Regular of language of all words of length 3

Question

Consider the language $$L = \{ x \in \{0,1\}^* \mid |x| = 3 \}.$$

I think the above language is regular. A DFA can be used to determine the above language.

Am I correct? Is the above language regular?

If this language $L$ is regular, then it should satisfy the pumping lemma. Then there exist $w = xyz$, where $xy^nz \in L$ for all $n \ge 0$.

But on the other hand, if we pump more letters then the resulting string will not be in the language. The language $L$ only contains words of length 3.

The pumping lemma states that for every regular language there exists an integer $p$, such that string $w$ of length at least $p$ can be written as $w = xyz$ and $y$ can be pumped.

Here are my doubts.

1. Is this language $L$ regular?
2. If so, does it satisfy the pumping lemma?
3. The pumping lemma states that every regular language has a pumping length $p \ge 1$. Does this language not have one?

Answer 1

Every finite language is regular. If $L$ is a finite language and $p$ is larger than the length of all words in $L$, then $L$ satisfies the pumping lemma with the constant $p$. Indeed, every word in $L$ of length at least $p$ can be pumped (vacuously).

Answer 2

Take pumping length = 4. All strings in the language with length greater than the pumping length must meet some condition. Since there are no strings in the language with length > 4, all these strings, all zero of them, meet any condition, and therefore the language is regular.

That’s true for any finite language where you can just take the pumping length longer than the longest string in the language.