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| bio | website | |
|---|---|---|
| location | ||
| age | 22 | |
| visits | member for | 1 year, 4 months |
| seen | 2 hours ago | |
| stats | profile views | 409 |
Doing math.
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Mar 3 |
comment |
$x_n \in \mathbb R, \quad x_n \to A \implies \max \{x_n,A-x_n\} \to ?$ And with $B(x,\varepsilon)$ I of course mean the $\varepsilon$-radius ball around $x$. |
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Mar 3 |
revised |
$x_n \in \mathbb R, \quad x_n \to A \implies \max \{x_n,A-x_n\} \to ?$ deleted 7 characters in body |
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Mar 3 |
answered | $x_n \in \mathbb R, \quad x_n \to A \implies \max \{x_n,A-x_n\} \to ?$ |
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Mar 3 |
comment |
Let $ (x_n) $ be a divergent sequence in a compact subset of $ \mathbb{R}^n $. Prove there are two subsequences that converge to different limits. Note that we are in $\mathbb{R}^{n}$ and not in $\mathbb{R}$. How do you define limsup and liminf of a vector valued sequence? |
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Mar 3 |
answered | Are these subsets of $\mathbb{R}$ homeomorphic? |
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Mar 2 |
comment |
Is $f(x)= \cos(e^x)$ uniformly continuous? @julien. Ah, shoot. It's $1$ :) made a silly mistake when drawing the circle. Then their difference is $2$ as wanted. |
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Mar 2 |
revised |
Let $ (x_n) $ be a divergent sequence in a compact subset of $ \mathbb{R}^n $. Prove there are two subsequences that converge to different limits. added 35 characters in body |
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Mar 2 |
comment |
Is $f(x)= \cos(e^x)$ uniformly continuous? Are you sure the distance of the $f$-values of $x_n$ and $y_n$ is always $2$, and not $1$? Isn't $\cos (2n\pi)=0$ and $\cos(2n\pi+\pi)=-1$ for all $n$? |
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Mar 2 |
comment |
Is $f(x)= \cos(e^x)$ uniformly continuous? @MhenniBenghorbal. He only needs bound for the derivative of $f$. |
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Mar 2 |
revised |
Is $f(x)= \cos(e^x)$ uniformly continuous? added 4 characters in body; edited title |
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Mar 2 |
answered | Let $ (x_n) $ be a divergent sequence in a compact subset of $ \mathbb{R}^n $. Prove there are two subsequences that converge to different limits. |
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Mar 2 |
revised |
Let $ (x_n) $ be a divergent sequence in a compact subset of $ \mathbb{R}^n $. Prove there are two subsequences that converge to different limits. edited title |
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Mar 1 |
reviewed | Edit suggested edit on Limit and intersection of sets |
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Mar 1 |
revised |
Limit and intersection of sets added a space to one of the equations |
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Mar 1 |
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Sequence of Nonnegative Measurable Functions About the edited answer: What indices do you consider to the sum in $g_{n}$? Note that $f_{k,n}(x)$ was defined only for $k\leq x$, which may be none if e.g. $0<x<1$. If you sum over all $k$‚ then to me it seems that $\int g_{n}= +\infty$ for any $n$, as the intervals $[k,k+2^{2n+k}]$ cover $[0,\infty)$ and $g_{n}$ is increasing. |
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Mar 1 |
comment |
Sequence of Nonnegative Measurable Functions And in the end, for $\limsup_{n\to\infty} f_{n}(q_{n}+x)$ to equal $+\infty$‚ you would need the sequence $(q_{n}+x)_{n=1}^{\infty}$ to contain $0$ infinitely often, which is not possible as it only contains it once (or never if $x$ is irrational). This requirement is because the limsup of $f_{n}$ is $+\infty$ only in the origin. |
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Mar 1 |
comment |
Sequence of Nonnegative Measurable Functions Why does this $g_{n}$ do the job? You only get $+\infty$ with $f_{n}$ in the origin, and everywhere else $0$. How do you enumerate $\mathbb{Q}$ so that for any $x\in\mathbb{R}$ and $\varepsilon>0$ you find $k\in\mathbb{N}$ with $|q_{n}+x|<\varepsilon$ for all $n\geq k$. You can't have $0$ appearing infinitely often in the sequence $(q_{n}+x)_{n=1}^{\infty}$ for every $x\in\mathbb{R}$, so I don't really see why this example works. Could you elaborate more on it? |
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Mar 1 |
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Sequence of Nonnegative Measurable Functions To me it seems that $\limsup_{n\to\infty} f_{n}(x)=0$ for all $x\neq 0$ in this example. |
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Feb 28 |
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Finite intersection of open sets @Brian: With Hagen's notations do you mean when $n=2$? Which would be the same as $n=1$ if induction assumption would be for $k\leq n$ instead of $k<n$. Or did I misinterpret something? |
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Feb 28 |
reviewed | No Action Needed Moving on the surface of a cube |
