Is intersection of regular language and context free language is "always" context free language

Question

I have read that intersection of regular language and context-free language is always context-free. Most of the places an standard example has been used to prove this, e.g., \begin{align*} L_1 &= L(a^*b^*)\\ L_2 &= \{a^nb^n\mid n\geq 0\} \quad\text{(which is context free)}\\ L_1\cap L_2 &= \{a^nb^n\mid n\geq 0\}\,. \end{align*} But my question is what if the regular language is finite, such as \begin{align*} L_1 &= \{ab\}\\ L_2 &= \{a^nb^n\mid n\geq 0\}\\ L_1\cap L_2 = \{ab\}\,. \end{align*} Here intersection comes out to be finite and the language generated by intersection of both language is also finite which is regular (I know it's also context-free but regular is a stronger answer here).

What mistake am I making in understanding the concept?

Answer 1

The claim is that the intersection of a regular language and a context-free language is context-free. You've intersected a regular language ($\{ab\}$) and a context-free language ($\{a^nb^n\mid n\geq 0\}$) and the result was a context-free language ($\{a b\}$). Sure, that language is also regular but every regular language is also context-free. The statement isn't that the intersection of a regular language and a context-free language is a non-regular context-free language.

Analogously, the child of two mammals is a mammal. You've taken two mammals and you're saying, "Why is their child a cat? It's supposed to be a mammal."

Answer 2

Your mistake is in interpreting the meaning of the statement

The intersection of a regular language and a context-free language is a context-free language.

This statement means that if $R$ is regular and $L$ is context-free then $R \cap L$ is context-free. It doesn't state that $R \cap L$ is not regular; this is just not part of the claim.

Your first example shows that $R \cap L$ could be not regular, but it could also be regular, as your other example shows.