Home / National News / Protester who climbed Statue of Liberty charged in federal court

 

(WASHINGTON) — The woman who scaled Lady Liberty on the Fourth of July was formally charged in federal court Thursday.

Therese Okoumou, 44, of Staten Island, New York, was charged with trespassing, interference with agency functions and disorderly conduct, prosecutors said.

Each charge carries a maximum penalty of six months in prison or a fine.

“As alleged in the information, the defendant staged a dangerous stunt that alarmed the public and endangered her own life and the lives of the NYPD officers who responded to the scene. While we must and do respect the rights of the people to peaceable protest, that right does not extend to breaking the law in ways that put others at risk,” said U.S. Attorney Geoffrey S. Berman.

Okoumou appeared in Manhattan federal court Thursday and pleaded not guilty to all of the charges.

She was released on her own recognizance and posted bail. About 50 supporters packed the courtroom and cheered when court was adjourned.

Okoumou, who emigrated from Congo in 1994, was protesting the Trump administration’s so-called zero tolerance policy and the separation of children at the U.S.-Mexico border.

“Trump has ripped this country apart,” Okoumou said, referring to President Trump, at a press conference after the hearing. “It is depressing. It is outrageous.

“In a democracy, we do not put children in cages,” she added. “Period.”

Okoumou’s climb and subsequent arrest were preceded by the arrests of several protesters who hung a large banner that read “Abolish I.C.E.” on the monument, which is prohibited by the federal code of regulations.

Okoumou was protesting with Rise and Resist, a group committed “to opposing, disrupting and defeating any government act that threatens democracy, equality, and our civil liberties,” according to the group’s website.

She is scheduled to appear in court again Aug. 3.

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