Timeline for What is your favorite "strange" function?
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 13, 2020 at 15:09 | comment | added | LSpice | @MichaelGreinecker, here's the new nLab link: ncatlab.org/nlab/show/constant+morphism . | |
| Oct 5, 2017 at 18:37 | comment | added | Tom Goodwillie | What do you mean by "any"? | |
| Sep 8, 2015 at 0:23 | comment | added | Christopher King | @PeterLeFanuLumsdaine Have you computed it for any value in $\emptyset$? | |
| Feb 2, 2013 at 10:09 | comment | added | Michael Greinecker | There are different sensible notions of a constant morphism: nlab.mathforge.org/nlab/show/constant+morphism | |
| Jul 16, 2012 at 16:50 | comment | added | Martin Brandenburg | A morphism in a category with terminal object $t$ may be called constant if it factors through $t$. According to this definition, $\emptyset \to S$ is constant iff $S$ is nonempty. | |
| Nov 14, 2010 at 3:44 | comment | added | Tom Goodwillie | On the other hand, there should be no doubt that the function $\emptyset\to S$ is constant if the set $S$ is nonempty. | |
| Nov 14, 2010 at 3:29 | comment | added | Tom Goodwillie | Elsewhere I have raised the question of whether this function should be considered a constant function. On the one hand, $f(x_1)=f(x_2)$ for every $x_1$ and $x_2$ in the domain; on the other hand, there is no $y$ in the codomain such that for every $x$ we have $f(x)=y$. I consider it non-constant. | |
| Nov 13, 2010 at 7:53 | comment | added | Peter LeFanu Lumsdaine | @Tom: …but it can always be very easily computed! | |
| Apr 23, 2010 at 9:53 | comment | added | Tom Smith | Here is the Wikipedia page: en.wikipedia.org/wiki/Empty_function. The function has all kinds of seemingly-contradictory properties: for instance it's continuous, bijective, order-preserving (and order-reversing!), but can't ever be evaluated. | |
| Apr 23, 2010 at 9:34 | comment | added | vonjd | @Tom: Could you also please give a good reference. Wikipedia is not very exhaustive here... Thank you | |
| Apr 23, 2010 at 6:36 | comment | added | Qiaochu Yuan | And, of course, it gives combinatorial meaning to the fact that 0! = 1, since the empty function is a bijection! | |
| Apr 22, 2010 at 19:13 | comment | added | Nate Eldredge | For me, it makes a good argument that the "correct" value for $0^0$ is 1: it's the number of functions from a set with 0 elements to a set with 0 elements. | |
| Apr 22, 2010 at 18:57 | history | answered | Tom Smith | CC BY-SA 2.5 |