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

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Women's singles
1938 U.S. National Championships
Final
ChampionUnited States Alice Marble
Runner-upAustralia Nancye Wynne [1]
Score6–0, 6–3
Details
Draw64
Seeds16
Events
Singles men women
Doubles men women
1938 tennis event results

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

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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

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Final eight

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Quarterfinals Semifinals Finals
                             

References

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  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.
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