To make sure: Pumping lemma for infinite regular languages only?

Question

So this is not about the pumping lemma and how it works, it\'s about a pre-condition.

Everywhere in the net you can read, that regular languages must pass the pumpin

Answer 1

"IN CONTEXT OF PUMPING LEMMA FOR REGULAR LANGUAGES "

Yes we agree, All finite languages are regular language means we can have Finite automata as well as regular expression for any finite language.
Whereas a infinite language may be regular or not. Venn-Diagram is shown below. So we need to only check for infinite language L where its regular of not!

Think about FA:

• Any automata for a finite language can not contains loop! (also regular expressions for finite language will be without * and +operation).

• Any automata for a infinite language(regular) will contain the loop. We can't construct an automata for infinite language without loop; where loop may be a self loop of via some other state. {If its self loop then a single symbol repeats any number of time, if via other state a sequence of symbols comes in loop can be repeat any numbers of time}.

Pumping means repeating. In pumping lemma w can be break in three parts x, y, z. The 'y' is in part of w occurs in loop (that's y>=1 ). So pumping lemma is nothing finding looping part y and repeating this looping part any numbers of time.
You can see if you repeat loop any number of times and generates new string w' its still in language.

NOTE: Regular Expressions for infinite language can't be without * and +operation!

[answer] There is no loop in an automata for finite language so we can't pump(generate by repeating) new strings in language. And Pumping Lemma is not applicable for finite language.

Although some writers also explain pumping lemma for finite language where i in xyⁱz can be repeat boundedly (say k ≤ i ≤ m )

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In Venn-Diagram Every finite set is regular. Infinite set may be regular or not. Regular-Languages ⊆ Non-Regular Languages

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