Zverev and Rublev hit the streets in Rotterdam

the russian Andrey Rublev, number five in the world rankings, was eliminated from his debut at the ATP 500 tournament in Rotterdam (Netherlands), after losing this Wednesday 6-4, 6-4 against the Australian Alex de Minaur.

The match soon turned complicated for Rublev, winner of Rotterdam in 2021, which saw De Miñaur capitalize on his first break point in the third game to break the Russian tennis player’s serve.

It was enough for the Australian player, number 25 in the ATP ranking, to score 6-4 in the first set in 44 minutes of play.

A script that repeated itself in the second set where De Miñaur again broke the serve of a Rublev in the third game, who in the opening game of the second set saved up to three breaking balls.

Advantage that the Australian player maintained until the end to score the second and final set 6-4 that ensured his presence in the round of 16 where Alex De Miñaur will face the American Maxime Cressywho played in the final of the Montpellier tournament last week.

Even more surprising was the German defeat Alexander Zverev, Seeded number eight in the tournament, he was eliminated in the second round after losing 4-6, 6-3, 6-4 against Dutch Tallon Griekspoor, number 61 in the world ranking.

Zverev was helpless to win 6-4 in the first set, after seeing how Griekspoor, winner of this year’s tournament in Pune (India), forced a third and final set by winning 6-3 in the second set.

A set where Tallon Griekspoor, who added the fourth victory of his career to a “top” 20, all of them in Rotterdam, asserted the “break” he achieved in the seventh game to win the sleeve (6-4) and the ticket to the quarterfinals.