Can a cube of discontinuous function be continuous?












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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.






continuity







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  • 3




    The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
    – Mindlack
    4 hours ago










  • What is your domain? It matters really quite a lot.
    – user3482749
    4 hours ago












  • I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
    – MJD
    3 hours ago


















4














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.






continuity







share|cite|improve this question




















  • 3




    The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
    – Mindlack
    4 hours ago










  • What is your domain? It matters really quite a lot.
    – user3482749
    4 hours ago












  • I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
    – MJD
    3 hours ago
















4












4








4


1





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.






continuity







share|cite|improve this question















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.






continuity




continuity






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share|cite|improve this question













share|cite|improve this question




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edited 3 hours ago







J. Abraham

















asked 4 hours ago









J. AbrahamJ. Abraham

473314




473314








  • 3




    The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
    – Mindlack
    4 hours ago










  • What is your domain? It matters really quite a lot.
    – user3482749
    4 hours ago












  • I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
    – MJD
    3 hours ago
















  • 3




    The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
    – Mindlack
    4 hours ago










  • What is your domain? It matters really quite a lot.
    – user3482749
    4 hours ago












  • I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
    – MJD
    3 hours ago










3




3




The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
4 hours ago




The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
4 hours ago












What is your domain? It matters really quite a lot.
– user3482749
4 hours ago






What is your domain? It matters really quite a lot.
– user3482749
4 hours ago














I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
3 hours ago






I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
3 hours ago












1 Answer
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9














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.






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    1 Answer
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    9














    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.






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      9














      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.






      share|cite|improve this answer


























        9












        9








        9






        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.






        share|cite|improve this answer














        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 3 hours ago

























        answered 3 hours ago









        Paul FrostPaul Frost

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