# I have a confusion in regards to DFA and NFA

Every language is regular if there exists a DFA or NFA that accepts it. Let's consider just the languages with finite alphabets and let's construct a language $T$ with this alphabet. My confusion is that I think that every language with finite alphabet has a DFA or NFA no matter what. We can do something like the following:

And that would accept any language defined in $\sum$

Your NFA accepts any word on $\Sigma$, not any language. Indeed, it accepts the (one) language that contains all finite words on $\Sigma$. This is usually written $\Sigma^*$.
Every finite-word language $L$ on $\Sigma$ is a subset of $\Sigma^*$, which is regular, but that does not mean that $L$ is regular, just like a subset of an open set does not need to be open and a subset of an infinite set does not need to be infinite (or finite, for that matter).
Finally, it should be noted that being regular, for a language, is rather the exception than the norm. As long as $\Sigma$ in not empty, there are uncountably many subsets of $\Sigma^*$, but only countably many finite-state automata.
• Yes correct, but let see that we have a subset of $\sum^*$. That machine will accept any word of that subset, so that subset is a regular language. My confusion is that. Maybe, the original statement needs to be restated to a machine that just accepts that specific language. – TheMathNoob Feb 8 '17 at 4:00
• If an automaton accepts all the words in language $L$ and then some, then it does not accept $L$, but a superset of it. That's the definition. – Fabio Somenzi Feb 8 '17 at 4:03