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15
Math.TechQA.Club
2016-09-17 19:53:56
74
Views
Proving $x+y =\sup\left\{ r+s: r,s \in \mathbb{Q}, r\leq x, s\leq y\right\}$
Published on
17 Sep 2016 - 19:53
#real-analysis
#supremum-and-infimum
50
Views
Calculate the upper boundary of the given sum
Published on
18 Sep 2016 - 17:17
#integration
#limits
#functions
#summation
#supremum-and-infimum
184
Views
Using Least Upper Bound Property to prove that $n>r$ for any $r\in \mathbb{R}$
Published on
19 Sep 2016 - 4:46
#real-analysis
#supremum-and-infimum
2.4k
Views
Limit is less than the supremum
Published on
19 Sep 2016 - 9:50
#real-analysis
#limits
#inequality
#convergence-divergence
#supremum-and-infimum
189
Views
Inner Product of Bounded Self Adjoint Linear Operator
Published on
27 Mar 2026 - 19:31
#normed-spaces
#linear-transformations
#inner-products
#supremum-and-infimum
#adjoint-operators
71
Views
Proof of one of the operations on Random Variables
Published on
22 Sep 2016 - 23:36
#probability-theory
#random-variables
#proof-explanation
#supremum-and-infimum
44
Views
Let $B(x) = \left\{b^t \ | \ t \in \mathbb{Q} ,\ x \in \mathbb{R} \land t \leq x \ \right\}$. Prove that $b^r = \sup B(r)$ if $r$ is rational.
Published on
26 Sep 2016 - 2:23
#real-analysis
#proof-verification
#proof-writing
#supremum-and-infimum
990
Views
Prove that the function is bounded: $f(x) = \frac{1}{x^{2}+1}$
Published on
26 Sep 2016 - 9:58
#calculus
#analysis
#functions
#convergence-divergence
#supremum-and-infimum
100
Views
Prove or disprove: The function is bounded: $f(n) = n + \frac{1}{n}$
Published on
26 Sep 2016 - 11:19
#calculus
#sequences-and-series
#analysis
#supremum-and-infimum
401
Views
Find the supremum and the infimum of $A$
Published on
26 Sep 2016 - 21:41
#real-analysis
#supremum-and-infimum
1k
Views
find the supremum and infimum of a set S
Published on
27 Sep 2016 - 20:18
#real-analysis
#supremum-and-infimum
63
Views
Prove that $\inf_{t>0}\frac{ω(t)}t=\lim_{t\to\infty}\frac{ω(t)}t$ if $ω:[0,\infty)\toℝ$ be bounded above on every finite interval and subadditive
Published on
29 Mar 2026 - 5:58
#limits
#limsup-and-liminf
#supremum-and-infimum
44
Views
For bounded $S\subseteq\Bbb{R}$, $\alpha = \sup(S)$. For all $\varepsilon>0$, $\exists x\in S$ s.t. $\alpha-\varepsilon< x\leq \alpha$.
Published on
29 Sep 2016 - 16:07
#real-analysis
#supremum-and-infimum
27
Views
find the sup of the following sequence
Published on
29 Sep 2016 - 21:14
#real-analysis
#convergence-divergence
#supremum-and-infimum
3.4k
Views
Proving $\inf F = - \sup E$
Published on
30 Sep 2016 - 4:59
#real-analysis
#general-topology
#supremum-and-infimum
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