Two lost tiebreakers didn't prevent Rublev from beating de Minaur and advancing to the quarterfinals

In Melbourne, Australia, the last match of the fourth stage of the Australian Open concluded today.

As a reminder, three matches among men and four among women had taken place earlier.

In the final match of the day, the tenth-ranked player in the world from Australia, Alex de Minaur, faced the fifth-ranked player in the ATP rankings, Andrey Rublev. The match lasted for 4 hours and 15 minutes.

In the first set, Alex served two double faults and made two errors, losing two games on his serve. Despite his opponent also making a fair number of mistakes, the Australian lost the first set 4-6 in favor of Rublev.

In the second set, the Australian got into the game, constantly catching up with his opponent. He brought the set to a tiebreak, where he won (7-5), leveling the sets at 1-1 with a 7-6 victory.

In the third set, De Minaur initially led 3-0, then 5-2, but Rublev caught up, and the set again went to a tiebreak. De Minaur emerged stronger in the tiebreak (7-4) and took another set 7-6 in his favor.

In the fourth set, with a record for himself, De Minaur showed some signs of fatigue. He made four winners, eight unforced errors, one double fault, and one serve into the net, resulting in a 3-6 set loss.

By the fifth and final set, both players approached the court quite tired as the time in Melbourne approached midnight. De Minaur struggled to play his usual game and move quickly across the court, losing all games on his serve and failing to win a single game on Rublev's serve. This led to a 0-6 set loss and an overall defeat for De Minaur.

Australian Open
Fourth Stage
Men's Singles
Alex de Minaur vs. Andrey Rublev 4-6, 7-6, 7-6, 3-6, 0-6

Alex de Minaur once again failed to advance to the fifth stage of a Grand Slam tournament. However, Andrey Rublev's victory secured his spot in the quarterfinals, where he will face the fourth-ranked player in the world from Italy, Jannik Sinner.