Question

Why is $\sup_{n \in \mathbb{N}}\Big|\frac{n-1}{n}z_n\Big|\le\sup_{n \in \mathbb{N}}\Big|\frac{n-1}{n}\Big|\sup_{n \in \mathbb{N}}\Big|z_n\Big|$

score 29 · Answer 1 · 2018-08-26 10:21:34

Why is $\sup_{n \in \mathbb{N}}\Big|\frac{n-1}{n}z_n\Big|\le\sup_{n \in \mathbb{N}}\Big|\frac{n-1}{n}\Big|\sup_{n \in \mathbb{N}}\Big|z_n\Big|$

score 219 · Answer 2 · 2018-08-30 04:27:26

219

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$S = \{1,1.1,0.9,1.01,\cdots\}$

Published on 30 Aug 2018 - 4:27

score 1.5k · Answer 3 · 2018-08-30 13:57:40

1.5k

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score 207 · Answer 4 · 2018-08-31 09:46:08

207

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score 66 · Answer 5 · 2018-09-01 17:43:05

66

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score 1.2k · Answer 6 · 2018-09-01 18:28:02

1.2k

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score 44 · Answer 7 · 2018-09-01 19:11:14

44

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score 110 · Answer 8 · 2018-09-02 14:44:00

110

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score 779 · Answer 9 · 2018-09-05 23:22:52

779

Views

$\inf$ and $\sup$ of $K = \bigg\{ \frac{m}{n}+\frac{4n}{m} : m,n \in \mathbb N\bigg\}$

Published on 05 Sep 2018 - 23:22

score 427 · Answer 10 · 2018-09-07 08:40:15

427

Views

Let $f$ be continuous real-valued function on $[0,1]$. Then, $F(x)=\max\{f(t):0\leq t\leq x\}$ is continuous

Published on 07 Sep 2018 - 8:40

score 335 · Answer 11 · 2018-09-08 20:46:00

335

Views

Let $P \subset Q \subset ℝ$, $P\neq\emptyset$ and $P$ and $Q$ are bounded above. Show that $\sup P\leq \sup Q$.

Published on 08 Sep 2018 - 20:46

score 366 · Answer 12 · 2018-09-08 23:23:31

366

Views

In Baby Rudin's Theorem 1.1, why is it important that $S$ has the least upper bound property?

Published on 08 Sep 2018 - 23:23

score 57 · Answer 13 · 2018-09-09 18:41:28

57

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Supremum and Infimum of the following sets

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score 289 · Answer 14 · 2018-09-10 02:20:31

289

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score 64 · Answer 15 · 2018-09-11 14:12:39

64

Views

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List Question

Why is $\sup_{n \in \mathbb{N}}\Big|\frac{n-1}{n}z_n\Big|\le\sup_{n \in \mathbb{N}}\Big|\frac{n-1}{n}\Big|\sup_{n \in \mathbb{N}}\Big|z_n\Big|$

$S = \{1,1.1,0.9,1.01,\cdots\}$

Least-Upper-Bound Property $\implies$ Greatest-Lower Bound Property (Rudin)

Bounds of a set with two variables

Proving the supremum

Proving the supremum of a set in the general case

Alternative proof for chains

Questions about the Proof that $\inf A = - \sup\left(-A\right)$ (Rudin)

$\inf$ and $\sup$ of $K = \bigg\{ \frac{m}{n}+\frac{4n}{m} : m,n \in \mathbb N\bigg\}$

Let $f$ be continuous real-valued function on $[0,1]$. Then, $F(x)=\max\{f(t):0\leq t\leq x\}$ is continuous

Let $P \subset Q \subset ℝ$, $P\neq\emptyset$ and $P$ and $Q$ are bounded above. Show that $\sup P\leq \sup Q$.

In Baby Rudin's Theorem 1.1, why is it important that $S$ has the least upper bound property?

Supremum and Infimum of the following sets

Supremum and infimum of $\sum_{k=1}^{n}\frac{a_k}{a_{k}+a_{k+1}+a_{k+2}}$

Problem with $\sup A, \min A, \max A$ and $\inf A$.

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