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| bio | website | |
|---|---|---|
| location | ||
| age | 22 | |
| visits | member for | 1 year, 4 months |
| seen | 2 mins ago | |
| stats | profile views | 407 |
Doing math.
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Mar 5 |
revised |
The space of continuous, bounded functions from a metric space $X$ to $\mathbb R$ added 23 characters in body |
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Mar 5 |
comment |
Determining if these sets are compact? I believe this question is a duplicate of something that was asked just few days ago. Just can't find it. |
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Mar 5 |
comment |
Simplifying $\{ x \in \mathbb{R} : x^2 - x - 6 \geq 0 \}$ when possible @DominicMichaelis. I believe the question asks to simplify the current expression to interval notation. |
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Mar 5 |
comment |
Continuity Properties @OC89: In your proof, what exactly are $\varepsilon$ and $\delta$? How did you choose them? |
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Mar 5 |
revised |
Continuity Properties deleted 24 characters in body |
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Mar 5 |
comment |
Continuity Properties By $B(x,\varepsilon)$ I of course mean the $\varepsilon$-radius open ball around $x$. |
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Mar 5 |
answered | Continuity Properties |
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Mar 5 |
revised |
Suppose that $x$ is a fixed nonnegative real number such that for all positive real numbers $E$ , $0\leq x\leq E$. Show that $x=0$. added 12 characters in body; edited title |
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Mar 5 |
answered | Suppose that $x$ is a fixed nonnegative real number such that for all positive real numbers $E$ , $0\leq x\leq E$. Show that $x=0$. |
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Mar 5 |
comment |
How to show that the vector subspaces of $\mathbb{R}^{n}$ are closed in $\mathbb{R}^{n}$? @Oliver: You use the fact that $\mathbb{R}^{m}$ is complete and not that it is closed. The closedness of $\mathbb{R}^{m}$ alone implies via the map $T:U\to \mathbb{R}^{m}$ that $U$ is closed in its own subspace topology, but not necessarily in the ambient space $\mathbb{R}^{n}$. Similarly it is also open, again in its subspace topology, since $\mathbb{R}^{m}$ is open. And if you change the word closedness to completeness, then this is basically just the answer of Seirios with some details opened up. |
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Mar 3 |
reviewed | Looks Good Rational numbers form a field. |
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Mar 3 |
revised |
Are these subsets of $\mathbb{R}$ homeomorphic? Edited the body. |
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Mar 3 |
reviewed | Edit suggested edit on projection - linear alebra |
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Mar 3 |
revised |
projection - linear alebra edited body |
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Mar 3 |
comment |
Prove $\operatorname{dist}(\overline{A},\overline{B}) = \operatorname{dist}(A, B)$ @vermiculus. Great. Let me know if some points need further clarification. |
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Mar 3 |
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Prove $\operatorname{dist}(\overline{A},\overline{B}) = \operatorname{dist}(A, B)$ @vermiculus. I'm not quite sure what you are after with that implication. You would have to justify it by using the definition, and thus infimum, anyways. And yes, the easiest way I see is just to show that $d(\overline{A},\overline{B})\leq d(a,b)$ for all $a\in A$, $b\in B$ to conclude that $d(\overline{A},\overline{B})\leq d(A,B)$. And for the reverse inequality, you should show that $d(A,B)\leq d(a,b)$ for all $a\in \overline{A}$, $b\in\overline{B}$, to conclude that $d(A,B)\leq d(\overline{A},\overline{B})$. |
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Mar 3 |
comment |
Prove $\operatorname{dist}(\overline{A},\overline{B}) = \operatorname{dist}(A, B)$ @vermiculus. The point is to show that $d(\overline{A},\overline{B})\leq d(a,b)$ is true for all $a\in A,\,b\in B$. Thus, by taking infimum over all $a\in A$ and $b\in B$ in this inequality, we obtain $d(\overline{A},\overline{B})\leq d(A,B)$. But out of curiosity, how do you justify $d(\overline{A},\overline{B})\leq d(A,B)$ without showing that $d(\overline{A},\overline{B})\leq d(a,b)$ for all $a\in A$, $b\in B$? |
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Mar 3 |
answered | Prove $\operatorname{dist}(\overline{A},\overline{B}) = \operatorname{dist}(A, B)$ |
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Mar 3 |
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Prove $\operatorname{dist}(\overline{A},\overline{B}) = \operatorname{dist}(A, B)$ You probably want inf over $x\in B$ in the definition of dist$(A,B)$. |
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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 $x$. |
