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source : https://en.wikipedia.org/wiki/Sign_function#/media/File:Signum_function.svg

As you can see in the graph that 1 and -1 are open circles and aren't included. But the range of the signum function is {1,0,-1} so I'm unable to understand why 1 and -1 are open circles in the graph and whereas 0 isn't because the function is discontinuous at 0.

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    $\begingroup$ How would you draw the graph in such a way that the value at $0$ is $0$, but everywhere else it's either $-1$ or $1$? $\endgroup$
    – Arthur
    Apr 24, 2019 at 8:34
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    $\begingroup$ Possibly because the Fourier series converges to 0 at 0. $\endgroup$
    – Dayton
    Apr 24, 2019 at 8:36
  • $\begingroup$ @Dayton This question has nothing to do with Fourier series. $\endgroup$
    – M. Winter
    Apr 24, 2019 at 8:50
  • $\begingroup$ Review the definition of a function. $\endgroup$
    – user65203
    Apr 24, 2019 at 8:54

1 Answer 1

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The function has to take one value at zero only, so you have to choose between $-1, 0, 1$ when defining it. Here $0$ is chosen

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