[proofplan] We prove the containment by taking an arbitrary language decided in polynomial space and bounding the number of configurations of its deciding Turing machine. A deterministic decider cannot visit the same nonhalting configuration twice, since doing so would force an infinite loop. The number of configurations using polynomially many tape cells is exponential in a polynomial, and the polynomial factors can be absorbed into a bound of the form $2^{n^m}$ for a larger natural number $m$. [/proofplan] [step:Choose a polynomial space decider for the language] Let $\Sigma$ be a finite alphabet and let $L \subseteq \Sigma^*$ satisfy $L \in \operatorname{PSPACE}$. By the definition of $\operatorname{PSPACE}$, there exist a deterministic Turing machine $M$, a natural number $k \in \mathbb{N}$, and a constant $A \geq 1$ such that $M$ decides $L$ and, for every input word $x \in \Sigma^*$ of length $n = |x|$, the computation of $M$ on $x$ uses at most \begin{align*} s(n) := A(n+1)^k \end{align*} tape cells. Let $Q$ denote the finite state set of $M$, and let $\Gamma$ denote the finite tape alphabet of $M$. Set \begin{align*} q := |Q| \end{align*} and \begin{align*} g := |\Gamma|. \end{align*} Since $M$ is a Turing machine, $q \geq 1$ and $g \geq 2$. [/step] [step:Bound the number of configurations available on inputs of length $n$] Fix an input word $x \in \Sigma^*$ with $|x| = n$. During the computation of $M$ on $x$, a configuration is determined by the current state, the head position among the at most $s(n)$ used tape cells, and the tape contents on those cells. Hence the number $N(n)$ of possible configurations reachable while respecting the space bound satisfies \begin{align*} N(n) \leq q\,s(n)\,g^{s(n)}. \end{align*} [guided] Fix an input word $x \in \Sigma^*$ and write $n := |x|$. The point of introducing configurations is that a deterministic computation is completely determined by its current configuration. For the fixed input $x$, a configuration of $M$ consists of three pieces of data: the current state in $Q$, the current tape-head position among the cells used by the computation, and the symbol written in each used tape cell. The space bound says that at most $s(n) = A(n+1)^k$ tape cells are ever used. Therefore there are at most $q = |Q|$ choices for the state, at most $s(n)$ choices for the head position, and at most $g^{s(n)}$ choices for the tape contents, because each of the at most $s(n)$ cells contains one of the $g = |\Gamma|$ tape symbols. Multiplying these independent choices gives the configuration bound \begin{align*} N(n) \leq q\,s(n)\,g^{s(n)}. \end{align*} This estimate is the whole reason polynomial space implies exponential time: the tape contents contribute $g^{s(n)}$, and when $s(n)$ is polynomial in $n$, this is exponential in a polynomial in $n$. [/guided] [/step] [step:Use determinism and halting to convert configurations into a time bound] Because $M$ decides $L$, it halts on every input. Since $M$ is deterministic, if its computation on $x$ ever repeated a nonhalting configuration, the future computation from that configuration would repeat exactly and the machine would never halt. Therefore no nonhalting configuration appears twice in the computation on $x$. It follows that the number of computation steps on input $x$ is at most $N(n)+1$. Thus $M$ decides $L$ in time bounded by \begin{align*} q\,s(n)\,g^{s(n)} + 1. \end{align*} [/step] [step:Absorb the configuration bound into a single exponential polynomial] For every $n \geq 1$, \begin{align*} s(n) = A(n+1)^k \leq A(2n)^k. \end{align*} Define \begin{align*} B := A2^k. \end{align*} Then $s(n) \leq Bn^k$ for every $n \geq 1$, so \begin{align*} q\,s(n)\,g^{s(n)} + 1 \leq qB n^k g^{Bn^k} + 1. \end{align*} Define $D > 0$ large enough so that, for every $n \geq 1$, \begin{align*} qB n^k g^{Bn^k} + 1 \leq 2^{D n^{k+1}}. \end{align*} Such a constant exists because the logarithm base $2$ of the left-hand side is bounded above by a constant multiple of $n^k + \log n$, and $n^k + \log n \leq 2n^{k+1}$ for all $n \geq 1$. Choose $m \in \mathbb{N}$ with $m \geq k+2$ and large enough that \begin{align*} D n^{k+1} \leq n^m \end{align*} for all sufficiently large $n$. Enlarging the implicit constant in the time bound to cover the finitely many smaller values of $n$, the machine $M$ runs in time $O(2^{n^m})$. [/step] [step:Conclude membership in exponential time] By the definition of deterministic time classes, the existence of a deterministic decider for $L$ running in time $O(2^{n^m})$ means \begin{align*} L \in \operatorname{TIME}(2^{n^m}). \end{align*} Since $L \in \operatorname{PSPACE}$ was arbitrary, every language in $\operatorname{PSPACE}$ lies in $\operatorname{TIME}(2^{n^m})$ for some $m \in \mathbb{N}$. Therefore \begin{align*} \operatorname{PSPACE} \subseteq \bigcup_{m \in \mathbb{N}} \operatorname{TIME}(2^{n^m}). \end{align*} [/step]