CFLs are not closed under complement
For a deterministic finite automaton, complement was almost free: flip the accepting and rejecting states and you recognize all the OTHER strings. You might hope the same trick works for context-free languages. It does not. The context-free languages are NOT closed under complement: there are context-free languages whose complement (every string over the alphabet NOT in the language) escapes the family entirely.
The cleanest way to see it is to combine two facts you already trust. First, the CFLs ARE closed under union. Second, they are NOT closed under intersection. Now recall De Morgan's law from set theory: L1 ∩ L2 equals the complement of ( (complement of L1) ∪ (complement of L2) ). Suppose, for contradiction, the CFLs WERE closed under complement. Then starting from two CFLs L1 and L2 we could complement each (still CFL), union them (still CFL, since union is safe), and complement again (still CFL) — and that whole expression equals L1 ∩ L2. We would have built the intersection of two CFLs as a CFL, for ANY two CFLs. But we know intersection is not closed. Contradiction. So complement cannot be a safe operation.
The deeper reason is the same one-stack limitation behind the intersection failure, plus an asymmetry of nondeterminism: a PDA accepts if SOME computation accepts, and there is no easy way to flip that to 'NO computation accepts' the way you can with a deterministic finite automaton. (The smaller class of DETERMINISTIC context-free languages IS closed under complement — exactly because a DPDA, like a DFA, has a single computation you can negate — which is one concrete way the deterministic CFLs are a genuinely different, smaller family.)
If complement were safe: take two CFLs, complement both (safe by assumption), union (always safe), complement again (safe by assumption). The result is exactly their intersection — making EVERY intersection of CFLs a CFL. But a^n b^n c^n shows intersection is not closed. The assumption must be false.
Complement closure plus union closure would force intersection closure — which fails, so complement is not closed.
Not closed does NOT mean 'never closes' — many specific CFLs have context-free complements. And the smaller DETERMINISTIC CFL family IS closed under complement; the non-closure is a property of the full context-free class.