How to prove that if a language A is not regular then A* isn't regular either?

I have tried the usual methods with no result.

  • 4
    $\begingroup$ What makes you think that this claim is correct? $\endgroup$
    – Shaull
    Nov 4, 2013 at 11:01
  • 3
    $\begingroup$ If you don't find a proof, look for a counter-example... $\endgroup$
    – Denis
    Nov 4, 2013 at 11:09
  • 1
    $\begingroup$ Hint: try a non-regular language over a unary alphabet $\{a\}$. See where that gets you. $\endgroup$
    – Shaull
    Nov 4, 2013 at 11:30
  • 2
    $\begingroup$ For Kleene star for a language over unary alphabet, see here. $\endgroup$ Nov 4, 2013 at 13:38
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    $\begingroup$ Actually, you don't even need to consider languages over a unary alphabet. Consider, for instance, non-regular languages containing at least all strings of length one over any alphabet. $\endgroup$
    – Patrick87
    Nov 4, 2013 at 14:45

1 Answer 1


Hint: Suppose $L$ is any language over the alphabet $\Sigma$. If $L$ is not regular then so is $L+\Sigma$, yet $(L+\Sigma)^* = \Sigma^*$ is regular.

  • 1
    $\begingroup$ Excellent answer; although I fail to see how this is a hint, so much as a complete proof that the claim is false. $\endgroup$
    – Patrick87
    Nov 4, 2013 at 19:31
  • $\begingroup$ @YuvalFilmus Does + mean union here? $\endgroup$
    – Turbo
    Nov 3, 2015 at 20:43
  • $\begingroup$ @Turbo. That's right. $\endgroup$ Nov 3, 2015 at 20:55

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