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

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Women's Singles
1938 U.S. National Championships
ChampionUnited States Alice Marble
Runner-upAustralia Nancye Wynne[1]
Final score6–0, 6–3
Details
Draw64
Seeds16
Events
Singles men women
Doubles men women
← 1937 · U.S. National Championships · 1939 →

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]

Seeds[edit]

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[edit]

Final eight[edit]

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

References[edit]

  1. ^ a b 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.