Victoria Azarenka, Aryna Sabalenka and Aliaksandra Sasnovich broke into the third round of a Grand Slam tournament at this year’s US Open.
You go, girls!
The former world number one and her younger compatriots won their first-round matches and are not going to stop at it.
This is for the fird time in history when four Belarusian women featured in the third round of Grand Slam.
Recall that Aliaksandra Sasnovich, Aryna Sabalenka and Vera Lapko announced themselves as Belarus’ representatives at Fed Cup final.
— US Open Tennis (@usopen) August 30, 2018
Meanwhile, their longtime leader remained in California during a child-custody battle with her former husband.
Since then, both Sabalenka and Sasnovich have shot up the rankings.
Sabalenka, who is the youngest of the elite group, has jumped from number 78 to top 20.
Besides rocketing in the world ranking last weekend, the Brising star claimed her first-ever WTA singles title.
Sasnovich, world number 35, and Lapko, world number 72, also had a breakout summer.
All three Belarusians are ranked higher than Azarenka, who is up to number 79 after starting the year number 209.
Azarenka is to face defending champion Sloane Stephens in the third round Friday.
Unfortunately, Vera Lapko and Aliaksandra Sasnovich were eliminated by their rivals this Wednesday.