# Thomas E.

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Doing math.

|             |                  | bio     | visits   |            |                  |
|-------------|------------------|---------|----------|------------|------------------|
|             | 3,197 reputation | website |          | member for | 1 year, 4 months |
| 2724 badges | location         |         | seen     | 2 mins ago |                  |

# 1,086 Actions

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| Mar5 | revised  | The space of continuous, bounded functions from a metric space $X$ to $\mathbb R$added 23 characters in body                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                             |
| Mar5 | 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.                                                                                                                                                                                                                                                                                                                                                                                                                                                                          |
| Mar5 | 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.                                                                                                                                                                                                                                                                                                                                                                                                                                           |
| Mar5 | comment  | Continuity Properties@OC89: In your proof, what exactly are $\varepsilon$ and $\delta$? How did you choose them?                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                         |
| Mar5 | revised  | Continuity Propertiesdeleted 24 characters in body                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       |
| Mar5 | comment  | Continuity PropertiesBy $B(x,\varepsilon)$ I of course mean the $\varepsilon$-radius open ball around $x$.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                               |
| Mar5 | answered | Continuity Properties                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    |
| Mar5 | 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                                                                                                                                                                                                                                                                                                                                                                                                                                             |
| Mar5 | 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 $x=0$.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      |
| Mar5 | 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.        |
| Mar3 | reviewed | Looks Good Rational numbers form a field.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                |
| Mar3 | revised  | Are these subsets of $\mathbb{R}$ homeomorphic?Edited the body.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                          |
| Mar3 | reviewed | Edit suggested edit on projection - linear alebra                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        |
| Mar3 | revised  | projection - linear alebraedited body                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    |
| Mar3 | comment  | Prove $\operatorname{dist}(\overline{A},\overline{B}) = \operatorname{dist}(A, B)$@vermiculus. Great. Let me know if some points need further clarification.                                                                                                                                                                                                                                                                                                                                                                                                                                                             |
| Mar3 | comment  | 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 $d(A,B)\leq d(\overline{A},\overline{B})$. |
| Mar3 | 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 $a\in A$, $b\in B$?                                                                                                      |
| Mar3 | answered | Prove $\operatorname{dist}(\overline{A},\overline{B}) = \operatorname{dist}(A, B)$                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       |
| Mar3 | comment  | Prove $\operatorname{dist}(\overline{A},\overline{B}) = \operatorname{dist}(A, B)$You probably want inf over $x\in B$ in the definition of dist$(A,B)$.                                                                                                                                                                                                                                                                                                                                                                                                                                                                  |
| Mar3 | 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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