Prove NLOGSPACE$\subset$PSPACE Condidering the proof, NLOGSPACE$\subset$PSPACE
I wrote following proof:
NLOGSPACE = NSPACE$(\log n)$ $\hspace{15pt} \because$ Definition of NLOGSPACE
NSPACE$(\log n)$ $\subseteq$ DSPACE$(\log^2 n)$ $\hspace{15pt} \because$ Theorem(1)
We let $m = \log n$, then,
DSPACE$(m^2)$ $\subset$PSPACE $\hspace{15pt} \because$ Definition(2)
Therefore, NLOGSPACE$\subset$PSPACE.
Theorem(1):
If $S$ is a space constructible function and $S(n) \ge \log n$, then NSPACE$(S(n)) \subseteq$ DSPACE$(S^2(n))$.
definition(2): PSPACE = DSPACE$(n)$$\cup$DSPACE$(n^2)$$\cup$DSPACE$(n^3)$$\cup$...$\cup$DSPACE$(n^k)$$\cup$...
But I am still not that sure if that &quot;We let $m = \log n$&quot; is ok, is this proving NLOGSPACE$\subset$PSPACE?
What would you think about a proof that says the following?

$NP \subset DTIME(2^n)$
let $m = \log n$, then $DTIME(2^n) = DTIME(m) \subset P$
Therefore $P = NP$

Here, you have indeed $DPSACE(log^2n)\subset PSPACE$ but the argument is weird.
Also please note that Savitch's theorem already proves that $NPSPACE \subset PSPACE$, and you have the result you want, since $NLOGSPACE \subset NPSPACE$.