# Concluding $SPACE(n^2) \neq SPACE(n^7)$ from universal turing machine running time

Let $M_U$ be an universal Turing machine which fulfills the following condition:

If $M$ running $x$ takes $f(x)$ space, then $M_U$ running on $\langle \langle M\rangle,x\rangle$ takes $(f(|x|))^3+2\cdot|x|+|\langle M\rangle|$ space.

Why can we conclude from the above that ${\sf Space}(n^2) \neq {\sf Space}(n^7)$?

-

I assume with ${\sf SPACE}$ you mean ${\sf DSpace}$. As usual for these types of hierarchy question you can use a diagonalization argument.
So build a machine $D$ that takes its input $w$ and then executes $M_u(\langle w ,w \rangle)$. While executing $M_u$ it checks also how much space is used and how many steps the machine $M_u$ has done. Then the result of $D$ is the following:
• if $D$ used more than $n^2$ space than reject,
• if $D$ does more than $2^{cn^2}$ steps then accept (in this case the simulation detected a cycle if $\langle w \rangle$ would be a $n^2$ space bounded TM),
• if none of the above occurred, then accept if $M_u$ rejected, and reject if $M_u$ accepted.
You can say that $D(w)$ anti-simulates the machine $M_u(\langle w ,w \rangle)$, if $\langle w \rangle$ is in ${\sf DSpace}(n^2)$. Clearly, $\langle D\rangle$ is in ${\sf DSpace}(n^7)$. However, it cannot be in ${\sf DSpace}(n^2)$. If it would be then you get a contradiction at $D(\langle D \rangle)$.
Is there a reason why this doesn't also apply for $\mathsf{NSPACE}$? (i.e. $\mathsf{NSPACE}(n^{2}) \neq \mathsf{NSPACE}(n^{7})$)? –  Luke Mathieson Dec 15 '12 at 15:11