Oklahoma City’s Shai Gilgeous-Alexander, Denver’s Nikola Jokic and Milwaukee’s Giannis Antetokounmpo were the only players to appear on every MVP ballot this season.
It only made sense that they would be unanimous All-NBA picks as well.
Gilgeous-Alexander — the league’s MVP — along with Jokic and Antetokounmpo were unveiled Friday night as first-team All-NBA players, along with Boston’s Jayson Tatum and Cleveland’s Donovan Mitchell.
Tatum was another unanimous first-team pick. Mitchell made the first team for the first time.
Antetokounmpo has seven first-team selections and nine appearances on the All-NBA team overall. Jokic is a five-time first-teamer and seven-time All-NBA pick, Tatum is first-team for the fourth time (fifth overall), Gilgeous-Alexander has been first-team in all three of his All-NBA appearances, and Mitchell is All-NBA for the second time in his career.
Gilgeous-Alexander, Jokic, Antetokounmpo and Tatum were all first-teamers last season as well.
LeBron James of the Los Angeles Lakers is an All-NBA player for the 21st time in 22 seasons. He made the second team.
Also on the second team: Minnesota’s Anthony Edwards, Cleveland's Evan Mobley, Golden State’s Stephen Curry and New York's Jalen Brunson.
Curry made All-NBA for the 11th time, something only 21 players in NBA history have done. Edwards and Brunson are two-time All-NBA players, and Mobley made a team for the first time.
Detroit's Cade Cunningham, Indiana's Tyrese Haliburton, the Los Angeles Clippers' James Harden, New York's Karl-Anthony Towns and Oklahoma City's Jalen Williams were named to the third team.
Harden is an eight-time selection, Towns is a three-time pick, Haliburton made a team for the second time and Cunningham and Williams both
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