2020 Bangkok Challenger – Doubles
Gong Maoxin and Zhang Ze were the defending champions[1] but lost in the first round to Treat Huey and Nathaniel Lammons.
Doubles | |
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2020 Bangkok Challenger | |
Champions | ![]() ![]() |
Runners-up | ![]() ![]() |
Final score | 3–6, 7–6(7–1), [10–5] |
Andrey Golubev and Aleksandr Nedovyesov won the title after defeating Sanchai Ratiwatana and Christopher Rungkat 3–6, 7–6(7–1), [10–5] in the final.
Seeds
Gonzalo Escobar / Miguel Ángel Reyes-Varela (Semifinals) Romain Arneodo / Andre Begemann (Quarterfinals) Tomislav Brkić / Ante Pavić (First round) Purav Raja / Adil Shamasdin (First round)
Draw
Key
- Q = Qualifier
- WC = Wild Card
- LL = Lucky Loser
- Alt = Alternate
- SE = Special Exempt
- PR = Protected Ranking
- ITF = ITF entry
- JE = Junior Exempt
- w/o = Walkover
- r = Retired
- d = Defaulted
First Round | Quarterfinals | Semifinals | Final | ||||||||||||||||||||||||
1 | ![]() ![]() | 7 | 7 | ||||||||||||||||||||||||
![]() ![]() | 5 | 5 | 1 | ![]() ![]() | 6 | 6 | |||||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 4 | 4 | ||||||||||||||||||||||
![]() ![]() | 3 | 4 | 1 | ![]() ![]() | 63 | 3 | |||||||||||||||||||||
4 | ![]() ![]() | 65 | 6 | [6] | WC | ![]() ![]() | 77 | 6 | |||||||||||||||||||
![]() ![]() | 77 | 3 | [10] | ![]() ![]() | 5 | 1 | |||||||||||||||||||||
WC | ![]() ![]() | 6 | 3 | [10] | WC | ![]() ![]() | 7 | 6 | |||||||||||||||||||
![]() ![]() | 2 | 6 | [6] | WC | ![]() ![]() | 3 | 77 | [10] | |||||||||||||||||||
![]() ![]() | 3 | 3 | ![]() ![]() | 6 | 61 | [5] | |||||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 77 | 6 | ||||||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 63 | 1 | ||||||||||||||||||||||
3 | ![]() ![]() | 4 | 4 | ![]() ![]() | 4 | 77 | [7] | ||||||||||||||||||||
WC | ![]() ![]() | 4 | 3 | ![]() ![]() | 6 | 62 | [10] | ||||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 6 | 68 | [10] | |||||||||||||||||||||
WC | ![]() ![]() | 4 | 4 | 2 | ![]() ![]() | 4 | 710 | [8] | |||||||||||||||||||
2 | ![]() ![]() | 6 | 6 |
gollark: pls latex \lim_{x \rightarrow \infty} \left (\frac{1}{x} \right) = 0
gollark: Something like
gollark: No, it means "as x goes to infinity, 1/x goes arbitrarily close to the result of that (if one exists)".
gollark: Also no. It's the limit of it as x goes to infinity, which is defined via ???? epsilon-delta ?????? calculululus ??????? something something Cauchy sequence.
gollark: No.
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