Proof Attempt: non Context free languages and non regular language with some words added stay non regular/context-free.

Question

Background: I was in class and heard this sentence from my professor "Well if you take a non context free language and add $2$ words to it then it will stay non context free".
Now I was trying to understand why does this happen always and how can I prove something like this, If we take the language $L=\{ a^nb^nc^n:n\ge0\}$ which isn't context free, and add $a,b$ to it this way: $\{a^nb^nc^n:n\ge0\}\cup\{a\}\cup\{b\}$, I can intuitively see that there's just a simple difference where this language accepts $a$ and $b$ words.

But my problem is how do I prove it formally for all languages, if this was about regular languages, since they're closed under intersection, I could assume that the new language with a couple more words is regular, take the intersection and get the irregular language I started with and reach a contradiction proving it's irregular.
But context free languages aren't closed under intersection, so I can't just say lets take $L\cap \{a^nb^nc^n:n\ge0, a, b\}=L$ it's non context free so $\{a^nb^nc^n:n\ge0, a, b\}$ is not context free.

I would appreciate any help of how to prove this formally for non context-free languages, and any feedback on my attempts is really appreciated.
Thanks in advance.

Answer 1

Indeed you cannot use the general closure under intersection, because it does not hold for context-free languages. So we proceed in a different way. Suppose you add a finite language $F$ to a non-context-free language $L$ and the result $K= L\cup F$ is context-free. Here $$L= (K- F) \cup (L\cap F).$$ $(K-F)$ is context-free, because $K$ is and $F$ is finite. $(L\cap F)$ might look more complicated at first sight, because $L$ is non-context-free; however, since $F$ is finite, also $(L\cap F)$ is finite and thus it is context-free. So we obtain $L$ as the union of a context-free and a finite language, which means that $L$ should also be context-free. But this contradicts our assumption.

In case you do not see why $(K-F)$, the difference of a context-free and a finite language must again be context-free: take a PDA accepting $K$. You can modify its control such that it additionally checks the input against the finite list of words in $F$ while it does the computation for $K$. If it accepts a word via the computation for $K$ but this word is on the list, the modified PDA rejects.

Answer 2

The key idea is that you can still take apply the Pumping Lemma for CFLs to $\{ a^{n}b^{n}c^{n} : n \geq 0\}$ to get some word outside of $L = \{ a^{n}b^{n}c^{n} : n \geq 0 \} \cup \{a,b\}$.

Note that if $L$ is not a CFL and we add infinitely many words, we may obtain a CFL (and in fact, a regular language). Let $\Sigma$ be our alphabet. As an example, $L \cup (\Sigma^{*} \setminus L) = \Sigma^{*}$, which is regular.

So the assumption about adding a finite number of words to remain a non-CFL is critical here.