enter image description here

In the picture above what is the domain of the piecewise-continous function?

The first option is that the domain of the piecewise-continous function is [1,6]

The second option is that the domain is $\left [ 1,2 \right )\cup \left ( 2,3 \right ) \cup \left ( 3,4 \right )\cup \left ( 4,6 \right ]$

I could find the next definitions in web pages:

" The function Y(t) is piecewise continous on the closed interval [a,b], if there exists values $t_0,t_1,..,t_n$ with $a=t_0<t_1< ...<t_n=b$ such that Y(t) is continous in each of the open intervals $t_{k-1}<t<t_k$, for $k=1,2,...,n$ and has left-hand and right-hand limits at each of the values $t_k$, for k=0,1,2,...,n "

And the next one is from a student's course notes : " A function is said to be piecewise continous if its graph consists of a finite number of continous pieces on any part of its domain.The endpoints of the pieces must be well-defined"

Here "well-defined" is kind of ambiguous, as it can be interpreted that the endpoints of the pieces of the function must exist or that the existance of the limits of the endpoints of the pieces it's enough in order to be piecewise continous.

If the domain is [1,6], a second question arises. How can a function of any kind(here a piecewise-continous function) with domain [a,b] have points inside its domain where is not defined?

EDIT: enter image description here

Then, this graph would be from a piecewise continous function in [1,6]?


The statement that the endpoints need to be well-defined doesn't mean that the function values are defined at the endpoints, just that we know what the endpoints are. To say that the function is piecewise continuous on $[a,b]$ in the first definition is not to say that the domain of the function is $[a,b]$ That would be contradictory, as you have pointed out.

I don't agree with either of the proposed answers. I don't agree with the first option because $2$ is not in the domain. I don't agree with the second option because $1$ is in the domain. My answer would be

$\left [ 1,2 \right )\cup \left ( 2,3 \right ) \cup \left ( 3,4 \right )\cup \left ( 4,6 \right ]$

  • $\begingroup$ I've just corrected it I meant $\left [ 1,2 \right )\cup \left ( 2,3 \right ) \cup \left ( 3,4 \right )\cup \left ( 4,6 \right ]$ $\endgroup$
    – roy212
    Feb 3 '19 at 18:38
  • $\begingroup$ Good, then that's my answer. $\endgroup$
    – saulspatz
    Feb 3 '19 at 18:41
  • $\begingroup$ I think I'm mixing domain with piecewise-continuity. I added a picture in my question. Please, can you you tell me if this function would be piecewise-continous in [1,6]too? $\endgroup$
    – roy212
    Feb 3 '19 at 18:58
  • $\begingroup$ No it isn't, because the interval fro $4$ to $5$ isn't covered. Look at the first definition of piecewise continuity carefully. How could you define the $t_k$ in this case? Answer-- you can't. For future reference, according to the rules of the site, you shouldn't ask new questions in comments. If you have another question, post it as a different question. $\endgroup$
    – saulspatz
    Feb 3 '19 at 19:03
  • $\begingroup$ I understand, but could you please answer my last doubt? So a function is piecewise-continous even if the endpoints are not defined? $\endgroup$
    – roy212
    Feb 3 '19 at 19:19

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