This question is language agnostic but is inspired by these c/c++ questions.
How to convert a single char into an int
Char to int conversion in C
Is
I don't know about all encodings, but at least in ASCII and <shudder> EBCDIC, the digits 0-9 all come consecutively and in increasing numeric order. Which means that all ASCII- and EBCDIC-based encodings should also have their digits in order. So for pretty much anything you'll encounter, barring Morse code or worse, I'm going to say yes.
Both the C++ Standard and the C standard require that this be so, for C++ and C program text.
All text encodings I know of typically order each representation of digits sequentially. However, your question becomes a lot broader when you include all of the other representations of digits in other encodings, such as Japanese: 1234567890. Notice how the characters for the numbers are different? Well, they are actually different code points. So, I really think the answer to your question is a hard maybe, since there are so many encodings out there and they have multiple representations of digits in them.
A better question is to ask yourself, why do I need to count on digits to be in sequential code points in the first place?
According to K&R ANSI C it is.
Excerpt:
..."This particular program relies on the properties of the character representation of the digits. For example, the test
if (c >= '0' && c <= '9') ...
determines whether the character in
c
is a digit. If it is, the numeric value of that digit is
c - '0'
This works only if
'0', '1', ..., '9'
have consecutive increasing values. Fortunately, this is true for all character sets...."
You're going to find it hard to prove a negative. Nobody can possibly know every text encoding ever invented.
All encodings in common use today (except EBCDIC, is it still in common use?) are supersets of ASCII. I'd say you're more likely to win the lottery than you are to find a practical environment where the strict ordering of '0' to '9' doesn't hold.