List questions from supremum-and-infimum

Published on 24 Sep 2018 - 2:10

797

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Is the set $A =\{x: x^2 < 3x\}$ bounded, find $\sup A$ and $\inf A$ if they exist?

Published on 24 Sep 2018 - 3:12

196

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Prove that this set $A=\{ \frac{m}{n} + \frac {4n}{m} : m,n \in N\}$ unbounded

Published on 24 Sep 2018 - 11:01

85

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How to show $(x_n)$ defined $x_1 = a$, $x_{n+1} = 2x_n - x^2_ny$ is increasing if $ay < 2$

Published on 24 Sep 2018 - 18:02

1k

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Prove that there exists a number $x$ such that $x^3 = 6$

Published on 25 Sep 2018 - 7:36

61

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Supremum in context of Dynamic Programming

Published on 26 Mar 2026 - 1:01

1.4k

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Let a be a real number and let $S = \{ x \in Q : x <a \}.$ prove that $a = sup S.$

Published on 27 Sep 2018 - 7:09

159

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How to show that these two supremums are equal?

Published on 01 Oct 2018 - 15:43

50

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For a bounded set A, there is a sequnce contained in A that converges to inf(A)

Published on 03 Oct 2018 - 3:41

44

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For a positive function, can the length of gradient get arbitrarily close to 0?

Published on 03 Oct 2018 - 4:51

46

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$x<\sup Y$ if and only if $\exists y\in Y: x<y$

Published on 04 Oct 2018 - 10:51

176

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Is this a Legitimate Proof Regarding the Infimum of a Set?

Published on 05 Oct 2018 - 22:27

185

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Prove that if $B\subset \mathbf{R}$ is Lebesgue measurable, then $|B|=\sup\{|A|: A \text{ is a closed bounded subset of $B$}\}$.

Published on 07 Oct 2018 - 3:57

65

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Can't proove that a function is convex

Published on 07 Oct 2018 - 17:50

417

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How to prove that infimum and limit inferior commute?

Published on 07 Oct 2018 - 20:23

List Question

prove that a set a can have at most one supremum.

Is the set $A =\{x: x^2 < 3x\}$ bounded, find $\sup A$ and $\inf A$ if they exist?

Prove that this set $A=\{ \frac{m}{n} + \frac {4n}{m} : m,n \in N\}$ unbounded

How to show $(x_n)$ defined $x_1 = a$, $x_{n+1} = 2x_n - x^2_ny$ is increasing if $ay < 2$

Prove that there exists a number $x$ such that $x^3 = 6$

Supremum in context of Dynamic Programming

Let a be a real number and let $S = \{ x \in Q : x <a \}.$ prove that $a = sup S.$

How to show that these two supremums are equal?

For a bounded set A, there is a sequnce contained in A that converges to inf(A)

For a positive function, can the length of gradient get arbitrarily close to 0?

$x<\sup Y$ if and only if $\exists y\in Y: x<y$

Is this a Legitimate Proof Regarding the Infimum of a Set?

Prove that if $B\subset \mathbf{R}$ is Lebesgue measurable, then $|B|=\sup\{|A|: A \text{ is a closed bounded subset of $B$}\}$.

Can't proove that a function is convex

How to prove that infimum and limit inferior commute?

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