Is there a need for $L\subseteq \Sigma^*$ to be infinite to be undecidable?I mean what if we choose a language $L'$ be a bounded finite version of $L\subseteq \Sigma^*$, that is $|L'|\leq N$, ($N \in \mathbb{N}$), with $L' \subset L$.  Is it possible for $L'$ to be an undecidable language?  I see that there is a problem of How to choose the $N$ words that $\in$ $L' $ for which we have to establish a rule for choosing which would be the first $N$ elements of $L'$, a kind of finite Kleene star operation. The aim is to find undecidability language without needing an infinite set, but I can't see it.EDIT Note: Although I chose an answer, many answers and all comments are important.
Is there an undecidable finite language of finite words?
formal languages;computability;undecidability
Yes, there is a need for $L$ to be infinite in order to be undecidable.To add up on the answers of Raphael and Sam, you should think about decidable as things that a computer-program can solve. The program required is very simple, it just needs to output Yes for elements in $L$, or otherwise, say no.So the more complex $L$ is, the longer the program you are required to write. In other words, the longer the program you run, you can check more things... So if someone gives a language $L$ which is finite, say $L=\{ a_1, a_2, \ldots, a_n\}$, you can write the following program:if INPUT = $a_1$ output Yes;if INPUT = $a_2$ output Yes;...if INPUT = $a_n$ output Yes;output No;Now, if some one gives you a larger $L$ (yet finite), you will just write a longer program. This is always true, and any finite $L$ will have it's own program. The only interesting case is what happens when $L$ is infinite - your program cannot be infinite. The issue of undecidability is even more interesting: its those (infinite) $L$'s that have no program that works correctly for them. We know that such languages must exists since there are way more (infinite) languages $L$ than the number of programs of finite (but unbounded) length.