1938 U.S. National Championships – Women's Singles

1938 U.S. National Championships – Women's Singles

Women's Singles
1938 U.S. National Championships
Champion Alice Marble
Runner-up Nancye Wynne[1]
Final score 6–0, 6–3

Second-seeded Alice Marble defeated Nancye Wynne 6–0, 6–3 in the final to win the Women's Singles tennis title at the 1938 U.S. National Championships.[1][2][3]

Contents

  • Seeds 1
  • Draw 2
    • Final eight 2.1
  • References 3

Seeds

The tournament used two lists of eight players for seeding the women's singles event; one for U.S. players and one for foreign players. Alice Marble is the champion; others show in brackets the round in which they were eliminated.[4]

Draw

Final eight

  Quarterfinals Semifinals Finals
                                       
  (6)   Margot Lumb 4 7 1  
(4)   Nancye Wynne 6 5 6  
  (4)   Nancye Wynne 5 6 8  
  3   Dorothy Bundy 7 4 6  
3   Dorothy Bundy 3 6 0
  (2)   Simonne Mathieu 6 3 6  
    (4)   Nancye Wynne 0 3
  2   Alice Marble 6 6
  2   Alice Marble 6 6 6  
(3)   Kay Stammers 8 3 0  
  2   Alice Marble 5 7 7
  4   Sarah Fabyan 7 5 5  
4   Sarah Fabyan 6 6  
  (1)   Jadwiga Jędrzejowska 1 4  

References

  1. ^ a b Collins, Bud (2010). The Bud Collins History of Tennis (2nd ed.). [New York City]: New Chapter Press. p. 470.  
  2. ^ Henry McLemore (September 25, 1938). "Budge Makes 'Grand Slam'; Miss Marble Gains Title".  
  3. ^ Henry McLemore (September 25, 1938). "Budge Completes Slam; Miss Marble Also Wins".  
  4. ^