[SOLVED] Can a cube of discontinuous function be continuous?

Can a cube (meaning $$g(x) = f(x)^3 = f(x) \cdot f(x) \cdot f(x)$$) of discontinuous function $$f: D \to \mathbb{R}$$ ($$D$$ is subset of $$\mathbb{R}$$) be continuous? I think it can't, since $$x^3$$ is injective, but I am not able to prove it or find a counterexample. @Tanner Swett 2019-01-08 05:51:24

If a function $$f(x)$$ is continuous, then its cube root $$g(x) = f(x)^{1/3}$$ is also continuous.

So the contrapositive is also true, which is:

If a function $$g(x)$$ is not continuous, then its cube $$f(x) = g(x)^3$$ is not continuous either.

(Strictly speaking, the contrapositive is actually "if the cube root $$f(x)^{1/3}$$ of a function $$f(x)$$ is not continuous, then the function $$f(x)$$ is not continuous either". But this is equivalent.) @Paul Frost 2019-01-08 01:05:45

Since $$\phi : \mathbb{R} \to \mathbb{R}, \phi(x) = x^3$$, is a homeomorphism, you see that $$f$$ is continuous iff $$\phi \circ f$$ is continuous. @Paul Sinclair 2019-01-08 18:09:44

To lower the level of this answer, note that $\phi^{-1}(x) = \sqrt x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $\phi\circ f$ is continuous, so is $f = \phi^{-1}\circ \phi \circ f$.

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