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

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]

Women's Singles
1938 U.S. National Championships
Champion Alice Marble
Runner-up Nancye Wynne[1]
Final score6–0, 6–3
Draw64
Seeds16

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 6 3 6
(2) Simonne Mathieu 3 6 0
(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. Collins, Bud (2010). The Bud Collins History of Tennis (2nd ed.). [New York City]: New Chapter Press. p. 470. ISBN 978-0942257700.
  2. Henry McLemore (September 25, 1938). "Budge Makes 'Grand Slam'; Miss Marble Gains Title". The Pittsburgh Press. UPI. p. 5 (sports section) – via Google News Archive.
  3. Henry McLemore (September 25, 1938). "Budge Completes Slam; Miss Marble Also Wins". The Milwaukee Journal. UP. p. 1 (sports) – via Google News Archive.
  4. Irving Wright, ed. (1939). Wright & Ditson Official Lawn Tennis Guide 1939. New York: American Sports Publishing. pp. 29, 31–33.
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