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LUXEMBOURG : Switzerland's Timea Bacsinszky will on Sunday attempt to win her first WTA tour title when she faces German sixth seed Sabine Lisicki in the final of the 222,000-dollar Luxembourg Open.
Bacsinszky, the world number 70, put out Belgian fifth seed Yanina Wickmayer, last week's Linz champion, 3-6, 6-2, 7-5.
Lisicki, meanwhile, saw off Israel's Shahar Peer 6-3, 4-6, 7-6 (7/5) in a 2hr 20min semi-final marathon.
Tournament favourites, US Open winner Kim Clijsters and Danish world number six Caroline Wozniacki, were eliminated earlier in the tournament.
- AFP /ls
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