\vert a_{n}-a\vert\), if we have \(|\vert a_{n}\vert-\vert a\vert\vert<\varepsilon\)​

   But we can not ensure\(|a_n-a|\)​ \(<\varepsilon\)

   Also a counter example: \(a_n = (-1)^n\). Then \(\lim_{n \to \infty} |a_n| = 1\), but \(\lim_{n \to \infty} a_n\) doesn't exist

   Thus the converse is not true.

‍