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Activity for Snoopy‭

Type On... Excerpt Status Date
Edit Post #290528 Post edited:
4 months ago
Edit Post #290527 Post edited:
4 months ago
Edit Post #290528 Post edited:
4 months ago
Edit Post #290528 Initial revision 4 months ago
Answer A: $\liminf (a_n+b_n) = \liminf(a_n)+\liminf(b_n)$ provided that $\lim a_n$ exists
We will use the following facts in the proof: - (1) If the limit of $(xn)$ exists, so does that of $(-xn)$ - (2) If the limit of $xn$ as $n\to\infty$ exists, then $$\liminfnxn=\limsupnxn=\limnxn$$ - (3) One obvious direction of the inequality is proved by definition: $$ \liminf{n\to\infty}(an...
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4 months ago
Edit Post #290527 Initial revision 4 months ago
Question $\liminf (a_n+b_n) = \liminf(a_n)+\liminf(b_n)$ provided that $\lim a_n$ exists
> Question. Suppose $(an)$ and $(bn)$ are two sequences of real numbers such that $\displaystyle \lim{n\to\infty}an=a.$ Show that $$ \liminf{n\to\infty}(an+bn)=\liminf{n\to\infty}(an)+\liminf{n\to\infty}(bn). $$ > To avoid tedious discussion with infinity, assume in addition that both $an$ an...
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4 months ago
Edit Post #290512 Post edited:
fixed typos: [I couldn't make the Tex code `A\cdot B = \{ ab\mid a\in A, b\in B\}` parsed in the editor correctly. Now it works with `A\cdot B = \\{ ab\mid a\in A, b\in B\\}`]
4 months ago
Comment Post #290515 It is interesting to know the categorical perspective of the problem.
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4 months ago
Edit Post #290512 Post edited:
4 months ago
Edit Post #290512 Post edited:
4 months ago
Edit Post #290514 Post edited:
4 months ago
Edit Post #290514 Initial revision 4 months ago
Answer A: $\sup(A\cdot B) = (\sup A)(\sup B)$ where $A$ and $B$ bounded sets of positive real numbers
Another way to write the proof is to use an alternative equivalent definition of supremum: > For a bounded set of real numbers $A$, $L=\sup A$ if >- $L$ is an upper bound for $A$, i.e., for all $a\in A$: $a\le L$; >- any number smaller than $L$ is not an upper bound of $A$: for every $\epsil...
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4 months ago
Edit Post #290513 Post edited:
4 months ago
Edit Post #290513 Initial revision 4 months ago
Answer A: $\sup(A\cdot B) = (\sup A)(\sup B)$ where $A$ and $B$ bounded sets of positive real numbers
Let $L:=(\sup A)(\sup B)$. By definition of "supremum", one needs to show the following two things, - $L$ is an upper bound for $A\cdot B$; - $L$ is the smallest upper bound for $A\cdot B$. The first statement is very easy to prove: since for every $a\in A$ and every $b\in B$, one has $a\...
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4 months ago
Edit Post #290512 Initial revision 4 months ago
Question $\sup(A\cdot B) = (\sup A)(\sup B)$ where $A$ and $B$ bounded sets of positive real numbers
> Problem. Suppose $A$ and $B$ are two subsets of positive real numbers. In addition, assume that $A$ and $B$ are both bounded. Show that $$ (\sup A)(\sup B) = \sup A\cdot B$$ where the set of the right-hand side is defined as $$ A\cdot B = \\{ ab\mid a\in A, b\in B\\} $$ The problem abo...
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4 months ago
Edit Post #290509 Post edited:
4 months ago
Edit Post #290508 Post edited:
fixed typos
4 months ago
Edit Post #290508 Post edited:
added more
4 months ago
Edit Post #290509 Post edited:
added a figure
4 months ago
Edit Post #290509 Initial revision 4 months ago
Answer A: Why is $ \int_0^{\infty}\left|\frac{\sin x}{x}\right|\ dx=\infty$?
Since, $\displaystyle \lim{x\to 0}\frac{\sin x}{x}=1$, the singularity of the integral is not at $x=0$. On the other hand, one can rewrite the integral as $ \int{0}^\infty\frac{1}{x}\cdot |\sin(x)|\ dx\. $ It suffices to analyze the sum $$ \int0^\pi\frac{\sin(x)}{x}\ dx+\int{\pi}^{N\pi}\frac{1...
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4 months ago
Edit Post #290508 Initial revision 4 months ago
Question Why is $ \int_0^{\infty}\left|\frac{\sin x}{x}\right|\ dx=\infty$?
> Question: Why is $$ \int0^{\infty}\left|\frac{\sin x}{x}\right|\ dx=\infty\quad ? $$ There are several other ways to state the fact in the question depending on the contexts. For examples: - The function $\displaystyle f(x)=\frac{\sin(x)}{x}$ is not absolutely integrable on $0,\infty)$. ...
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4 months ago
Edit Post #287850 Post undeleted about 1 year ago
Edit Post #287849 Post undeleted about 1 year ago
Edit Post #287849 Post deleted about 1 year ago
Edit Post #287850 Post deleted about 1 year ago
Edit Post #287850 Post edited:
about 1 year ago
Edit Post #287850 Post edited:
about 1 year ago
Comment Post #287849 @#53398 Thank you for your comments. I have edited the post by adding the reference. The sketchy proof in the post is intended as a summary of the standard textbook proof. For the sake of clarity, I add the complete argument explicitly. For the last part of your comment, I think I'm using the def...
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about 1 year ago
Edit Post #287849 Post edited:
about 1 year ago
Edit Post #287850 Post edited:
about 1 year ago
Edit Post #287849 Post edited:
about 1 year ago
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about 1 year ago
Edit Post #287850 Post edited:
about 1 year ago
Edit Post #287850 Initial revision about 1 year ago
Answer A: Universal property of quotient spaces
I figured out an answer after posting the question for a while. I would like to record it here. The problem with the mentioned argument is that one attempted to prove the continuity of $g$ at one point $[x]$ using merely the continuity of $f$ at $x$ instead of using the global continuity of $f$ ...
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about 1 year ago
Edit Post #287849 Post edited:
about 1 year ago
Edit Post #287849 Post edited:
about 1 year ago
Edit Post #287849 Post edited:
about 1 year ago
Edit Post #287849 Post edited:
about 1 year ago
Edit Post #287849 Initial revision about 1 year ago
Question Universal property of quotient spaces
A typical textbook theorem about quotient space is as follows: >Theorem (Gamelin-Greene Introduction to Topology p.106): Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equiva...
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about 1 year ago
Edit Post #287846 Post edited:
about 1 year ago
Edit Post #287846 Post edited:
The condition (2') edited.
about 1 year ago
Comment Post #287846 Yes, thanks for that. I will edit it into the post.
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about 1 year ago