2007 ASB Classic – Doubles

Doubles
2007 ASB Classic
Champion Janette Husárová
Paola Suárez
Runner-up Hsieh Su-wei
Shikha Uberoi
Score6–0, 6–
Draw16
Seeds4

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

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

Seeds

  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

Key

Draw

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

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