Is it possible to select by range on leftmost multiple-index?

Question

I have an index m_idx on (k1,k2,k3)
If I do
SELECT c1,c2,c3... FROM tb FORCE INDEX (m_idx) WHERE k1=500 AND k2 IN(...) AND k2>2000 ORDER BY k1 LIMIT 1000;
OR
SELECT c1,c2,c3... FROM tb FORCE INDEX (m_idx) WHERE k1 IN (500,1000,1500 ...) AND k2 IN(...) AND k2>2000 ORDER BY k1 LIMIT 1000;
Handler_read_next = 999

BUT if I try to use a range on k1:
SELECT c1,c2,c3... FROM tb FORCE INDEX (m_idx) WHERE k1>=500 AND k2 IN(...) AND k2>2000 ORDER BY k1 LIMIT 1000;
Handler_read_next = 58035
In all cases EXPLAIN says that the key used is m_idx
But I think that in the third case m_idx it's not used (I also have an index only on k1).
Otherwise I don't understand why it'is reading more than 1000 rows.
I was expecting to scan m_idx index, and ONLY the first 1000 rows that meet the conditions to be read from table.
But in fact I think that for the third case it scans the index and those rows who meet k1 condition are read from the tb and k2 and k3 conditions are checked after the rows are read from tb.
I use: MySql with MyISAM, WINDOWS 7 64, the tb has 1 mil rows;
So my questions are:
Is it possible to select by range on leftmost multiple-index?
OR
I'm doing something else wrong?
Thank you.

Answer 1

1. No it is not.
2. No, you're doing everything right

http://dev.mysql.com/doc/refman/5.1/en/range-optimization.html ("7.3.1.3.2. The Range Access Method for Multiple-Part Indexes" part)

So you cannot help optimizer to perform this query faster.

And avoid using FORCE INDEX, since optimizer knows better what index to use.

Also:

k1>=500 AND k2 IN(...) AND k2>2000

Depending of which part returns fewer records, either k1>=500 or k2 IN(...) AND k2>2000 (I'm not sure why do you need > 2000 here, due to you can compare it manually, before add to IN()), you could also try to create index k2 (if k2 part returns less amount of records).