1988 Lipton International Players Championships – Women's doubles
1988 tennis event results
Martina Navratilova and Pam Shriver were the defending champions but did not compete that year.
Steffi Graf and Gabriela Sabatini won in the final 7–6(8–6), 6–3 against Gigi Fernández and Zina Garrison.[1]
Seeds
Champion seeds are indicated in bold text while text in italics indicates the round in which those seeds were eliminated.
Draw
Key
Finals
| Semifinals
| | | Final
| | | | | | | | | | | | | | | | |
|
| 1
| 6
| 3
| | | | | 4
|
| 6
| 3
| 6
| | | | 4
|
| 66
| 3
|
| | |
| | | | | 2
|
| 78
| 6
|
| | | 3
|
| 1
| 6
|
| | | | | 2
|
| 6
| 7
|
| |
Top half
Section 1
Section 2
Bottom half
Section 3
Section 4
References
External links
|
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Grand Slam events | |
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Category 5 tournaments | |
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Category 4 tournaments |
- Dallas (S, D)
- Oakland (S, D)
- Houston (S, D)
- Tokyo Indoor (S, D)
- Eastbourne (S, D)
- Cincinnati (S, D)
- New Orleans (S, D)
- Filderstadt (S, D)
- Brighton (S, D)
- Chicago (S, D)
|
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Category 3 tournaments | |
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Category 2 tournaments |
- Brisbane (S, D)
- Oklahoma City (S, D)
- Kansas (S, D)
- Tokyo Outdoor (S, D)
- Geneva (S, D)
- Strasbourg (S, D)
- Birmingham (S, D)
- Nice (S, D)
- Aix-en-Provence (S, D)
- San Diego (S, D)
- Sofia (S, D)
- Phoenix (S, D)
- Brentwood (S, D)
- Indianapolis (S, D)
|
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Category 1 tournaments |
- Buenos Aires (S, D)
- Guarujá (S, D)
- Auckland (S, D
- Wellington (S, D)
- Singapore (S, D)
- Taiwan (S, D)
- Taranto (S, D)
- Barcelona (S, D)
- Båstad (S, D)
- Brussels (S, D)
- OTB Open (S, D)
- Aptos (S, D)
- Athens (S, D)
- Paris (S, D)
- San Juan (S, D)
- Guarujá II (S, D)
|
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Team events | |
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