Prove PSPACE is closed under complement?

Question

How would you prove PSPACE is closed under complement? So far, my thought process is that we can create an algorithm to show that P is closed under complement. I'm struggling with how I can connect that to PSPACE...

Answer 1

Let $P$ be an arbitrary language in PSPACE. In other words, there exists a Turing machine $T$ that decides (using polynomial space) whether an arbitrary input $x$ is in $P$.

Now we can write an algorithm to decide whether $x$ is in $\neg P$:

• Run $T$ on $x$
• If $T$ accepts, reject
• If $T$ rejects, accept

This takes however much space $T$ takes, plus a constant. A polynomial plus a constant is still a polynomial. Therefore, $\neg P$ is in PSPACE.