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Let's say that we have a language $L$ that is finite.
How can I prove that the complement of $L$, i.e., $\bar{L}$, is always an infinite language?

Obs.: infinite language in this case means that is possible to construct an infinite set of words that are acceptable by $\bar{L}$.

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The set $\Sigma^*$ is an infinite set (you can build a bijection between the naturals and the words in $\Sigma^*$ if you wish to prove this, or you can observe that, for every $k \in \mathbb{N}$, $\Sigma^*$ contains at least one word of length $k$).

Let $L$ be a finite language and suppose towards a contradiction that $\overline{L} = \Sigma^* \setminus L$ is finite. Then $\Sigma^* = L \cup \overline{L}$ must also be finite (since it's a finite union of finite sets). This provides the sought contradiction.

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Here is a formal proof for the statement (warning, this really is a formal proof, and hence is not very intuitive):

Let $\Sigma$ be a non-empty alphabet (either finite or infinite), and let $L\subset \Sigma^*$, with $|L|\in \mathbb{N}$ ($L$ is finite).

Let $\alpha \in \Sigma$. By definition of the klenee-star operator, we can construct a function $g:\mathbb{N}\rightarrow \Sigma^*$ by $g(k)=\alpha^k$ ($\alpha$ repeated $k$ times). This function is one-to-one, since $|g(k)|=|\alpha^k|=k$ (where $|w|$ denotes the length of the string $w$), and thus for any $k_1, k_2$ with $g(k_1)=g(k_2)$ we must have $k_1=|g(k_1)|=|g(k_2)|=k_2$.

Thus, we proved that $\aleph_0=|\mathbb{N}|\le |\Sigma^*|$.

Now, let us begin showing that $\aleph_0\le |\Sigma^* \setminus L|$. Denote by $n_{max}:=\max\{|w|\mid w\in L\}+1$, and define $f:\mathbb{N}\rightarrow (\Sigma^*\setminus L)$ by $f(k)=\alpha^{n_{max}+k}$.

First, we have to show that this really is a function (by showing that $f(k)\in \Sigma^*\setminus L$ for all $k\in\mathbb{N}$). Indeed, if we assume towards contradiction that $f(k)\in L$ for some $k$ - we get that $|f(k)|<n_{max}$ by the definition of $n_{max}$. But $|f(k)|=|\alpha^{n_{max}+k}|=n_{max}+k \ge n_{max}$ which is clearly a contradiction (we got that $n_{max}<n_{max}$).

Now we have to show that $f$ is one-to-one. By a similar argument to what we did before to prove that $\Sigma^*$ is infinite, we can conclude that $f$ is one-to-one, and hence $\aleph_0=|\mathbb{N}|\le |\Sigma^*\setminus L|=|\bar L|$ which is exactly what we wanted to show.

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Lets denote $n:=|L|\in \mathbb{N}$.

For any nonempty $\Sigma$, it is clear that $\Sigma^*$ is infinite.

By definition, $\bar L = \Sigma^*\setminus L$. Therefore, $|\bar L| = |\Sigma^*\setminus L| = |\Sigma^*|-|L| = \infty - n=\infty$.

It is not a formal proof (neither a valid one), but it should suffice for you as intuition to understand why $\bar L$ is infinite.

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  • $\begingroup$ Sorry, in your answer you mean that $\Sigma$ is a finite alphabet? In this case, what do you mean by $\Sigma^*$ ? $\endgroup$
    – igorkf
    Nov 19 '21 at 17:44
  • $\begingroup$ -1, I find subtracting cardinalities is handwavy at best and wrong at worst (and in all cases potentially confusing to beginners). For example, $|\mathbb{Z}\setminus\mathbb{N}| = \infty - \infty = 0$, despite negative integers existing. $\endgroup$
    – ComFreek
    Nov 19 '21 at 18:18
  • $\begingroup$ Yes, I agree with @ComFreek. This is not a valid proof. $\endgroup$
    – igorkf
    Nov 19 '21 at 18:25
  • $\begingroup$ I never said this is a valid proof. Its just intuition. Substracting cardinalities is only well-defined when the RHS of the subtraction is finite (as in this case) $\endgroup$
    – nir shahar
    Nov 19 '21 at 18:36
  • $\begingroup$ Ok, you didn't said that this was a valid proof. $\endgroup$
    – igorkf
    Nov 19 '21 at 18:42

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