Mary Washington completed an unbeaten CAC women's tennis regular season for the 13th year in a row, claiming the No. 1 seed in the upcoming CAC Women's Tennis Tournament. The eight-team CAC Tournament field is now set following the final conference regular season matches on Saturday.

Top-seeded UMW (7-0 CAC) will play host to No. 8 seed St. Mary's (Md.) on Saturday (April 27) in first round action, while No. 2 Christopher Newport will tangle with No. 7 Frostburg State to open league tournament play. 

The third-seeded Salisbury Sea Gulls will welcome the sixth-seeded Penn State Harrisburg on Saturday with the No. 4 seed Southern Virginia Knights battling No. 5 York (Pa.) to round out the first round pairings. 

Teams to the higher remaining institutions on Wednesday, May 1 for the CAC semifinals. The CAC Championship match is slated for Saturday, May 4 at the home courts of the highest remaining seed. 

2019 CAC Women's Tennis Tournament

Saturday, April 27 - First Round
Match 1: #8 St. Mary's (Md.) at #1 Mary Washington
Match 2: #5 York (Pa.) at #4 Southern Virginia
Match 3: #7 Frostburg State at #2 Christopher Newport
Match 4: #6 Penn State Harrisburg at #3 Salisbury

Wednesday, May 1 - Semifinal Round 
Match 5 - Winner Match 1 vs. Winner Match 2 (at highest remaining seed)
Match 6 - Winner Match 3 vs. Winner Match 4 (at highest remaining seed)

Saturday, May 4 - Championship Match
Match 7 - Winner Match 5 vs. Winner Match 6 (at highest remaining seed)