10.25 Supremum and Infimum
-
Compute the supremum and infimum of the following sets. Justify
-
\(\left\{ \frac{n^2 + 2}{n^2 + 1} : n \in \mathbb{N} \right\}\)
\(sup=\frac32\), \(inf=1\)
Also can use equivalence to prove yesterday
\(\frac{1}{n^2+1} \to 0 \quad n \to \infty\)
\(1 + \frac{1}{n^2+1} > 1 \implies 1\) is a lower bound
Since \(A\neq\emptyset\), \(1\) is a lower bound \(\implies A\) has an infimum.
We have to prove that \(1 = \inf A\).
Let \(b\) be another lower bound for \(A\) \((b \leq a, \forall a \in A)\).
We have to show that \(b \leq 1\) \((b < 1\) or \(b = 1)\).
Assume \(b > 1 \implies (b - 1) > 0\), by Archimedean property \(\exists N \in \mathbb{N}\) such that \(\frac{1}{N} < b - 1\).
\(N < N^2 < N^2+1\)
\(\frac{1}{N} > \frac{1}{N^2} > \frac{1}{N^2+1}\)
\(\frac{1}{N^2+1} < \frac{1}{N} < b - 1\)
\(1 + \frac{1}{N^2+1} < b - 1 + 1\)
\(1+\frac{1}{N^2+1}<b\)
Contradiction \(b\) is a lower bound.
\(\therefore b \leq 1\)
Therefore \(1 = \inf A\) 2. \(\left\{ \frac{n}{3n + 1} : n \in \mathbb{N} \right\}\)
\(inf=\frac14\), \(\sup=\frac13\) 3. \(\left\{ \frac{1}{n} + (-1)^n : n \in \mathbb{N} \right\}\)
\(inf=-1\), \(sup=\frac32\) 4. \(\left\{ x \in \mathbb{Q} : q^3 \leq 3 \right\}\)
\(sup=\sqrt[3]{3}\), no \(inf\) 5. \(\left\{ x \in \mathbb{Q} : x \geq 0 \text{ and } x^2 \leq 10 \right\}\)
\(inf=0\), \(sup=\sqrt{10}\)
\(E = \{x \in \mathbb{Q} : x \geq 0 \text{ and } x^2 \leq 10 \}\)
\(0\in E\implies E\neq\emptyset\)
\(0\) is a lower bound \(\implies 0 = \inf E\)
\(x^2 \leq (\sqrt{10})^2 \implies x \leq \sqrt{10} \implies \sqrt{10}\) is an upper bound for \(E\)
By Supremum Axiom, \(E\) has a supremum
\(E=\{x\in\mathbb{Q}:0\leq x\leq\sqrt{10}\}\)
\(\sqrt{10} \notin \mathbb{Q}, \ \sqrt{10} \notin E\)
To prove \(\sup E = s\), we need to prove \(\forall \varepsilon > 0, \ \exists a \in E\) such that \(s - \varepsilon < a\)
Let \(\max \{0, \sqrt{10} - \varepsilon\} < \sqrt{10}\)
Note that consider two cases that \(\varepsilon\) is big or small
By density of \(\mathbb{Q}\), \(\exists q \in \mathbb{Q}\) such that \(\max \{0, \sqrt{10} - \varepsilon\} < q < \sqrt{10}\)
\(\implies 0 < q < \sqrt{10}\) and \(\sqrt{10} - \varepsilon < q\)
\(\implies q \in E\) and \(\sqrt{10} - \varepsilon < q \leq \sqrt{10} \implies \sqrt{10} = \sup E\) 6. \(\left\{ x \in \mathbb{R} : x^2 + x + 1 \geq 0 \right\}\)
no sup and inf 7. \(\left\{ x \in \mathbb{R} : x^2 + x + 1 < 0 \right\}\)
no sup and inf 8. \(\left\{ x \in \mathbb{R} : |2x + 1| < 5 \right\}\)
\(-3<x<2\)
\(sup=2\), \(inf=-3\)
-
-
Determine if the following sets are dense in \(\mathbb{R}\).
-
\(\left\{ x \in \mathbb{R} : x < 100 \right\}\)
No 2. \(\mathbb{R} - [0, 2]\)
No 3. \(\mathbb{Q} - \{0\}\)
Yes
-
-
-
Show that if \(\begin{aligned}\lim_{n \to \infty} |a_n| = 0,\end{aligned}\) then \(\begin{aligned}\lim_{n \to \infty} a_n = 0.\end{aligned}\)
Since \(\begin{aligned}\lim_{n\to\infty}|a_{n}|=0\end{aligned}\), then \(\forall\varepsilon>0,\exists n_0\in\mathbb{N},\) such that for \(n\geq n_0\), \(||a_{n}|-0|<\varepsilon\Rightarrow||a_{n}||<\varepsilon\Rightarrow|a_{n}|<\varepsilon\Rightarrow|a_{n}-0|<\varepsilon\)
Thus \(\begin{aligned}\lim_{n\to\infty}a_{n}=0\end{aligned}\) 2. Use (1) to prove that \(\begin{aligned}\lim_{n \to \infty} \frac{(-1)^n}{n} = 0. \end{aligned}\)
Since \(\begin{aligned}\lim_{n\to\infty}\frac{(-1)^{n}}{n}=0=|\lim_{n\to\infty}\frac{(-1)^{n}}{n}|=\lim_{n\to\infty}\frac{1}{n}=0\end{aligned}\)
-