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I read this question:what is a function

Here comes my question, Can we call following a increasing function (or may not even a function):

$$ f:[0,1]\rightarrow\mathbb{R}\\ f(x)= \begin{cases} x,& x\in[0,1/3]\\ 2x,& x\in[2/3,1]\\ \end{cases} $$

For my understanding, I would say this it not even a function since there's no image of domain $(1/3,2/3)$, am I right?

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  • $\begingroup$ Well it is a function, but certainly not one defined over $[0,1]$, rather $[0, 1/3] \cup [2/3, 1]$ (the image domain is also smaller than $\mathbbm (R) $). That being said, it is an increasing function on both sub-intervals, hence on the entire definition domain as well. $\endgroup$ – Quantuple Mar 24 '16 at 23:55
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Yes, this is a relation which is not a function. However, if we say that $f:[0,1/3]\cup[2/3,1]\to\mathbb R$, this is a perfectly good function, and it is increasing.

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  • $\begingroup$ so the discontinuous points of an increasing function cannot be interval,right? $\endgroup$ – DuFong Mar 24 '16 at 23:57
  • $\begingroup$ Yes. An increasing function cannot be discontinuous over an entire interval; it can only have a countable number of discontinuities. $\endgroup$ – Alex S Mar 24 '16 at 23:58

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