# In the proof that $DSPACE(S)\subseteq DTIME(2^{O(S)})$, why precisely do we require that $S=\Omega(\log n)$

I have read and understood various proofs, but have not been able to understand precisely why we require $$S=\Omega(\log n)$$.

We believe that $$DSPACE(X) \subseteq DTIME(X)$$ is false. As a consequence, $$X$$ must be chosen so that $$X \neq 2^{O(X)}$$. Additionally, considering $$DSPACE(X) \subseteq DTIME(Y)$$ we want $$Y = 2^{O(X)}$$ to allow the Turing machine requiring $$X$$ space to halt on all of its inputs: this is because the number of configurations is $$O(2^{O(X)})$$ and any accepting computation must halt within $$O(2^{O(X)})$$ steps, i.e. this is $$2^{O(X)}$$ if $$X = \Omega(\lg n)$$. Moreover, the function $$X$$ must be space constructible as well, and $$\lg n$$ is space constructible. This is also consistent with the definition of the complexity classes $$L$$ and $$P$$. The former is the class of the languages (or problems) requiring at most $$O(\lg n)$$ space, the latter the class of languages requiring at most polynomial time. The relationship $$L \subset P$$ holds because a Turing machine requiring at most logarithmic space can use at most linear time, so polynomial time.