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According to the intermediate value theorem, we can infer that a single point function, for example $f(x) = \sqrt x + \sqrt(-x)$, which has a range of only $0$ and a domain of only $0$ is not a continuous function. Is my conclusion right or there is actually exception when it is continuous? or outside of the calculus realm of thinking single point function CAN BE continuous?

MysteryGuy
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Firman Nugraha
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    What (precise) definition are you using? See also: https://math.stackexchange.com/questions/455296/can-a-function-with-just-one-point-in-its-domain-be-continuous. – StackTD Nov 13 '18 at 14:26
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    The intermediate value theorem does not imply that $f(x) = \sqrt x+\sqrt{-x}$ is not continuous, since it states that $f$ takes all values between $f(0) = 0$ and... $f(0) = 0$. – krirkrirk Nov 13 '18 at 14:28
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    No, we certainly cannot infer what you claim from the IVT. Why not? Nobody can say what the error in your reasoning is since you don't explain how you think IVT implies this... – David C. Ullrich Nov 13 '18 at 15:48

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Although there are similar duplicates here a quick reasoning:

  • You consider a real-valued function with domain $D = \{0\}$, so you can use the limit definition of continuity: $$f \mbox{ is continuous at } x_0 = 0 $$ $$\Leftrightarrow \lim_{n\to\infty}f(x_n) = f(0) \mbox{ for all sequences } \{x_n\}_{n \in D}\subset D \mbox{ with } \lim_{n\to\infty}x_n = 0$$
  • The only possible sequence converging to $0$ in the domain $\{0\}$ is the constant sequence $x_n = 0$ for all $n \in \mathbb{N}$. $$\lim_{n\to\infty}f(x_n) = f(0) = 0$$ So, $f$ is contiuous.
trancelocation
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