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Azarenka leads Australian Open seeding

Victoria Azarenka, shown during the 2012 U.S. Open women's singles final, was named the No. 1 seed for the 2013 Australian Open. Azarenka, the defending tournament champion, is ranked No. 1 in the world. UPI/John Angelillo
Victoria Azarenka, shown during the 2012 U.S. Open women's singles final, was named the No. 1 seed for the 2013 Australian Open. Azarenka, the defending tournament champion, is ranked No. 1 in the world. UPI/John Angelillo | License Photo

MELBOURNE, Jan. 7 (UPI) -- Victoria Azarenka was announced as the No. 1 women's seed for the Australian Open, where last year she won her first Grand Slam title.

Azarenka defeated Maria Sharapova in straight sets in winning the 2012 women's singles title in Melbourne in 2012.

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In addition to the first major title, the win allowed Azarenka to reach No. 1 in the rankings for the first time. She's been No. 1 for 46 of the 50 weeks since.

The women's singles seedings, announced Monday, follow the WTA rankings. The draws for the event are to be announced Friday. The tournament opens next Monday and the women's singles final is scheduled for Jan. 26.

Sharapova is seeded second and much attention will be placed Friday in which side of the draw organizers place third-seeded Serena Williams, who has lost just two matches since last March. Sharapova won the French Open last year but Williams took Wimbledon and the U.S. Open. Williams has won the Australian Open five times, most recently in 2010.

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Agnieszka Radwanska is the No. 4 seed, with No. 5 Angelique Kerber, sixth-seeded Li Na, seventh-seed Sara Errani and No. 8 Petra Kvitova in the next tier.

Seeds 9-16 are, in order: Samantha Stosur, Caroline Wozniacki, Marion Bartoli, Nadia Petrova, Ana Ivanovic, Maria Kirilenko, Dominika Cibulkova and Roberta Vinci.

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