2011 BGL Luxembourg Open – Doubles
Timea Bacsinszky and Tathiana Garbin were the defending champions but chose not to participate this year.
Doubles | |
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2011 BGL Luxembourg Open | |
Champions | ![]() ![]() |
Runners-up | ![]() ![]() |
Final score | 7–5, 6–3 |
Iveta Benešová and Barbora Záhlavová-Strýcová won the title beating Lucie Hradecká and Ekaterina Makarova in the final, 7–5, 6–3.
Seeds
Iveta Benešová / Barbora Záhlavová-Strýcová (Champions) Lucie Hradecká / Ekaterina Makarova (Final) Klaudia Jans-Ignacik / Jasmin Wöhr (Quarterfinals) Anabel Medina Garrigues / Anastasia Pavlyuchenkova (Semifinals)
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
Draw
First Round | Quarterfinals | Semifinals | Final | ||||||||||||||||||||||||
1 | ![]() ![]() | 6 | 6 | ||||||||||||||||||||||||
![]() ![]() | 3 | 4 | 1 | ![]() ![]() | 3 | 6 | [10] | ||||||||||||||||||||
![]() ![]() | 7 | 3 | [7] | ![]() ![]() | 6 | 3 | [4] | ||||||||||||||||||||
![]() ![]() | 5 | 6 | [10] | 1 | ![]() ![]() | 6 | 7 | ||||||||||||||||||||
3 | ![]() ![]() | 7 | 6 | ![]() ![]() | 2 | 5 | |||||||||||||||||||||
![]() ![]() | 5 | 4 | 3 | ![]() ![]() | 4 | 4 | |||||||||||||||||||||
![]() ![]() | 1 | 0 | ![]() ![]() | 6 | 6 | ||||||||||||||||||||||
![]() ![]() | 6 | 6 | 1 | ![]() ![]() | 7 | 6 | |||||||||||||||||||||
![]() ![]() | 6 | 6 | 2 | ![]() ![]() | 5 | 3 | |||||||||||||||||||||
![]() ![]() | 3 | 3 | ![]() ![]() | 77 | 4 | [5] | |||||||||||||||||||||
![]() ![]() | 4 | ![]() ![]() | 63 | 6 | [10] | ||||||||||||||||||||||
4 | ![]() ![]() | w/o | 4 | ![]() ![]() | 2 | 6 | [11] | ||||||||||||||||||||
![]() ![]() | 6 | 5 | [10] | 2 | ![]() ![]() | 6 | 3 | [13] | |||||||||||||||||||
![]() ![]() | 2 | 7 | [4] | ![]() ![]() | 5 | 3 | |||||||||||||||||||||
![]() ![]() | 1 | 5 | 2 | ![]() ![]() | 7 | 6 | |||||||||||||||||||||
2 | ![]() ![]() | 6 | 7 |
gollark: For example, they have the "fundamental theorem of algebra", where a polynomial of degree n *always* has n roots.
gollark: Anyway, by defining an answer to sqrt(-1) you can attain the complex numbers, which are a very powerful extension to the real number line.
gollark: Bold of you to only accept the reals as "numbers". Complex numbers have rights too.
gollark: In what way?
gollark: Yes, I'm aware.
References
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