Is $a$ < $b$ if $f(x)$ is differentiable on $(a,b)$ and continuous in $[a,b]$? $f(x)$ is differentiable on $(a,b)$ and continuous in $[a,b]$.

In theory $a$ can be equal to $b$. But then if $f(x)$ is defined on one point it's not differentiable because by definition derivative is the rate of change between 2 points on an interval. 

By conventional notation we write the left-most coordinate of an interval adjacent to the left bracket and the right-most coordinate of the interval adjacent to the right bracket which in such case would mean that $a$ is before $b$.

So can we claim that $a&lt;b$?

Or maybe if given $(a,b)$ the convention pre-supposes that $a \neq b$?
In my opinion (and these three words imply that this is just some very subjective thinking) it depends on how you define differentiability. If your definition of differentiability requires $f$ to be defined on an open interval (an interval of the form $(a,b)$ with $a&lt;b$), then you are right, because then if $f$ is defined only on a set $[a,a]=\{a\}$ with one point, talking about differentiability of $f$ doesn't make any sense (because for this case it is not defined, and - as you have pointed out already - cannot tell anything about the rate of change, cause $f$ cannot change anyhow since there is only one value at which $f$ is defined).

However, it is sometimes convenient to define differentiability for those functions by just defining them to be differentiable (From a logical point of view this also makes sense and guides you sooner or later to the question how statements about the empty set are handled, see the comments above). This is, at first glance, pure convention. That way, one can replace a statement like "Let $f:[a,b]\to\mathbb R$ $\color{red}{\text{for }a&lt;b}$ be continuous and differentiable in $(a,b)$ such that ..." by something like "Let $f:[a,b]\to\mathbb R$ be continuous and differentiable in $(a,b)$ such that ...". The second statement is easier to read because there are no redundant information. But as you can see, it is only a matter of taste.

Let's take a closer look at the definition of differentiability. What does
\begin{align*}
\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}\quad\text{exists}
\end{align*}
really mean? It means that for each sequence of numbers $(x_n)$ contained in the domain of definition of $f$ and so that $x_n\neq x_0$ for each $n$, the limit
\begin{align*}
\lim_{n\to\infty}\frac{f(x_n)-f(x_0)}{x_n-x_0}
\end{align*}
exists and is the same, no matter how the sequence $(x_n)$ actually looks like. Thus, $f$ is differentiable at $x_0$, if this works for each such sequence. But if there is no such sequence, then there is nothing to check, there is nothing that can go wrong, so, well, then we say, $f$ is indeed differentiable. But, again, this is our choice. We also could say, that $f$ is not differentiable in this case, and we are again talking about empty sets and how to treat them.

From a porgrammers point of view: Imagine that you would like to write a program that checks whether all elements of a set of objects (a list) $L$ satisfy a given property $P$, say. Thus, your function yields true, if all elements of $L$ satisfy $P$, otherwise false. You probably know that functions (depends on the language) have to return something, in any case! So we would probably write our function like this:

bool function check(List L, Property P) {
   for each (x in L) {
       if (!P(x))
           return false;
   }
   return true;
}


In your case, $P(x)=f$ is differentiable at $x$, and $L=(a,b)$. In case $L=\emptyset$, check(L, P) returns true. Try to rewrite this function so that it returns false if $L=\emptyset$. You would need a couple more lines. In summary, it is just handy to say that any property for elements of the empty set is true.
Again (see comments) from a logical point of view, this is the only reasonable way to go. If the statement "Each element of the empty set is red" was false, then its negation "There is one element of the empty set which is not red" had to be true, which is clearly impossible.

To avoid all those discussions, my former Analysis Prof. just defined intervals to contain at least two elements. But again, just definition/convention. To cite my former Algebra Prof.: "When it comes to mathematical definitions there is no true or false, just convenient or inconvenient."
Note that if $a=b$ then $(a,b)$ is an empty set thus "f has a derivative on $(a,b)$" is a true statement.

Thus, by convention of noting lower bound first we can only claim $a \leq b$.

EDIT:
About the derivative on empty set.
In logic, a statement is either true or its logical negation is true (which is the same as saying that the statement is false).

Now, the logical negation of "f has a derivative $\forall x \in \emptyset$" is "for atleast one $x \in \emptyset$ $f(x)$ has no derivative".

Since there are no points in the empty set, one can't find a point for which $f(x)$ has no derivative.  Thus logical negation is false, making our statement true.

Conclusion: "f has a derivative $\forall x \in \emptyset$".