As an alternative to the selected answer, this statement will remove spans of contiguous space characters from a column and replace the spaces with a single space character. This operates ONLY on the space characters, and does not require any special "token" character (and does not require that the the special "token" character NOT appear within the source column.
UPDATE mytable
SET mycol =
REPLACE(
REPLACE(
REPLACE(
REPLACE(
REPLACE(
REPLACE(
mycol
,SPACE(21),' ')
,SPACE(13),' ')
,SPACE(8),' ')
,SPACE(5),' ')
,SPACE(3),' ')
,SPACE(2),' ')
WHERE mycol LIKE '% %'
To run this expression as a test, use it in a SELECT statement:
SELECT REPLACE(
REPLACE(
REPLACE(
REPLACE(
REPLACE(
REPLACE(
mycol
,SPACE(21),' ')
,SPACE(13),' ')
,SPACE(8),' ')
,SPACE(5),' ')
,SPACE(3),' ')
,SPACE(2),' ') AS new_mycol
FROM mytable t
WHERE t.mycol LIKE '% %'
This could be extended with larger spans of SPACE characters, for fewer replacements. The replacements start with the largest spans of contiguous space characters, and then the replacements are progressively smaller, to catch all of the occurrences of spaces as efficiently as possible. (This isn't the most "efficient" algorithm, but it works suitably.)