Timeline for What is the most useful non-existing object of your field?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Dec 3, 2023 at 5:43 | comment | added | Toby Bartels | For example, given any infinite binary sequence $\alpha$ (so $\alpha=(\alpha_i)_{i=1}^\infty$ where each $\alpha_i\in\{0,1\}$, we can define a real number $y$ as the limit of the Cauchy sequence $\beta_i=2+\sum\limits_{j=1}^i2^{-j}$ of rational numbers. Then $y\leq2$ iff $\alpha_i=0$ for all $i$, while $y>2$ iff $\alpha_i=1$ for at least one $i$. But we have no way to decide this if we only know finitely many terms (and those all happen to be $0$). | |
| Dec 3, 2023 at 5:43 | comment | added | Toby Bartels | @Vincent : The ordering extends to $\mathbb R$, but it doesn't have the same properties. In particular, you can't prove $\forall\,x,y\in\mathbb R,\;x\geq y\;\vee\;x<y$. | |
| Dec 1, 2023 at 20:49 | comment | added | Vincent | I was thinking of suggesting $f(y) = 0$ if $y \leq 2$ and $f(y) = 1$ if $y > 2$, but I guess that this requires a similar theorem stating that the ordering on $\mathbb{Q}$ extends to all of $\mathbb{R}$. But isn't this the definition of $\mathbb{R}$, the closure of $\mathbb{Q}$ with respect to the ordering? | |
| Dec 1, 2023 at 17:07 | comment | added | Toby Bartels | @Vincent : Give an example then! Perhaps you'd suggest $f(y):=\cases{1&$x=y$\\0&$x\ne y$,}$ but this is a partial function whose domain is $\{y\;|\;x=y\;\vee\;x\ne y\}$. So you need to prove the classical theorem to conclude that this is really a function from $\mathbb R$ to $\{0,1\}$. | |
| Dec 1, 2023 at 12:16 | comment | added | Vincent | Why can't you prove the existence of such a function constructively? What is more constructive than giving an example? | |
| S Sep 30, 2014 at 18:59 | history | answered | Toby Bartels | CC BY-SA 3.0 | |
| S Sep 30, 2014 at 18:59 | history | made wiki | Post Made Community Wiki by Toby Bartels |