2007 ASB Classic – Doubles

Elena Likhovtseva and Vera Zvonareva were the defending champions, but neither chose to participate this year.

Doubles
2007 ASB Classic
Final
ChampionSlovakia Janette Husárová
Argentina Paola Suárez
Runner-upChinese Taipei Hsieh Su-wei
India Shikha Uberoi
Score6–0, 6–
Details
Draw16
Seeds4
Events
Singles Doubles
← 2006 · WTA Auckland Open · 2008 →

Janette Husárová and Paola Suárez won the title, defeating Hsieh Su-wei and Shikha Uberoi 6–0, 6–2 in the final.[1]

Seeds

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  1.   Janette Husárová /   Paola Suárez (champions)
  2.   Gisela Dulko /   Meilen Tu (first round; retired due to Dulko's right quad strain)
  3.   Eleni Daniilidou /   Jasmin Wöhr (semifinals)
  4.   Marta Domachowska /   Jelena Kostanić (first round)

Draw

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Draw

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First round Quarterfinals Semifinals Final
1   J Husárová
  P Suárez
6 6  
WC   M Erakovic
  S Errani
1 2   1   J Husárová
  P Suárez
6 6  
    É Loit
  Á Szávay
6 6       É Loit
  Á Szávay
2 1  
    C Ji
  T Li
4 4   1   J Husárová
  P Suárez
6 7  
3   E Daniilidou
  J Wöhr
77 2 6 3   E Daniilidou
  J Wöhr
4 5  
    C-w Chan
  V Uhlířová
65 6 3 3   E Daniilidou
  J Wöhr
6 6  
    E Birnerová
  J Craybas
5 3       J Janković
  T Križan
2 3  
    J Janković
  T Križan
7 6   1   J Husárová
  P Suárez
6 6  
    S Sun
  T Sun
7 6       S-w Hsieh
  S Uberoi
0 2  
    R Oprandi
  J Schruff
5 0       S Sun
  T Sun
7 6  
    L Baker
  N Kriz
3 6 6     L Baker
  N Kriz
5 2  
4   M Domachowska
  J Kostanić
6 2 3     S Sun
  T Sun
     
    L Granville
  C Gullickson
6 6       S-w Hsieh
  S Uberoi
w/o    
    E Bychkova
  M Müller
2 3       L Granville
  C Gullickson
5 2  
    S-w Hsieh
  S Uberoi
1       S-w Hsieh
  S Uberoi
7 6  
2   G Dulko
  M Tu
2 r  

References

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  1. ^ "2007 Auckland - ITF tournament draws". International Tennis Federation (ITF).
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