Timeline for Is a function piecewise continuous if it has a left-limit and a right-limit at every point in its interval domain and equals at least one of these? [closed]
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Dec 20, 2020 at 18:20 | review | Reopen votes | |||
| Dec 20, 2020 at 19:03 | |||||
| Dec 20, 2020 at 18:01 | history | edited | immeasurable | CC BY-SA 4.0 |
Added the requirement that at each point, the function must equal the left- or right-limit.
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| Dec 20, 2020 at 17:50 | comment | added | immeasurable | Addendum: I had intended to add the following to my comment: "And one can say the function is piecewise continuous." | |
| Dec 20, 2020 at 17:41 | comment | added | immeasurable | Thanks. I was not aware of Thomae's function or Young's theorem. Please comment on my current understanding. If a real-valued function, whose domain is an interval, has left- and right-limits and equals at least one of them at each point in the domain, then it is continuous, except possibly at a countable number of points. | |
| Dec 19, 2020 at 18:36 | history | closed |
Nate Eldredge LSpice Christian Remling user44191 abx |
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| Dec 19, 2020 at 17:12 | review | Close votes | |||
| Dec 19, 2020 at 18:38 | |||||
| Dec 19, 2020 at 16:58 | answer | added | Gerald Edgar | timeline score: 2 | |
| Dec 19, 2020 at 16:23 | comment | added | Bill Johnson | Are you a student? Here is an exercise for you. Prove that a function on a closed interval has the property you want if and only if it is a uniform limit of a sequence of step functions. | |
| Dec 19, 2020 at 16:10 | comment | added | Alessandro Della Corte | On question 1. this can help: mathoverflow.net/a/231462/167834 | |
| Dec 19, 2020 at 15:59 | comment | added | Alessandro Della Corte | The answer to 2. and 3. is no. In fact it is no even if the two one-sided limits agree. See for instance: math.stackexchange.com/q/698448/787383 | |
| Dec 19, 2020 at 15:58 | comment | added | Nik Weaver | Here's an easy example: enumerate the rationals in $[0,1]$ as $(a_n)$. Then for $t \in [0,1]$ define $f(t) = \sum \{2^{-n}: a_n \leq t\}$. This is an increasing function which is discontinuous at every rational. | |
| Dec 19, 2020 at 15:47 | review | First posts | |||
| Dec 19, 2020 at 18:38 | |||||
| Dec 19, 2020 at 15:45 | history | asked | immeasurable | CC BY-SA 4.0 |