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1948 U.S. National Championships – Women's singles

1948 tennis event results From Wikipedia, the free encyclopedia

Third-seeded Margaret Osborne duPont defeated Louise Brough 4–6, 6–4, 15–13 in the final to win the women's singles tennis title at the 1948 U.S. National Championships.[1][2]

Runner-upUnited States Louise Brough
Score4–6, 6–4, 15–13
Quick facts Women's singles, Final ...
Women's singles
1948 U.S. National Championships
Final
ChampionUnited States Margaret Osborne duPont[1]
Runner-upUnited States Louise Brough
Score4–6, 6–4, 15–13
Events
Singles men women
Doubles men women
Close

Seeds

The seven seeded U.S. players are listed below. Margaret Osborne duPont is the champion; others show in brackets the round in which they were eliminated.[3]

  1. United States Louise Brough (finalist)
  2. United States Doris Hart (quarterfinals)
  3. United States Margaret Osborne duPont (champion)
  4. United States Pat Canning Todd (semifinals)
  5. United States Shirley Fry (third round)
  6. United States Beverly Baker (quarterfinals)
  7. United States Gussy Moran (semifinals)

Draw

Key

Final eight

Quarterfinals Semifinals Finals
               
7 United States Gussy Moran 6 6
2 United States Doris Hart 4 4
7 United States Gussy Moran 8 4
3 United States Margaret Osborne duPont 10 6
3 United States Margaret Osborne duPont 1 6 6
United States Beverly Baker 6 2 0
3 United States Margaret Osborne duPont 4 6 15
1 United States Louise Brough 6 4 13
1 United States Louise Brough 6 6
United States Virginia Kovacs 3 2
1 United States Louise Brough 6 6
4 United States Pat Canning Todd 3 3
4 United States Pat Canning Todd 6 6
United States Madge Vosters 3 1

References

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