A question regarding definition of Deterministic Subexponential Time (SUBEXP)

Question

First Look at the definition of SUBEXP from Complexity Zoo:

SUBEXP: (Deterministic Subexponential-Time) The intersection of DTIME($2^{n^\epsilon}$) over all $\epsilon$>0. (Note that the algorithm used may vary with $\epsilon$.) or it can be written as: SUBEXP = $\bigcap_{\epsilon>0}$DTIME$(2^{n^\epsilon})$.

So, I bring the definition of EXP which is:

EXP = $\bigcup_{k\geq 1}$DTIME$(2^{n^k})$

The definition of EXP is clear, since it includes all polynomial of n to the power of 2. (e.g. $2^{n^{30}}$ or $100^{n^{99}}$ etc.)

First question: what is domain of $\epsilon$? I guess it is between 0 and 1 but it didn't specify in the definition. Is it usual that when we have $\epsilon$ then it means between 0 and 1.

Second question: Now, in case of SUBEXP, it is not clear how the definition is about the intersection? I mean, Shouldn't be written as following: $\bigcup_{1>\epsilon>0}$DTIME$(2^{n^\epsilon})$. For example by definition above what is the intersection of: $2^{n^{0.01}} \bigcap 2^{n^{0.02}} ?$

Third question: There are two definition of SUBEXP in wikipedia, Is there definition that take over all subexponential or we don't since this is why we have two definitions.

Thank you!

Answer 1

In the definition of SUBEXP, $\epsilon$ ranges over all positive reals. But you get the same definition if you ask that $\epsilon < \epsilon_0$, for an $\epsilon_0>0$ of your choice; if you ask that $\epsilon$ be rational; if you only go over $\epsilon = 1/n$; and so on. This is because DTIME is monotone: if $f \leq g$ then $\mathsf{DTIME}(f) \subseteq \mathsf{DTIME}(g)$.

An alternative definition of SUBEXP would be: $$ \mathsf{SUBEXP} = \bigcup_{g(n) = o(1)} \mathsf{DTIME}(2^{n^{g(n)}}), $$ often denoted simply by $\mathsf{DTIME}(2^{n^{o(1)}})$.

Some examples: $\mathsf{P} \subseteq \mathsf{SUBEXP}$; a function which can be computed in time $2^{n^{1/\log\log n}}$ is in $\mathsf{SUBEXP}$; and a function which can be computed in time $2^{\log^{10} n}$ is in $\mathsf{SUBEXP}$.

In contrast, a function which can be computed in time $2^{n^{1/10}}$ is not necessarily in $\mathsf{SUBEXP}$ (and by the time hierarchy theorem, there is such a function which lies outside $\mathsf{SUBEXP}$).

A function in $\mathsf{DTIME}(2^{n/\log n})$ lies in SUBEPT but not necessarily in SUBEXP.