Activity for Snoopyâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #290515 |
It is interesting to know the categorical perspective of the problem. (more) |
— | 11 months ago |
| Edit | Post #290512 |
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| Edit | Post #290514 |
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| Edit | Post #290514 | Initial revision | — | 11 months ago |
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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... (more) |
— | 11 months ago |
| Edit | Post #290513 |
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— | 11 months ago |
| Edit | Post #290513 | Initial revision | — | 11 months ago |
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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 $a\... (more) |
— | 11 months ago |
| Edit | Post #290512 | Initial revision | — | 11 months ago |
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$\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... (more) |
— | 11 months ago |
| Edit | Post #290509 |
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| Edit | Post #290508 |
Post edited: fixed typos |
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| Edit | Post #290508 |
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| Edit | Post #290509 |
Post edited: added a figure |
— | 11 months ago |
| Edit | Post #290509 | Initial revision | — | 11 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 $$ \int0^\pi\frac{\sin(x)}{x}\ dx+\int{\pi}^{N\pi}\frac{1... (more) |
— | 11 months ago |
| Edit | Post #290508 | Initial revision | — | 11 months ago |
| Question | — |
Why is $ \int_0^{\infty}\left|\frac{\sin x}{x}\right|\ dx=\infty$? > Question: Why is $$ \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 $0,\infty)$. ... (more) |
— | 11 months ago |
| Edit | Post #287850 | Post undeleted | — | almost 2 years ago |
| Edit | Post #287849 | Post undeleted | — | almost 2 years ago |
| Edit | Post #287849 | Post deleted | — | almost 2 years ago |
| Edit | Post #287850 | Post deleted | — | almost 2 years ago |
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| Edit | Post #287850 |
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— | almost 2 years 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... (more) |
— | almost 2 years ago |
| Edit | Post #287849 |
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| Edit | Post #287850 |
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| Edit | Post #287849 |
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| Edit | Post #287849 |
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| Edit | Post #287850 |
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| Edit | Post #287850 | Initial revision | — | almost 2 years ago |
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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 $f$ ... (more) |
— | almost 2 years ago |
| Edit | Post #287849 |
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| Edit | Post #287849 |
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| Edit | Post #287849 |
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| Edit | Post #287849 |
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| Edit | Post #287849 | Initial revision | — | almost 2 years ago |
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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... (more) |
— | almost 2 years ago |
| Edit | Post #287846 |
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— | almost 2 years ago |
| Edit | Post #287846 |
Post edited: The condition (2') edited. |
— | almost 2 years ago |
| Comment | Post #287846 |
Yes, thanks for that. I will edit it into the post. (more) |
— | almost 2 years ago |
| Edit | Post #287846 | Initial revision | — | almost 2 years ago |
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Criterion in terms of the bases for determining whether one topology is finer than another When topologies are given by bases, one has a useful criterion in terms of the bases for determining whether one topology is finer than another: > Lemma 13.3. (Munkres's Topology p.81) Let $\mathscr{B}$ and $\mathscr{B}^{\prime}$ be bases for the topologies $\mathcal{T}$ and $\mathcal{T}^{\prime... (more) |
— | almost 2 years ago |
| Edit | Post #287667 |
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| Edit | Post #287667 |
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| Edit | Post #287664 |
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| Edit | Post #287720 | Initial revision | — | almost 2 years ago |
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Proving $|{\bf R}^{\bf R}|=|2^{\bf R}|$ using the Schroeder-Bernstein Theorem Let $A$ be the set of all functions from ${\bf R}$ to ${\bf R}$ and $B$ the power set of ${\bf R}$. Then $|A|=|B|$. This is a well-known result in set theory. A quick search on Google returns answers in various places explicitly using cardinal arithmetic. Question: Can one prove the above res... (more) |
— | almost 2 years ago |
| Comment | Post #287667 |
Yes. Adding more text will push the mentioned expression down to the next line. If I change the expression to $f(x,0)$, then the scroll bar disappears. (more) |
— | almost 2 years ago |