Check if all of the delimiters in an expression are matched and closed

Problem

Write an algorithm to determine if all of the delimiters in an
expression are matched and closed.

{(abc)22}[14(xyz)2] should pass

[ { ] } should fail

{ (x) } [ should fail

Any advice of code functional bug, performance in terms of algorithm time complexity, code style are appreciated.

delimiters = ['[', ']', '(', ')', '{', '}']
delimiters_map = {']':'[', ')':'(', '}':'{'}

def check_match(source):
    working_set = []
    for i,v in enumerate(source):
        if v not in delimiters:
            continue
        if v in delimiters_map: # end of a delimiter:
            if working_set[-1] != delimiters_map[v]:
                return False
            else:
                working_set.pop(-1)
        elif v in delimiters:
            working_set.append(v)

    if len(working_set) > 0:
        return False
    else:
        return True

if __name__ == "__main__":
    print check_match('{(abc)22}[14(xyz)2]')
    print check_match('[ { ] }')
    print check_match('{ (x) } [')

Performance

Checking if a list contains a value is an \$O(n)\$ operation.
To make if v not in delimiters fast,
it would be better to make delimiters a set instead of a list.

The elif v in delimiters: is unnecessary, because at that point we already know that v is in delimiters, so you can replace with a simple else:.

Creating lists and maps of characters

This is a PITA to type:

delimiters = ['[', ']', '(', ')', '{', '}']
delimiters_map = {']':'[', ')':'(', '}':'{'}

An easier and less error-prone way is to call list to turn a string into a list of characters:

openers = list('[({')
closers = list('])}')
delimiters = set(openers + closers)
delimiters_map = dict(zip(closers, openers))

If it's a stack, call it a stack

working_set is a misleading name. It suggests a set, but it's a list.
And you're using it as a stack, so I suggest to call it a stack.

Use doctests

Doctests are awesome. You can run them with python -mdoctests script.py.

def check_match(source):
    """
    >>> check_match('{(abc)22}[14(xyz)2]')
    True

    >>> check_match('[ { ] }')
    False

    >>> check_match('{ (x) } [')
    False

    >>> check_match('')
    True

    """
    # ... (your implementation)

StackExchange Code Review Q#150771, answer score: 7