Question

Take $S$ to be a set of ordinals, show $\cup S=\sup(S)$

score 90 · Answer 1 · 2026-03-27 14:39:50

90

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Take $S$ to be a set of ordinals, show $\cup S=\sup(S)$

Published on 27 Mar 2026 - 14:39

score 2.5k · Answer 2 · 2014-12-31 17:30:59

2.5k

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Properties of the supremum and infimum.

Published on 31 Dec 2014 - 17:30

score 154 · Answer 3 · 2015-01-01 14:09:40

154

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Showing that $\inf\left|x-p\right|$ is continuous

Published on 01 Jan 2015 - 14:09

score 2.2k · Answer 4 · 2019-10-02 11:16:18

2.2k

Views

What is the meaning of $\inf$?

Published on 02 Oct 2019 - 11:16

score 56 · Answer 5 · 2019-10-02 21:27:50

56

Views

What's wrong with my proof about the largest element of a set?

Published on 02 Oct 2019 - 21:27

score 546 · Answer 6 · 2019-10-03 15:20:06

546

Views

Proving $\sup{A}=\inf{U_A}$, where $U_A$ denotes set of all upper-bounds of $A$

Published on 03 Oct 2019 - 15:20

score 39 · Answer 7 · 2019-10-03 20:02:33

39

Views

how to solve linear and exponential mixed equation $\frac{(n+1)^{2}}{2^{n}}$

Published on 03 Oct 2019 - 20:02

score 50 · Answer 8 · 2019-10-04 14:49:57

50

Views

Proof with supremum

Published on 04 Oct 2019 - 14:49

score 727 · Answer 9 · 2019-10-07 09:06:36

727

Views

Understanding the supremum of a r.v. with an example

Published on 07 Oct 2019 - 9:06

score 102 · Answer 10 · 2019-10-08 16:40:24

102

Views

Find $\sup_{a a^* + b b^* = 1} | a^2x + aby |$ for fixed $x,y \in \mathbb{C}$, $|x| = 1$, $y \ne 0$.

Published on 08 Oct 2019 - 16:40

score 2.3k · Answer 11 · 2019-10-08 21:34:23

2.3k

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Show that if A⊆B -knowing that B is bounded and non-empty set- , then inf B≤ inf A≤ sup A≤ sup B

Published on 08 Oct 2019 - 21:34

score 73 · Answer 12 · 2019-10-09 04:08:49

73

Views

Let $A\subseteq R$ be non-empty and bounded.

Published on 09 Oct 2019 - 4:08

score 49 · Answer 13 · 2019-10-10 10:44:25

49

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If $(A_n)_{n\epsilon \Bbb N}$ is increasing, prove $lim inf A_n = lim sup (A_n) = \bigcup^\infty_{n=1} (A_n)$?

Published on 10 Oct 2019 - 10:44

score 52 · Answer 14 · 2019-10-11 04:01:40

52

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Is $\sup\{|f(x')-f(x'')|:a\leq x',x''\leq b\}$=$\sup\{f(x):a\leq x\leq b\}-\inf\{f(x):a\leq x\leq b\}$

Published on 11 Oct 2019 - 4:01

score 40 · Answer 15 · 2019-10-12 13:36:35

40

Views

Comparison between $\sup W$ and $\sup \overline{W}$

Published on 12 Oct 2019 - 13:36

List Question

Take $S$ to be a set of ordinals, show $\cup S=\sup(S)$

Properties of the supremum and infimum.

Showing that $\inf\left|x-p\right|$ is continuous

What is the meaning of $\inf$?

What's wrong with my proof about the largest element of a set?

Proving $\sup{A}=\inf{U_A}$, where $U_A$ denotes set of all upper-bounds of $A$

how to solve linear and exponential mixed equation $\frac{(n+1)^{2}}{2^{n}}$

Proof with supremum

Understanding the supremum of a r.v. with an example

Find $\sup_{a a^* + b b^* = 1} | a^2x + aby |$ for fixed $x,y \in \mathbb{C}$, $|x| = 1$, $y \ne 0$.

Show that if A⊆B -knowing that B is bounded and non-empty set- , then inf B≤ inf A≤ sup A≤ sup B

Let $A\subseteq R$ be non-empty and bounded.

If $(A_n)_{n\epsilon \Bbb N}$ is increasing, prove $lim inf A_n = lim sup (A_n) = \bigcup^\infty_{n=1} (A_n)$?

Is $\sup\{|f(x')-f(x'')|:a\leq x',x''\leq b\}$=$\sup\{f(x):a\leq x\leq b\}-\inf\{f(x):a\leq x\leq b\}$

Comparison between $\sup W$ and $\sup \overline{W}$

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