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| bio | website | |
|---|---|---|
| location | ||
| age | 22 | |
| visits | member for | 1 year, 4 months |
| seen | 18 hours ago | |
| stats | profile views | 409 |
Doing math.
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May 19 |
awarded | Constituent |
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May 7 |
awarded | Caucus |
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Mar 21 |
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“Maximal” sets with cardinality $\aleph_0$? Few comments and questions: 1. What do you mean by being dense in itself? Dense in its own subspace topology? In this sense every set is dense in itself. 2. Denseness is a topological property. Saying that $\mathbb{Q}$ is more dense or less dense doesn't make sense: it either is dense (in some superset) or it isn't. |
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Mar 20 |
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Set of Continuous Functions, Functionals, and Equicontinuity What are your thoughts on this problem? Have you tried anything? |
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Mar 14 |
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I thought I proved the Limit here didn't exist… Note that $\sqrt{2x^{2}}=|x|\sqrt{2}$ and not $x\sqrt{2}$. |
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Mar 13 |
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Cantor Function Question @AsafKaragila: I'm guessing that he might mean the following. Some analysis textbooks define the Cantor function by defining it as constant pieces outside the Cantor set (in the open intervals that were removed at each step), and then inside the Cantor set by limits. In this case, you don't get any expression of the function inside the Cantor set, but only in its complement. |
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Mar 13 |
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Cantor Function Question Congrats for 1,000 answers. |
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Mar 13 |
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How is Hausdorff Distance sensitive to position? I only mentioned that condition in order to avoid writing concrete counter-examples. However, the shortest distance also fails to satisfy the triangle-inequality. Take $A=[0,1]$, $B=[2,3]$, and $C=[4,5]$ as a simple example. By shortest distance, $d(A,C)>d(A,B)+d(B,C)$, since $3>2$. I don't think the triangle inequality is negotiable when defining metric spaces. What is there left, the symmetry? |
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Mar 13 |
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limit points of sequence of real numbers What are your thoughts on these problems? Have you tried to solve them yet? |
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Mar 13 |
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How is Hausdorff Distance sensitive to position? A small note to the second sentence in the last paragraph: the shortest distance is not actually a metric. It fails to satisfy e.g. the condition $d(A,B)=0\Leftrightarrow A=B$. |
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Mar 13 |
revised |
condition $\mu ^*(B\cup C)=\mu ^*(B)+\mu ^*(C)$ added 45 characters in body |
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Mar 13 |
answered | condition $\mu ^*(B\cup C)=\mu ^*(B)+\mu ^*(C)$ |
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Mar 13 |
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Is there really no way to integrate $e^{-x^2}$? You probably mean that any continuous function is Riemann integrable on a compact interval. |
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Mar 12 |
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Trying to define $\mathbb{R}^{0.5}$ topologically Thanks! Interesting way of looking at it. |
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Mar 12 |
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Trying to define $\mathbb{R}^{0.5}$ topologically Hi Ittay. Maybe this question is just triviality, but could you clarify why $A\times A$ with a single point removed is still connected? I know if $A=\mathbb{R}$ then one can either produce arguments with topological dimensions or use path connectedness. But for any connected space $A$ I can't see how it follows immediately. |
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Mar 12 |
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Open sets in a subspace topology edited tags |
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Mar 12 |
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How is Hausdorff Distance sensitive to position? edited tags |
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Mar 12 |
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How is Hausdorff Distance sensitive to position? Note that the first definition given for Hausdorff distance is not actually even a distance; the real Hausdorff distance is the "generalized" version. Could you explain more in detail what you mean by the expression: sensitive to position? |
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Mar 12 |
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$\int fg = \lim \int f_n g$ for any $g \in L^q$ and for uniformly bounded $f_n$ in $L^p$ About your second last comment: If you would use DCT to argue the following: $\big| \int f_{n}g-\int fg\big|=\big| \int f_{n}g-fg\big|\leq \|f_{n}g-fg\|_{1}\to 0$, then you would have to show that $|f_{n}g-fg |\leq h$ for some integrable $h$. But do we know that $fg\in L^{1}$? |
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Mar 12 |
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$\int fg = \lim \int f_n g$ for any $g \in L^q$ and for uniformly bounded $f_n$ in $L^p$ It would make sense if $q$ was the Hölder conjugate of $p$. Are you sure there is no typo in the material? Because usually we say that $f_{n}\to f$ weakly in $L^{p}$ if the convergence you wrote applies for all $g\in L^{q}$, where $q$ is the Hölder conjugate of $p$. Also about your current proof, $\|f_{n}-f\|_{p}\to 0$ is not entirely trivial either. You could prove that $f_{n}\to f$ almost everywhere + $\|f_{n}\|\leq M$ for all $n$ implies $f_{n}\to f$ in $L^{p}$ (using dominated convergence theorem). Note that almost everywhere convergence is not sufficient alone to make it happen. |
