您可以使用CTE，然后使用滞后/超前函数来检查是否有并列关系：

with ranked as (
  SELECT name,
        points, 
        rank() OVER (ORDER BY points DESC) AS point_rank
  FROM teams
)
select name, points, point_rank, 
       (   point_rank = lag(point_rank, 1, -1::bigint) over (order by point_rank) 
        or point_rank = lead(point_rank, 1, -1::bigint) over (order by point_rank)
       ) as is_tie
from ranked;

第一行和最后一行需要lag和lead函数的默认值，以避免在那里检查空值。
示例：https://dbfiddle.uk/-01aFLr4

一种方法是将当前查询放入公用表表达式中，然后使用它来标识哪些排名重复：

WITH cte AS (
    SELECT name,
           points, 
           rank() OVER (ORDER BY points DESC) AS point_rank
    FROM teams;
)

SELECT cte.name,
       cte.points,
       cte.point_rank
       CASE WHEN t.point_rank IS NOT NULL THEN 'false' ELSE 'true' END AS tie
FROM cte
LEFT JOIN
(
    SELECT point_rank
    FROM cte
    GROUP BY point_rank
    HAVING COUNT(*) = 1
) t
    ON cte.point_rank = t.point_rank

SELECT name
   , points
   , rank() OVER (rrr) AS point_rank
   -- , count(*) OVER (ppp) AS ppp_cnt
   , rank() OVER (pp2) AS sub_rank
   , (COUNT(*) OVER (ppp) > 1) AS is_tie
FROM teams
WINDOW ppp AS (PARTITION BY points )
        , pp2 AS (PARTITION BY points ORDER BY ctid )
        , rrr AS (ORDER BY points DESC)
ORDER BY points DESC
        ;

结果（我额外添加了两行）：

DROP SCHEMA
CREATE SCHEMA
SET
CREATE TABLE
INSERT 0 6
  name  | points | point_rank | sub_rank | is_tie 
--------+--------+------------+----------+--------
 Team_1 |     14 |          1 |        1 | f
 Team_2 |     11 |          2 |        1 | f
 Team_3 |      9 |          3 |        1 | t
 Team_4 |      9 |          3 |        2 | t
 Team_5 |      5 |          5 |        1 | t
 Team_6 |      5 |          5 |        2 | t
(6 rows)