Alex De Minaur Stuns Kei Nishikori at the US Open 2019

By 1 year ago

Rising Australian star Alex De Minaur stunned the world No.7, Kei Nishikori 6-2, 6-4, 2-6, 6-3 in 2 hours and 43 minutes at the US Open 2019 to reach the round of 16 for the first time in a Grand Slam tournament.

Set 1

Alex De Minaur dominated the opening set right from the start as he broke Kei Nishikori in his first service game. The Australian once again broke Nishikori in the third game to get a comfortable 3-0 lead over the Japanese. However, Nishikori finally claimed his game after breaking De Minaur in the next game.

However, it was the 20-year-old who was controlling the set. He broke Nishikori’s service for the third time and held his serve while serving for the set to win the opener 6-2.

Set 2

The second set was a replica of the first one as Alex De Minaur broke the 29-year-old in the first game. It looked like the world No.38 will go through the set. However, Nishikori fought back hard to break back in the eighth game as he leveled back the set.

Kei Nishikori couldn’t hold his next service game as he was broken for the fifth time in the match. The Australian then served out the second set 6-4.

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Set 3

Alex De Minaur was looking to seal the match after a dominating opening two sets. However, Kie Nishikori retaliated to break De Minaur in the fourth game of the third set and took a lead for the first time in the match. The Japanese took the third set 6-2 after breaking De Minaur when the latter was serving to stay in the set.

Set 4

The fourth set was quite an equal one until 3-3 when Alex broke Nishikori to take a lead at 4-3. He then went onto win his first match against Kei Nishikori after securing the fourth set 6-3. He will face either Grigor Dimitrov or Kamil Majchrzak in the fourth round of the US Open 2019.

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